Physics Labs

Table of Contents 0. Graphical Analysis (Optional) 1. Experimental Error and Data Analysis – I 2. Experimental Error and Data Analysis – II 3. Free Fall 4. Force Table 5. Atwood’s Machine 6. Numerical and Symbolic Methods in One Dimensional Dynamics 7a. Friction (ask which) 7b. Friction (air track) (ask which) 8. Energy and Systems 9. Conservation of Linear Momentum 10a. Conservation of Angular Momentum (A-Frame Base) – (ask which) 10b. Conservation of Angular Momentum – (ask which) 11. Orbits and Kepler’s Laws 12. The Period of a Simple Pendulum 13. Boyle’s Law • Appendix – Notes on Graphing Lab 0: Graphical Analysis Equipment: Excel. Theory (Always READ this section before you do the lab!) This section will describe the basic procedures for creating scatter plots in Excel and finding a fit to those functions. We will break this down into four basic steps: Step 1: Entering the data in the proper order Step 2: Creating a scatter plot Step 3: Creating Axis Labels and a Graph Title Step 4: Creating a fit line or curve. Step 1: Entering the Data A common source of error when graphing occurs from two similarly related ideas. What is supposed to be on the vertical and horizontal axes, and which columns should be which in Excel. The standard in the US is to plot vertical versus horizontal. When entering data into excel, the LEFT column represents the horizontal data while the right column represents the vertical data. While this may at first seem backwards, consider this: we say 𝑦 versus 𝑥 and yet we plot coordinates (𝑥, 𝑦). Excel follows this notation. So in a graph of velocity versus time, for instance, the velocity should be represented on the vertical axis and time should be on the horizontal axis. To enter the data into Excel for a plot of velocity versus time, we need the left column to represent the time (horizontal data) and the right column to represent the velocity (vertical data). An example of this is shown here. Step 2: Creating a scatter Plot The next step is to have Excel create a plot of the data. The types of plots we will always create in these labs are called scatter plots. To create this you want to highlight all of the data in both columns that you wish to plot. You then want to choose the Insert tab, neat the top-left of Excel (shown in the red box). In the picture below, Graphical Analysis Page 2 of 3 you can see that we have highlighted the data we are selecting (green box) and we are selecting the buttons to create a scatter plot. (Please do NOT use the connected data points. This will obscure your fit curve.) Step 3: Creating Axis Labels and a Graph Title Once step 2 is complete, Excel will create a scatter plot for you which is shown in the figure below. Excel defaults to creating a title for the graph called "Chart Title." You should double click this and change it to something appropriate. To create axis labels, you need to click somewhere on the graph then click the plus sign in the upper right corner when it shows up. This brings up the list of options you can see in the figure. You want to select the "Axis Titles" checkbox which will cause text boxes to show up, which you can edit accordingly. Graphical Analysis Page 3 of 3 Step 4: Creating Axis Labels and a Graph Title It is important to note that this step CANNOT be completed when using Excel online. All Blinn students have access to download and install Excel free of charge. Excel is not installed on all physics lab computers. Graphical Analysis You may notice there is a Trend line option in the previous figure. We do not want to create a trend line that way because Excel defaults that to a linear function. Nature and graphs that represent nature's behavior are not always represented as lines, but sometimes as curves. To be able to select the type of function you want to, right-click on one of the data points in the plot. This will bring up a few options, one of which is "Add Trendline." When you click on that, it will bring up a panel on the right side of Excel which looks like the picture to the right. You'll notice it defaults to linear, which is fine if that is representative of the data, as in the example we've been discussing. However, if the data is quadratic, for instance, select Polynomial Order 2 (a cubic function would be Order 3). While not necessarily common in first semester physics, second semester physics sees a number of dependencies which are exponential or logarithmic. These would be what you would want to select if that was representative of your data. One vital step is to make sure that you select the "Display Equation on chart" box. This will provide the formula for the resulting fin on the graph. With all the steps complete the graph above graph would look like this. Finally, remember, whenever you are asked to create a graph in Excel, you must always: Include the graph with your lab report! Page 4 of 3 Graphing – Data Sheet Name________________ Group______ Part A. Velocity as a function of time with constant acceleration. Given the following data, plot a graph of 𝑣 versus 𝑡 and apply a linear fit. 𝑡(s) 𝑣 (m/s) 1.0 3.5 2.0 5.0 3.0 5.5 4.0 6.8 5.0 7.5 6.0 8.0 7.0 10.0 8.0 10.2 9.0 11.0 In Chapter 2 you will be introduced to an equation that relates 𝑣 and 𝑡 when there is constant acceleration. That equation is given below. 𝑣 = 𝑣! + 𝑎𝑡 Questions Based on the equation and the fit on your graph answer the following questions: A-1. What is the acceleration of the object? Be sure to include the appropriate units. A-2. What is the velocity at the time defined to be zero? Be sure to include the appropriate units. Part B. Density Suppose a beaker is placed on a balance (scale) and different volumes of a liquid are poured into it. Furthermore, suppose the mass of the beaker plus the liquid is recorded each time, and the following data are collected. 𝑉𝑜𝑙𝑢𝑚𝑒 (mL) 𝑀𝑎𝑠𝑠 (g) 23 192.8 34 204.0 45 219.6 49 236.5 68 246.3 Given this data, create a graph of mass versus volume and include a best-fit line. Questions B-1. What will the total mass be if there was 50 mL of liquid in the beaker? Show any calculations. Include the appropriate units. Graphical Analysis – Procedure and Data sheet Page 2 of 6 B-2. What is the mass of the beaker? Include appropriate units. B-3. What is the average density of the liquid? Include appropriate units. Part C. Fitting Non-linear Relationships. In the middle part of the semester we will deal with rotational motion. One of the common equations we will deal with in that unit is the following: " 𝜃 = 𝜃! + 𝜔! 𝑡 + #𝛼𝑡 # where 𝜃 is the angle in radians (rad) at a particular time, 𝜃! is the angle at the time defined to be zero, 𝜔! is the angular speed at the time defined to be zero, 𝛼 is the angular acceleration and 𝑡 is the time. In this case, time has units of seconds (s). You do not need to know anything else about this equation at this time to perform this section. Using the set of data below, create a plot of angle in radians versus time (in seconds) and fit the it with a quadratic. Angle (rad) Time (ms) 1.480 1.921 2.291 2.495 1.983 1.464 1.017 0.03642 –1.063 –2.517 5.2 50.8 96.4 142.0 187.6 233.2 278.8 324.4 370.0 415.6 Questions C-1. Based on your graph, what are the values for 𝜃! , 𝜔! , and 𝛼? Enter these values on the left blanks. What must the units for those variables be in order to stay dimensionally consistent if 𝜃 is supposed to be in (rad) and 𝑡 is in (s)? Enter units in on the right blanks. 𝜃! : __________________ _______________ 𝜔! : __________________ _______________ 𝛼: __________________ _______________ In the second semester of physics some of you may encounter the following equation. Graphical Analysis – Procedure and Data sheet Page 3 of 6 𝑉 = 𝑉! 𝑒 $%/' Here, 𝑉 is something called potential which has units of volts (V), 𝑉! is the potential at the time defined to be zero, 𝑡 is time in seconds (s) and 𝜏 is something called the time constant. Below is a set of data that is exponential, which means that you will want to use the exponential trend line with your fit. Create a plot of potential vs time and answer the following questions based on your plot. Voltage (V) 10.533 9.041 6.736 5.118 4.311 3.447 2.462 1.956 1.495 1.347 0.990 0.831 0.663 0.501 Time (s) 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5 12.5 13.5 C-2. Based on your graph, what are the values for 𝑉! and 𝜏? Enter these values on the left blanks. What must be the units for those variables in order to stay dimensionally consistent if 𝑉 is supposed to be in (V) and 𝑡 in (s)? Enter those units on the right blanks. 𝑉! : __________________ _______________ 𝜏: __________________ _______________ C-3. Let us assume the theoretical accepted value of 𝑉! was supposed to be 12.0 and the theoretical value for 𝜏 was supposed to be 4.20. Based on your results from your graph, calculate the percent error in your results. You can use the following formula for that calculation and the rest of the space to show your work. 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑒𝑟𝑟𝑜𝑟 = 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 − 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 × 100% 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑐𝑐𝑒𝑝𝑡𝑒𝑑 Graphical Analysis – Procedure and Data sheet Page 4 of 6 Part D. Decay of a Radioactive Nuclide Radioactive nuclei undergo a process called decay, during which the nucleus changes its identity by emitting one or more particles or by emitting a photon. During this process, the nucleus may change from one element to another. One common example is the nucleus of Americium-241 ( #("Am or Am-241) which is commonly used in household smoke detectors. It decays by the emission of an alpha-particle which has 2 protons and 2 neutrons. The remaining nucleus is now Neptunium-237. The decay of any particular nucleus is random and unpredictable; however, the rate of decay of a large quantity of the element follows a simple equation characterized by a well-defined “half-life” for the element. The halflife is the time required for one-half of the particles to decay. In the case of Am-241 the half-life is 432.2 years. Below is the table which shows the number of particles measured every 100 years. Using this data, you will attempt to determine the appropriate equation for the number of nuclei remaining as a function of time. N t (yrs) 1000 0 852 100 726 200 618 300 527 400 499 500 382 600 325 700 Make a graph of the data of N as a function of time and answer the questions below. D-1. Fit a linear function to the graph. Write the best-fit equation below it and use it to calculate the half-life of Am-241. To do so, assume 𝑁(𝑡) = 500 and solve for t. D-2. Now fit an exponential function to the graph. Write the equation for the best-fit curve in the space below and use it to calculate the half-life of Am-241. As before, set 𝑁(𝑡) = 500 and solve for t. Graphical Analysis – Procedure and Data sheet Page 5 of 6 D-3. Now fit a quadratic to the graph. Write the equation for the best-fit curve in the space below and use it to calculate the half life of Am-241. As before, set 𝑁(𝑡) = 500 and solve for t. D-4. Calculate the percent error between each of the three estimates for the half-life and the actual half-life of 432.2 years. Linear fit:_______________________ Exponential fit: ___________________ Quadratic fit:_____________________ Based on these results, which function would you assume best represents the nature of radioactive decay? _________________________ D-5. It turns out that the exponential function is the equation that actually describes the decay of a radioactive nuclide. Does this fact coincide with your conclusion above? Yes No If No, what may be responsible for the error? Graphical Analysis – Procedure and Data sheet Page 6 of 6 D-6. Go to the plot of the quadratic and exponential fits. Right-click on the fit equation and click “Format Trendline Label.” This will bring up a panel on the right. Click on the drop-down box that says General. Choose Scientific. You will notice that the numbers in your fit equations have changed. For both of these fits, recalculate the half lives and recalculate the percent error between your new calculated results and the actual half-life of 432.2 years. Use the rest of the page to do these calculations, but place your answers on the lines below. Exponential fit: Half life = _______________________ Quadratic fit: Half life = _______________________ Percent Error = _____________________ Percent Error = _____________________ Error and Data Analysis - I Section A: Propagation of Error I - Significant Figures Identifying Significant Digits When you make a measurement there are digits where we are certain of their measured value and estimated digits. We typically consider the certain digits and one estimated digit as significant. If a meter-stick measurement gives a length between 23.5 cm and 23.6 cm, but closer to the first then you could your best estimate for the next to be a 3; the measurement, with four significant figures, is 23.53 cm. When presented with a number you must be able to identify the significant digits. In scientific notation all digits are significant, assuming there are no leading zeroes. For example, 6.8040 × 10!" has five significant digits. Beyond scientific notation, the only subtlety is the digit zero; all non-zero digits are always significant. Zeros to the left of the first non-zero digit are not significant. This becomes an issue when there is a decimal expression; a number like 0.00005020 has four significant digits. All those leading zeros are just placeholders; writing this in scientific notation makes it clear: 5.020 × 10!# . In large numbers without decimal points, all zeros to the right are not significant. For example, 78000 has two significant digits but 78000., with the decimal point, has five; if the first zero is intended to be significant then you should use scientific notation and write 7.80 × 10$ . Multiplication, Division and Single-variable Functions In a later laboratory exercise, we will discuss a much more precise way to propagate errors. We have standard rules of thumb we use to keep track of the propagation of error using significant digits. When multiplying or dividing numbers or when applying single-variable functions to numbers we ascribe the smallest number of significant digits in the initial numbers to the final result. In your discussion of one-dimensional kinematics, you will learn (or have learned) that the time of fall for an object dropped from a height h with an acceleration due to gravity of g is 𝑡 = -2 ℎ/𝑔. If we with these values for h and g: ℎ = 15.4 m and 𝑔 = 9.806 m/s % where h has three significant figures and g has four, then we should round our result to three significant digits. Suppose that after punching these values into your calculator, it displays ten digits: 𝑡 = 1.772268073 s. Rounding this to three significant digits gives the desired result. 𝑡 = -2 ℎ/𝑔 = 1.77 s The square root function used here is a function of just one variable, so its presence does not affect the number of digits in the final result. We apply this rule to all single-variable functions. Addition or Subtraction Number Number of Significant Digits Last Significant Decimal Place 2.0150 5 10-4 28000 2 103 1200. 4 100 0.0023050 5 10-7 Error and Data Analysis - I Page 2 of 6 When adding or subtracting numbers, it is not the number of significant figures that matters; it is the last significant decimal place of each number. When adding or subtracting numbers we pay attention to the largest of the last significant decimal place and ignore digits beyond that. Consider the table above. If you are summing all these then the calculator gives 29202.0173, but we should round to the thousands or 103 place. 2.0150 + 28000 + 1200. +0.002305 = 29000 Adding the first and the last numbers gives a last significant decimal place of one ten-thousandth or 10-4. 2.0150 + 0.002305 = 2.0173 Section B: Accuracy, Precision and Types of Error Accuracy and Precision The words accuracy and precision are often used interchangeably but they have quite different meanings. Accuracy describes how close your result is to the accepted or theoretical value. Precision describes how close your data is to itself; a smaller spread is more precise. ��������� It is common to illustrate this with a target practice example. The diagram shows distributions of shots at a target. �������� The distributions that are more tightly packed are more precise and those with the center of the distribution closer to the center of the target are more accurate. Percent Difference Percent difference is a simple measure of precision for two experimental values. Label these values x1 and x2. The percent difference is the absolute value of the difference over the average. percent difference = |difference| |𝑥% − 𝑥& | × 100% = × 100% (𝑥& + 𝑥% ) average 2 Percent Error Percent error is a simple measure of accuracy, comparing an experimental value E to its theoretical or accepted value A. percent error = experimental − accepted 𝐸−𝐴 × 100% = × 100% accepted 𝐴 Often percent error is defined with an absolute value. We will use the convention that its sign indicates whether the experimental value is too large or too small. Error and Data Analysis - I Page 3 of 6 Systematic Errors Systematic errors affect accuracy. As a trivial example, suppose you need to measure a length with a meter stick. If the end of a meter stick had been repeatedly dropped to the floor, then it may be too short and thus read too long. If a measuring device is incorrectly calibrated, then there is a systematic error; every measurement will reflect that poor calibration. If a scale is not properly zeroed, then all measurements are systematically off. Sometimes instruments drift and need to be regularly recalibrated. If an analog meter is read from an angle then that introduces a parallax error; imagine looking at an analog meter, like an analog speedometer, from an angle. If an important factor is ignored then that introduces error, typically a systematic one. For instance, if the thermal expansion of a ruler or calipers is ignored, then small systematic errors are introduced. Note that if the measurements are made under varying conditions then this error could be random. Ignoring air resistance can introduce systematic error. Some measurements require a significant lag time that must be respected; for instance, in thermodynamics a system must be sufficiently near equilibrium before a temperature can be measured and that takes time. Random Errors Random errors affect precision. Often there is physical variation in the thing being measured. If one wants to find the mass of a penny then there will be considerable variation in the masses found; there is some small variation in the masses of pennies as they are minted and a much larger variation after being handled. Any instrument has a finite resolution, and this introduces random errors. Random physical effects like vibration, electronic noise and drafts and temperature fluctuations are often significant. Section C: Random Errors and Probability Distributions Probability and Probability Distributions Randomness is described by probability. Flip a coin and it has a 50% probability landing on either heads or tails. If a second coin is flipped it has the same 50% probability for each outcome; this results in a 50% probability of one head and one tail and a 25% chance for two heads and 25% for two tails. Flipping many coins results in a distribution of probabilities for getting different numbers of heads (or tails.) By repeating an experiment many times one can reproduce the probability distribution. If for the coin flip, one defines success as a coin landing on heads, then the probability of success is ½ = 50%. Flipping several coins will not result in half successes but as the number of flips becomes large, the number of successes divided by the number of flips will approach ½. This illustrates a fundamental general principle: For any probability distribution, the number of successes over the number of trials will approach the probability of success as the number of trials approaches infinity. A probability distribution is some function that gives probabilities of outcomes for some random variable; the sum of all probabilities must always be one. For a simple example let x be a random variable for a coin flip with 𝑥 = 1 for a head and 𝑥 = 0 for a tail. The probability distribution is then & 𝑃(0) = % = 𝑃(1) where 1 = 𝑃(0) + 𝑃(1) . As another example consider the sum of the faces of two dice, which can vary from 2 to 12. Here we get: Error and Data Analysis - I Page 4 of 6 x 2 3 4 5 6 7 8 9 10 11 12 P(x) 1 36 2 36 3 36 4 36 5 36 6 36 5 36 4 36 3 36 2 36 1 36 These two examples are discrete distributions. For a general discrete distribution, we have a random variable where value x has probability P(x). Here 1 = ∑' 𝑃(𝑥) where the sum is over all values of x. The expectation value of some function 𝑓(𝑥) of a random variable is written 〈𝑓(𝑥)〉 and is defined as: ⟨𝑓(𝑥)⟩ = U 𝑓(𝑥)𝑃(𝑥) (1) ' In addition to discrete distributions we also have continuous distributions. Here we have a continuous random variable x. The probability of x being between x and 𝑥 + 𝑑𝑥 is 𝑃(𝑥)𝑑𝑥. The condition that the probabilities sum to one becomes the “normalization” condition that: ∫ 𝑃(𝑥)𝑑𝑥 = 1, where the (definite) integral is over all possible values of the random variable x. The probability that x has a value between x1 and x2 is: '! X 𝑃(𝑥)𝑑𝑥 . '" In the continuous case, the expectation value becomes 〈𝑓(𝑥)〉 = X 𝑓(𝑥)𝑃(𝑥)𝑑𝑥 , where again the integral is over all possible values of x. For both discrete and continuous distributions, the mean µ and standard deviation s are defined as 𝜇 = 〈𝑥〉 (mean) and 𝜎 = -〈 (𝑥 − 𝜇)% 〉 = -〈𝑥 % 〉 − 𝜇% (standard deviation) The standard deviation is a measure of the spread of the distribution; a small standard deviation means the highest probability is that the random variable is close to the mean and a large one implies a larger spread. The square of the standard deviation 𝜎 % is known as the variance. The most important example of a continuous distribution is the normal distribution. 𝑃(,* (𝑥) = 1 𝜎√2𝜋 𝑒 ! . ('!()! %* ! where X 𝑃(,* (𝑥)𝑑𝑥 = 1 !. The factor multiplying the exponential is there to make it normalized, i.e. integrating to one. By evaluating the appropriate integrals, one can show that the µ and s in the expression are the mean and standard deviation of the distribution. ������ ������������ �(�) μ-σ μ μ+σ � This is just what is popularly referred to as a “bell-shaped” curve. This shape is also referred to as a Gaussian. Integrating Gaussians requires the definition of a new function as its antiderivative, the error function. Error and Data Analysis - I Page 5 of 6 erf(𝑥) = 2 √𝜋 ' ! X 𝑒 !/ 𝑑𝑡 0 Using this we can find the probability that a normal distribution lies between x1 and x2 as '! 1 𝑥% − 𝜇 1 𝑥& − 𝜇 X 𝑃(,* (𝑥)𝑑𝑥 = erf a b − erf a b 2 2 √2 𝜎 √2 𝜎 '" Independent Random Variables Two random variables x and y are independent when they are not correlated; this means that 𝑃(𝑥, 𝑦) = 𝑃(𝑥) 𝑃(𝑦), where 𝑃(𝑥, 𝑦) is the probability of both x and y occurring. As an example, smoking cigarettes and lung cancer are not independent, because they are significantly correlated, but left-handedness and lung cancer are most likely independent. For independent random variables 〈𝑥𝑦〉 = 〈𝑥〉〈𝑦〉. This is easy to show for discrete distributions. 〈𝑥𝑦〉 = U 𝑥𝑦 𝑃(𝑥, 𝑦) = U U 𝑥𝑦 𝑃(𝑥) 𝑃(𝑦) = dU 𝑥 𝑃(𝑥)e fU 𝑦 𝑃(𝑦)g = 〈𝑥〉〈𝑦〉 ',1 ' 1 ' 1 It is an important fact that the variance of the sum of two independent random variables is the sum of the variances. % 𝜎'21 = 𝜎'% + 𝜎1% This will be proved in the Appendix. The mean of the sum is also the sum of the means. 𝜇'21 = 〈𝑥 + 𝑦〉 = 〈𝑥〉 + 〈𝑦〉 = 𝜇' + 𝜇1 It is an important fact that when many of different or the same probability distributions are added (or averaged) the result tends toward a normal distribution with the same mean and standard deviation. This is known as the Central Limit Theorem. Random errors are usually the cause by the accumulation of many small effects; because of this they are modeled well by normal distributions. Section D: Data Analysis I - Mean and Standard Deviation Data Analysis Suppose you have randomly distributed data. xi is a random data point with a continuous distribution with the same probability P(x), mean µ and variance s2. Suppose there are N such data points in the data set: {𝑥3 } = {𝑥& , ⋯ , 𝑥4 }. The statistical mean, 𝑥̅ , of the data set is the average of the data points. 4 1 1 𝑥& + 𝑥% + ⋯ + 𝑥4 𝑥̅ = U 𝑥3 = U 𝑥3 = 𝑁 𝑁 𝑁 35& 3 The statistical mean is an estimate of the mean of the probability distribution; its expectation value is the mean of the distribution: 〈𝑥̅ 〉 = 𝜇. The statistical standard deviation s (or sx to be more explicit) is an estimate of s, the standard deviation of the distribution. The statistical standard deviation of the same data set {𝑥3 } is defined in terms of the statistical mean 𝑥̅ . Error and Data Analysis - I 𝑠=n Page 6 of 6 (𝑥& − 𝑥̅ )% + (𝑥% − 𝑥̅ )% + ⋯ + (𝑥4 − 𝑥̅ )% 1 U(𝑥3 − 𝑥̅ )% = n 𝑁−1 𝑁−1 3 The factor of N – 1 seems odd, but it is needed to guarantee that the expectation of the 𝑠 % is 𝜎 % , the variance of the probability distribution: 〈𝑠 % 〉 = 𝜎 % . We will not prove this here. The statistical standard deviation is a measure of precision and will not reflect systematic errors. Appendix: Data Analysis I - Mean and Standard Deviation Proof of the Variance of the Sum % For independent random variables x and y, 𝜎'21 = 𝜎'% + 𝜎1% . This is not a difficult proof. Evaluate the two terms % % separately in 𝜎'21 = 〈(𝑥 + 𝑦)% 〉 − 𝜇'21 . 〈(𝑥 + 𝑦)% 〉 = 〈𝑥 % 〉 + 〈𝑦 % 〉 + 2〈𝑥𝑦〉 = 〈𝑥 % 〉 + 〈𝑦 % 〉 + 2〈𝑥〉〈𝑦〉 = 〈𝑥 % 〉 + 〈𝑦 % 〉 + 2 𝜇' 𝜇1 Here we have used 〈𝑥𝑦〉 = 〈𝑥〉〈𝑦〉 for independent x and y. Using 𝜇'21 = 𝜇' + 𝜇1 we can evaluate the second term. % % 𝜇'21 = o𝜇' + 𝜇1 p = 𝜇'% + 𝜇1% + 2 𝜇' 𝜇1 Subtracting cancels the 2 𝜇' 𝜇1 terms, giving the desired result. % % 𝜎'21 = 〈(𝑥 + 𝑦)% 〉 − 𝜇'21 = 〈𝑥 % 〉 − 𝜇'% + 〈𝑦 % 〉 − 𝜇1% = 𝜎'% + 𝜎1% Name ________________ Group ______ Worksheet: Error and Data Analysis - I Equipment and Setup: Mathematica file – DataAnalysis-I.nb Section A: Significant Figures Questions A-1. Complete the table showing the number of significant digits and the last significant decimal place for each number given. Number Number of Significant Digits Last Significant Decimal Place 3.14159 590 590. 0.0003854 2.6´1021 A-2. Give your answers with appropriate significant figures. 3.14159 + 0.0003854 = ______________________________ 3.14159 × 0.0003854 = ______________________________ A-3. Using 𝐴 = 0.02528 m, 𝜔 = 364.6 s !& and 𝑡 = 0.00243 s, calculate the result with appropriate significant figures and units. (Note that to evaluate such a function it is essential to use radians, so put your calculator in its radians mode.) 𝐴 sin(𝜔𝑡) = ______________________________ Section B: Target Practice Computer Setup for Section B 1. The interactive panel shows a target. At the top are controls: there is a Setter Bar with four buttons, one for each trial. 2. Start with Trial 1. Click on the panel to get cross-hairs for aiming. Drag the cross-hairs to the center. 3. Click “Fire” and a dot will appear. Repeat this many times to observe a pattern. Worksheet - Error and Data Analysis - I Page 2 of 6 4. Repeat the procedure for trials 2 through 4. 5. The small button in the upper right resets the panel. Data Recording for Section B ¾¾¾® Precision 1. Rank the four trials for accuracy and precision in the table below. (Write 1 through 4 into the appropriate boxes.) ¾¾¾® Accuracy Questions B-1. List some factors that could account for a lack of precision in this simulation of target practice. B-2. List some factors that could account for a lack of accuracy in this simulation. Section C: Probability Distributions Computer Setup for Section C.1 1. The interactive panel shows a Setter Bar with two options: “Coin Flip of Heads” and “Dice Rolls of One”. 2. Below that is a button “Random” that generates the coin flips or dice rolls. Next to the button there is an Input Field for n, the number of coins or dice. n must be a positive integer. 3. The small button in the upper right resets the panel. Worksheet - Error and Data Analysis - I Page 3 of 6 Data Recording for Section C.1 1. Start with coin flips and set 𝑛 = 10. For three different trials record the number of heads x and the fraction of heads, 𝑥/𝑛. Trial 1 2 3 x x/n 2. Now set 𝑛 = 1000 and similarly record the number of heads x and the fraction of heads, 𝑥/𝑛 for three trials. Trial 1 2 3 1 2 3 x x/n 3. Now do the same for 𝑛 = 1,000,000. Trial x x/n 4. Change the Setter Bar to dice rolls and set 𝑛 = 10. For three different trials record the number of ones x and the fraction of heads, 𝑥/𝑛. Trial 1 2 3 1 2 3 1 2 3 x x/n 5. Repeat with 𝑛 = 1000 Trial x x/n 6. And for 𝑛 = 1,000,000. Trial x x/n Worksheet - Error and Data Analysis - I Page 4 of 6 Questions C-1. The probability of a coin flipping to heads is 1/2. Discuss what happens to 𝑥/𝑛 as n increases in steps 1 through 3. C-2. The probability of a die rolling one is 1/6. Discuss what happens to 𝑥/𝑛 as n increases in steps 4 through 6. Computer Setup for Section C.2 1. This simple interactive panel has two sliders for the mean µ and the standard deviation s. The reset button is at the upper right. Data Recording for Section C.2 1. Adjust the slider for the mean µ. What happens to the graph as µ increases? 2. Adjust the slider for the standard deviation s. What happens to the graph as s increases? Computer Setup for Section C.3 1. There is an Interval Slider that varies the values of x1 and x2. When at the edge of the Interval Slider the values are set to ±¥. 2. The area between these x-values is listed to the right. 3. A small reset button is at the upper right. Data Recording for Section C.3 1. For data from a normal distribution, what fraction of the values are less than the mean? Worksheet - Error and Data Analysis - I Page 5 of 6 2. What fraction of the values are within one standard deviation s of the mean? 2s ? 3s? 3. In particle physics they use the 5s standard. When they are certain of a result within five standard deviations, they are ready to publish their discovery. What is the probability of a normal distribution being more than 5s from the mean? Computer Setup for Section C.4 1. There are several controls at the top. At the left there are three tabs: Flip Coins, Binomial Distribution and Sum of Dice Roll. Below each tab are pop-up menus specific to that tab. For each n is the number of coins, trials or dice. For the Binomial Distribution p is the probability of success, where the coin flip is the special case of p = ½. 2. To the right is a Setter Bar, where you choose either a graph of the discrete probability distribution, that same graph with the normal distribution (Gaussian) with the same standard deviation and mean, or a probability table for the distribution. 3. The reset button is at the upper right. Data Recording for Section C.4 1. What is the probability of flipping 50 coins with exactly 25 heads? (Hint: read this directly off the table.) 2. The variance (the square of the standard deviation s) for a binomial distribution is 𝜎 % = 𝑛 𝑝 (1 − 𝑝). Find the standard deviation for n = 25 and p = 0.8, and compare with the result from the interactive panel. Now do the same for n = 60 and p = 0.4. 3. Now click Graph with Gaussian on the Setter Bar. Comment on the comparison between the discrete distribution and the Gaussian as n gets larger. Worksheet - Error and Data Analysis - I Page 6 of 6 Section D: Data Analysis Computer Setup for Section D 1. The interactive panel generates a random set of data. 2. Clicking “Generate Data” selects a value of µ and s, the mean and standard deviation of a normal distribution. It then generates 5 random points with that distribution. 3. The reset button is at the upper right. Data Recording for Section D 1. Click “Generate Data” and record the mean, standard deviation and the data. x µ = __________________ s = __________________ Questions D-1. Calculate the statistical mean 𝑥̅ and the statistical standard deviation s for this data. Show all work. D-2. The statistical mean 𝑥̅ and the statistical standard deviation s can be viewed as approximations to µ and s, the mean and standard deviation of the distribution. Calculate the percent errors in 𝑥̅ and s taking µ and s as the accepted values. % error 𝑥̅ = __________________ % error s = __________________ Error and Data Analysis - II Section A: Mathematical Preliminary – Partial Derivatives and the Total Differential Partial Derivatives In Calculus I, the derivative of functions of one variable was discussed; using the rules of differentiation (sum rule, product rule, quotient rule, chain rule, etc.) you should now be able to evaluate the derivative of any function built up from known functions. Calculus I and also Calculus II discuss only single variable calculus. All of multivariable calculus is put off to the third course, Calculus III. The key subtlety of multivariable calculus is integration. We will now see that taking derivatives of functions of more than one variable, partial differentiation, is no more difficult than the ordinary derivatives learned in Calculus I. !" If 𝑓(𝑥) is a function of one variable then we write its derivative, the ordinary derivative, as !#. For functions of more than one variable, we use partial derivatives. For notation, we replace the d with ¶ and then we write the $" $" partial derivatives of 𝑓(𝑥, 𝑦) as $# and $%. Suppose you have a function 𝑓(𝑥, 𝑦, ⋯ ) of several independent variables, x, y, … . To evaluate a partial derivative with respect to one independent variable you just take the ordinary derivative with respect to that variable, while treating the other independent variables as constants. As an example, consider 𝑓(𝑥, 𝑦) = 𝑎𝑥𝑦 & − 𝑏𝑥 ' Where x and y are the independent variables and a and b are constants. The partial derivatives become 𝜕𝑓 𝜕𝑓 = 𝑎𝑦 & − 3𝑏𝑥 & and = 2𝑎𝑥𝑦 . 𝜕𝑥 𝜕𝑦 Total Differential The total differential of some function can be interpreted as the infinitesimal change in the function due to infinitesimal changes in each of its independent variables. First, consider a function of one variable 𝑓(𝑥). If x changes by the infinitesimal amount 𝑑𝑥, then 𝑓(𝑥) will shift by (𝑑𝑓/𝑑𝑥)𝑑𝑥. The total differential is written 𝑑𝑓 𝑑𝑥 . 𝑑𝑥 For functions of more than one variable 𝑓(𝑥, 𝑦, ⋯ ) we write the total differential with partial derivatives 𝑑𝑓 = 𝑑𝑓 = 𝜕𝑓 𝜕𝑓 𝑑𝑥 + 𝑑𝑦 + ⋯ . 𝜕𝑥 𝜕𝑦 We now find the total differential for the previous example function. 𝑓(𝑥, 𝑦) = 𝑎𝑥𝑦 & − 𝑏𝑥 ' ⟹ 𝑑𝑓 = (𝑎𝑦 & − 3𝑏𝑥 & ) 𝑑𝑥 + 2𝑎𝑥𝑦 𝑑𝑦 Section B: Propagation of Error II - General Functions of Random Variables Uncertainty and Notation In the earlier experiment, Error and Data Analysis – I, we discussed the mean and standard deviation for a probability distribution and for a statistical sample of data. For some variable x we will take its uncertainty 𝛿𝑥, to be the statistical standard deviation of a data, an estimate of the error in x or a combination of both. Error and Data Analysis - II Page 2 of 5 The standard notation for writing such uncertainty is to use the ± notation. A value x with uncertainty 𝛿𝑥 is written as 𝑥 ± 𝛿𝑥. An alternative notation, which is common when listing values of constants, is to write digits as xxxx(x) or xxxx(xx). The expression in brackets gives the uncertainty in the last one or two digits, respectively. For example, consider the listed values of the Earth’s mass 5.9722(6) × 10&( kg ⟺ (5.9722 ± 0.0006) × 10&( kg and the electron’s mass. 9.1093837015(28) × 10)'* kg ⟺ (9.1093837015 ± 0.0000000028) × 10)'* kg Propagation of Uncertainty In the earlier experiment, we discussed the mean, variance and standard deviation for a probability distribution. The key thing we need from that discussion is that the variance (the square of the standard deviation) of the sum of two (or more) independent random variables is the sum of the variances. & 𝜎#+%+⋯ = 𝜎#& + 𝜎%& + ⋯ We will consider the uncertainty 𝛿𝑥 to our best estimate of 𝜎# , the standard deviation of the underlying probability distribution for x. It follows from this, that for independent random variables x and y with uncertainties uncertainty 𝛿𝑥 and 𝛿𝑦, that the uncertainty in the sum is 𝛿(𝑥 + 𝑦 + ⋯ ) = F𝛿𝑥 & + 𝛿𝑦 & + ⋯ . We now want to general this for arbitrary function of multiple independent random variables. We make the assumption that uncertainties are small, so they may be viewed as infinitesimal. The total differential describes the infinitesimal change in f due to the infinitesimal changes in variables. 𝜕𝑓 𝜕𝑓 𝑑𝑥 + 𝑑𝑦 + ⋯ 𝜕𝑥 𝜕𝑦 When x and y are independent, then each term in the above sum is independent. 𝑑𝑓 = It follows that the uncertainty in the function f is given by 𝜕𝑓 & & 𝜕𝑓 & & 𝛿𝑓(𝑥, 𝑦, ⋯ ) = GH I 𝛿𝑥 + H I 𝛿𝑦 + ⋯ 𝜕𝑥 𝜕𝑦 Special Case: Function of One Variable For a function of one variable the partial derivatives become a single ordinary derivative and the square-root gives an absolute value. 𝑑𝑓 J 𝛿𝑥 𝑑𝑥 As an example of this let is find 𝑢 = sin 𝜃, where 𝜃 ± 𝛿𝜃 = 58.32° ± 0.07°. First find the uncertainty. (Note that angles are dimensionless quantities and here the 𝛿𝜃 must be converted to radians using 1° = *./.) 𝛿𝑓(𝑥) = J 𝑑 sin 𝜃 𝜋 J 𝛿𝜃 = |cos 𝜃| 𝛿𝜃 = |cos 58.32°| × 0.07 = 0.00064 𝑑𝜃 180 Using sin 58.32° = 0.85099, we get: 𝛿𝑢 = J sin(58.32° ± 0.07°) = 0.8510 ± 0.0006 Error and Data Analysis - II Page 3 of 5 Special Case: Products, Quotients and Powers Now we will consider a function of the form 𝑢 = 𝑓(𝑥, 𝑦, ⋯ ) = 𝑎 ∙ 𝑥 0 ∙ 𝑦 1 ∙ ⋯ Where a is a constant and the powers 𝑚2 can have any real value. Note that you can use negative powers for division. There is a standard trick that is used here that involves “logarithmic derivatives”. Take the natural log of both sides and then evaluate the total differential. ln(𝑢) = ln(𝑎) + 𝑚 ln(𝑥) + 𝑛 ln(𝑦) + ⋯ Taking the differential of a logarithm gives a simple result: 𝑑 ln(𝑢) = expression then gives !3 3 . The total differential of the 𝑑𝑢 𝑑𝑥 𝑑𝑦 =𝑚 +𝑛 +⋯ 𝑢 𝑥 𝑦 This allows us to read off the partial derivatives 𝜕𝑢 𝑢 =𝑚 𝜕𝑥 𝑥 and then write an expression for the uncertainty. 𝛿𝑢 𝛿𝑥 & 𝛿𝑦 & = G𝑚 & H I + 𝑛 & H I + ⋯ 𝑢 𝑥 𝑦 4# There is a simple interpretation of this. The relative uncertainty of a variable x is # . Significant figures are a crude measure of relative uncertainty. We can see that the variable with the largest relative uncertainty has the largest effect on the relative uncertainty of the result. Section C: Data Analysis II – The Method of Least Squares Data and Curve Fitting We will consider a set of data points {(𝑥* , 𝑦* ), (𝑥& , 𝑦& ), ⋯ } and a function 𝑦 = 𝑓(𝑥) that fits the data. Generally, the function 𝑓(𝑥) will have some parameters and we find the optimal set of parameters to get a bestfit. Here, we will consider the simplest case of a best-fit line. The function has the form: 𝑦 = 𝑓(𝑥) = 𝑚 𝑥 + 𝑏 and the parameters are m and b. We will use a method known as least squares to find the optimal values of m and b and thus the best-fit line. Error and Data Analysis - II Page 4 of 5 � (�� � �� ) δ� (�� � �� ) (�� � �� ) (�� � � (�� )) δ� (�� � � (�� )) δ� (�� � � (�� )) δ� (�� � � (�� )) (�� � �� ) δ� (�� � � (�� )) (�� � �� ) �� �� � (�) = � � + � �� �� � �� The error for a particular data point (𝑥2 , 𝑦2 ) is 𝛿2 = |𝑓(𝑥2 ) − 𝑦2 |. The idea of the least squares method is to sum the squared errors and then minimize this sum over the parameters m and b. Label the sum of these squared errors as 𝑆(𝑚, 𝑏). 𝑆(𝑚, 𝑏) = \ 𝛿2& = \(𝑓(𝑥2 ) − 𝑦2 )& = \(𝑚 𝑥2 + 𝑏 − 𝑦2 )& 2 2 2 Each sum goes from 1 to N, where N is the number of data pairs. Finding the Parameters In Calculus I you learned that to find a local maximum or minimum of a function, set the derivative to zero. Here since 𝑆(𝑚, 𝑏) is a function of two variables, we will minimize it by setting both partial derivatives to zero. 0= 𝜕 𝑆(𝑚, 𝑏) = 2 \(𝑚𝑥2 + 𝑏 − 𝑦2 ) 𝑥2 = 2 ]𝑚 \ 𝑥2& + 𝑏 \ 𝑥2 − \ 𝑥2 𝑦2 ^ 𝜕𝑚 2 0= 2 2 2 𝜕 S(𝑚, 𝑏) = 2 \(𝑚𝑥2 + 𝑏 − 𝑦2 ) = 2 ]𝑚 \ 𝑥2 + 𝑁𝑏 − \ 𝑦2 ^ 𝜕𝑏 2 2 2 The reason for the N in the second expression is that the sum of a constant gives that constant multiplied by number of data pairs N, ∑ 𝑏 = 𝑁𝑏. Dividing a sum by N gives an average. 𝑥̅ = 1 1 1 ccc& = 1 \ 𝑥2& and 𝑥𝑦 \ 𝑥2 , 𝑦c = \ 𝑦2 , 𝑥 ccc = \ 𝑥2 𝑦2 𝑁 𝑁 𝑁 𝑁 2 2 2 2 Dividing the two expressions (from the derivatives) by 2𝑁 gives: 0 = 𝑚 ccc 𝑥 & + 𝑏 𝑥̅ − ccc 𝑥𝑦 and 0 = 𝑚 𝑥̅ + 𝑏 − 𝑦c . To eliminate b multiply the second expression by 𝑥̅ and subtract it from the first expression. ccc& − 𝑥̅ & e − (𝑥𝑦 0 = 𝑚 d𝑥 ccc − 𝑥̅ 𝑦c) . Error and Data Analysis - II Page 5 of 5 With this we can find m and using the second expression we can solve for b in terms of m. 𝑥𝑦 ccc − 𝑥̅ 𝑦c 𝑚= & and 𝑏 = 𝑦c − 𝑚 𝑥̅ ccc 𝑥 − 𝑥̅ & Let us summarize with an outline of the procedure for finding the best-fit line. Start with the data ccc& and ccc, {(𝑥* , 𝑦* ), (𝑥& , 𝑦& ), ⋯ }, next evaluate the averages 𝑥̅ , 𝑦c, 𝑥 𝑥𝑦 and then find m and b. Quality of Fit – R2 The smaller the value of 𝑆(𝑚, 𝑏) = ∑2 𝛿2 & , the better the fit. This value has the units of the y-values and rescales with the y-values; if we multiply all the y-values by 100 then the 𝑆(𝑚, 𝑏) value increases by a factor 1002. We want a dimensionless (without units) and scale invariant measure of quality of fit. From our data set we can define D by: 𝐷 = \(𝑦2 − 𝑦c)& 2 Using D as a denominator we can get 𝑆(𝑚, 𝑏)/𝐷 as dimensionless and scale-invariant. With 𝑆/𝐷, a smaller value is a better fit. We define R2 to vary typically from 0 to 1 with a perfect fit being 𝑅& = 1, but negative values are possible ∑2 𝛿2 & 𝑆(𝑚, 𝑏) 𝑅 =1− =1− ∑2(𝑦2 − 𝑦c)& 𝐷 & Name ________________ Group ______ Worksheet: Error and Data Analysis - II Equipment and Setup: Mathematica file – DataAnalysis-II.nb Section A: Partial Derivatives and the Total Derivative Computer Setup for Section A 1. The interactive panel shows a 3-dimensional plot of 𝑧 = 𝑓(𝑥, 𝑦) = 0.02𝑥 & 𝑦 + 0.06𝑥𝑦 − 0.06𝑦 & + 1.2. Note that the x, y and z directions are labeled. 2. The control is a 2-dimensional slider that varies the x and y coordinates. 3. The small button in the upper right resets the panel. The coordinates are reset to their initial values of 𝑥 = −2 and 𝑦 = 2. 4. The red dot is the point 𝑃 = (𝑥, 𝑦, 𝑧), where 𝑧 = 𝑓(𝑥, 𝑦). The blue plane is the surface of constant x passing through P and the green plane is the surface of constant y through P. Curves of constant x and y are drawn onto these planes. 5. At the right side these two planes are redrawn as graphs of z versus y and z versus x, with colors that match the planes. The curves are also drawn on the side graphs and the line tangent to P is also included, which has a slope corresponding to the partial derivatives. The numerical values of the partial derivatives at P are included above the graphs. Data Recording for Section A 1. Move the slider to find some position where both partial derivatives are positive, 𝜕𝑓⁄𝜕𝑥 > 0 and 𝜕𝑓 ⁄𝜕𝑦 > 0. Record the values of (𝑥, 𝑦), (There are many possible answers but find some position.) x = _________________ and y = ___________________ 2. Do the same for, 𝜕𝑓⁄𝜕𝑥 < 0 and 𝜕𝑓⁄𝜕𝑦 < 0. x = _________________ and y = ___________________ 3. Hit the reset button to return to the initial values of 𝑥 = −2 and 𝑦 = 2 and record the values of the partial derivatives at that position. You will calculate these values later in a question. $" $# $" = ________________ and $% = ________________ Questions A-1. Follow the discussion in the theory section to evaluate the partial derivatives of the function below, as functions of x and y and also find its total differential. 𝑓(𝑥, 𝑦) = 𝑎𝑥 ( + 𝑏𝑥𝑦 & − 𝑐𝑦 $" $# = ___________________ , $" $% = ________________ and 𝑑𝑓 = _____________________ Worksheet - Error and Data Analysis - II Page 2 of 4 A-2. Do the same for the function used in the interactive panel: 𝑓(𝑥, 𝑦) = 0.02𝑥 & 𝑦 + 0.06𝑥𝑦 − 0.06𝑦 & + 1.2 . $" $# = ___________________ , $" $% = ________________ and 𝑑𝑓 = _____________________ A-3. Now evaluate these partial derivatives at 𝑥 = −2 and 𝑦 = 2 and verify that they are the same as you recorded in part 3 of the data above. $" $# = ___________________ , $" $% = ________________ and 𝑑𝑓 = _____________________ Section B: Propagation of Error Questions B-1. Suppose that 𝑦 = ln(𝑥), where 𝑥 = 1.0385 ± 0.0003. Find y with the appropriate uncertainty. Worksheet - Error and Data Analysis - II Page 3 of 5 B-2. Consider 𝑦 = 𝐴 sin(𝜔𝑡), which is a function of A, w and t, where 𝐴 = (2.054 ± 0.012)m, 𝜔 = (2735 ± 8) s )* and 𝑡 = (0.000341 ± 0.00015)s. Find y with the appropriate uncertainty. B-3. In simple kinematics, the time of fall for an object dropped from a height h is 𝑡 = r2ℎu𝑔. Suppose the values of g and h where experimentally determined to have the values and uncertainty 𝑔 = (9.8066 ± 0.0021)m/s & and ℎ = (25.631 ± 0.015)m. Find 𝛿𝑡/𝑡, the relative uncertainty in t and also find the t value with its uncertainty. Worksheet - Error and Data Analysis - II Page 4 of 5 Section C: The Method of Least Squares Computer Setup for Section C 1. When the “Generate Data” button is pressed, the interactive panel generates a set of 7 random (𝑥, 𝑦) data points. We are trying to find the best-fit line for this data. 2. There are two sliders that vary the values of the slope m and intercept b of a line; the values of m and b are shows to the right of each slider. 3. The method of least squares minimizes ∑2 𝛿2& , the squared error discussed in the theory section. There is a check box “Show Squared Error and R2”, where R2 is the quality of fit parameter also discussed. Checking the box gives these values below. The “Snap to Best-fit” check box will be used at the end. 4. The small button in the upper right resets the panel. Data Recording for Section C 1. Press the “Generate Data” button and record the values. x y 2. Adjust the sliders for the slope and intercept to get what, by your eye, is the best-fit. You will have to alternate varying m and b to get your best values. Record these values. m = __________________ and b = __________________ 3. Now check the “Show Squared Error and R2” box and record their values from your by-eye estimate. Squared Error = __________________ and R2 = __________________ 4. Repeat the procedure from part 1, varying the sliders, to get values of m and b that give the smallest squared error and largest R2. (This is a bit tedious and sensitive to the controls. You have to make small adjustments, alternating between both sliders. You do not have to do this perfectly.) Record m and b to and also the squared error and R2 values. m = __________________ and b = __________________ Squared Error = __________________ and R2 = __________________ Worksheet - Error and Data Analysis - II Page 5 of 5 5. Now check “Snap to best-fit”. This changes the values of m and b and the squared error and R2 values. Record these values. You will calculate these m and b values in the question that follows. m = __________________ and b = __________________ Squared Error = __________________ R2 = __________________ and Questions C-1. Use the table below to guide you through the calculation of the m and b values. The x and y columns come from your data table. The xy and x2 columns are evaluated from the x and y values to the left. At the bottom, for the row with summations, you sum the numbers above. For the row with averages, you divide the sums by n, the number of data points (7 in this case) to get the averages. x y x2 xy \ 𝑥2 = \ 𝑦2 = \ 𝑥2 𝑦2 = \ 𝑥2& = 𝑥̅ = 𝑦c = ccc = 𝑥𝑦 ccc 𝑥& = 2 2 𝑚= 5555)#̅ %5 #% 55 ! )#̅ ! #55 2 2 = _________________ and 𝑏 = 𝑦c − 𝑚 𝑥̅ = ______________________ C-2. Your values should be the same as from part 5. Are they? Free Fall Experiment Equipment and Setup: Picket fence, Photogate, Capstone file – Free Fall.cap Theory Free fall is an example of one-dimensional motion with constant acceleration. When under the influence of only gravity, meaning there is no friction or any other force, bodies will fall with a constant downward acceleration of g, the accepted value of g is: m 𝑔accepted = 9.80 ' s In this experiment we will measure g. The tools we will use to measure acceleration will be repeated in several later experiments. One of the goals of this experiment is to understand how the Capstone program calculates velocity from a falling object that is referred to as a picket fence. x The Picket Fence and the Photogate A photogate uses a photocell with an infrared light source and detects whether the light beam is blocked or not. The picket fence we will use is a clear strip of plastic with black stripes. When the picket fence is dropped through the photogate, the Capstone program records when the light is blocked. Specifically, it is measuring the time between each time the light beam is blocked; this is when the leading edge of the black stripes passes through the photogate. The distance between leading edges is a constant Δ𝑥, which you will measure. The times between leading edges: Δ𝑡( , Δ𝑡' ,… get smaller with time because the picket fence is speeding up. Next leading edge passes photogate at t3 = t2 +Δt3 . x 3 = x 2 + Δx Δx Next leading edge passes photogate at t2 = t1 +Δt2 . x 2 = x 1 + Δx Δx Next leading edge passes photogate at t1 = t0 +Δt1 . x 1 = x 0 + Δx Δx x0 = 0 First leading edge passes photogate at t0 = 0. Δt1 Δt2 Δt3 A running total of times 𝑡) , 𝑡( , 𝑡' ,… can then be found, where we start with 𝑡) = 0. By adding the next Δ𝑡* t0 = 0 t1 = t0 +Δt1 t2 = t1 +Δt2 t3 = t2 +Δt3 to the previous time give the next time 𝑡* = 𝑡*+( + Δ𝑡* Getting a running total of 𝑥 values is similar, but easier because the Δ𝑥 is constant; start with 𝑥) = 0 and add the Δ𝑥 to the last x to get the next x value: 𝑥* = 𝑥*+( + Δ𝑥. Thus, we can get position as a function of time from the data. This is shown in the diagram to the right and above. Velocity and Time Calculation In addition to position as a function of time, the Δ𝑡 data combined with the measured value of Δ𝑥 also gives a way to get velocity as a function of time. This is a bit more subtle. For motion with constant acceleration, we will see that the average velocity over some time interval is the average of the initial and final velocities for that interval. With constant acceleration, the velocity as a function of time graph is a straight line, since 𝑣(𝑡) = 𝑣) + 𝑎𝑡. We also know that since velocity is the time derivative of position x, it follows that the displacement Δ𝑥 is the area under the velocity versus time graph. The plot below shows this. Because the graph is a line, the area under the graph is the area of a trapezoid. The area of a trapezoid is the same as the area of a rectangle with the same base b, with a height that is the average of the trapezoid’s heights ℎ( and ℎ' . t Free Fall Experiment Page 2 of 4 𝑣* + 𝑣, ℎ( + ℎ' ⟹ Area = Δ𝑥 = Δ𝑡 2 2 With this it follows that, for constant acceleration, the average velocity is the average of the initial and final velocities. Area = 𝑏 v vf tf v(t )t = Area = vi + v f ti 2 Δt v(t )= v0 +at vi + v f Δ𝑥 𝑣* + 𝑣, 𝑣ave = = Δ𝑡 2 This shows how to get the velocity as a function of time from the raw Δ𝑡 data. Associate the average velocity Δ𝑥/Δ𝑡 with the instantaneous velocity at 𝑡 = 𝑡mid , the midpoint of the time interval. 𝑡 = 𝑡mid Δ x = 2 vi ti 1 Δ𝑥 = <𝑡* + 𝑡, = ⟹ 𝑣 = 𝑣ave = 2 Δ𝑡 tmid = ti + t f tf 2 Δt The discussion in the previous section showed how to get a running total of times from the Δ𝑡 data. For each time interval we find 𝑡mid and 𝑣ave using the expressions above, remembering that the measured Δ𝑥 is constant. That then gives the experimental calculation of velocity as a function of time. Statistics As a follow-up to the previous experiments on data analysis, we will use calculate the mean and standard deviation of your experimental measurements of the acceleration g. To review the relevant formulas, if you have a list of N experimental values xi, where in your case the x values are the g values. The mean or average 𝑥̅ is given by: " 1 𝑥 = % 𝑥! 𝑁 !#$ The percent error is a measure of accuracy, comparing the experimental value to the accepted value. It is given by: % Error = experimental − accepted × 100% accepted The standard deviation sd is a measure of precision. It measures the spread of the data or how close the data is to itself; the expression for standard deviation is: " 1 𝑠𝑑 = ( %(𝑥! − 𝑥 )% 𝑁−1 !#$ t Free Fall Experiment Page 3 of 4 Procedure Setup 1. Turn on the Pasco interface at your lab station and plug the photogate into Digital Channel 1. 2. In the “Physics Folder” on your computer desktop open Free Fall.cap . 3. The photogate needs to be placed near the edge of the lab table so the picket fence can fall freely through it without hitting the photogate or table. Have something soft under the picket fence so it doesn’t crash into the hard floor. It could crack or break. (A) Data Recording – Finding Acceleration 1. Hold the picket fence vertically slightly above the photogate. In Capstone, click Record and drop the picket fence. It should not hit anything or rotate while passing though the photogate. If it falls through on an angle that will introduce error. After the picket fence has passed through the photogate, click the Stop button. 2. The times, Δt between leading edges should be listed in the table in Capstone. The program automatically makes a plot of velocity versus time. The run number is listed in the information box on the graph. 3. Click on the graph to make it active. Maximize the graph by clicking the “scale axes” tool (above graph, on left). If the data does not look linear then there was some error in dropping the picket fence; discard the and try again. To delete the last run click “Delete Last Run” at the bottom of the display (triangle with X), or delete specific runs using the drop down menu next to this control. You can switch between runs using the drop down menu on “Data Selector” tool above the graph (triangle). 4. Once you have a linear set of data, fit the data with a linear best-fit using the drop down menu on the “Curve Fits” tool above the graph. 5. From the equation displayed as the best-fit line, determine the acceleration of your picket fence and record it in Table 1 below as your measured acceleration for Trial 1. 6. Repeat this process until you have 8 good data runs and have recorded 8 measured values of acceleration on table 1. Delete any unwanted runs, but keep at least 8 good runs. 4. For the eight measured free-fall accelerations, calculate the average, the percent error of that average value using the given accepted value and then the standard deviation or your eight values. Record these in Table A. Free Fall Experiment Page 4 of 4 (B) Data Recording – Reproducing the Computer Calculation of Velocity versus Time 1. In Capstone, review your eight data sets and find the data set that gives the best agreement between the measured acceleration and the accepted value of the acceleration due to gravity, g. Select this data set using the dropdown menu on the “Data Selector” tool above the graph (triangle). 2. Use a ruler to measure the spacing between successive pairs of leading edges of the black stripes on your picket fence. Record that as Δ𝑥 above Table B. 3. Copy the Δt values of your selected run from the table in Capstone into the first column in Table B. 4. Next fill out the column in Table B called ‘Time at Each Leading Edge’. The first value is t = 0 s by default. Successively add Δt from the first column to get the next value. (See the instructions below the table and the discussion in the theory section.) 5. Now fill in the column called ‘Midpoint of Time Interval’. This is the average of the two times listed in the adjacent cells to the left. 6. Now fill in the column called ‘Average Velocity for Time Interval’ using the formula given below the table. Note that Δx is the same for each of these calculations, so this should be easy. 7. Now fill in the column called ‘Position of Each Leading Edge’. The first value is 0 m, then add Δx successively to get each value. (C) Questions and Graphs 1. Question 1 will have you plot position as a function of time from your calculations in Table B and answer questions from that plot. 2. Question 2 will similarly have you plot velocity as a function of time from your calculations in Table B and then answer questions from that plot. Free Fall – Data Sheet Name________________ Group______ Table A Trial Acceleration (m/s2) 1 2 3 4 5 6 Average: x 7 % Error of Average 8 Standard Deviation: sd Dx is the spacing between the leading edges. Dx = ___________________ Table B Raw Data (from computer) Dt (s) Time at each Leading Edge t (s) Midpoint of Time Interval Average Velocity for Time Interval 𝑡mid (s) 𝑣ave (m/s) 0 tf = ti + Dt x (m) 0 ti Dt Position of Each Leading Edge 𝑡mid 𝑡* + 𝑡, = 2 𝑣ave ∆𝑥 = ∆𝑡 xi x f = xi + Δx Free Fall – Data Sheet Page 2 of 2 Question 1: Include a graph made using software of x vs. t (from your calculations in the Table B. This is from the x column and the t column.) The graph should be a parabola. From the equation of the best-fit parabola find the acceleration and the initial velocity. Question 2: Also include a graph v vs. t (from your calculations of 𝑣ave and tmid,.) draw the best-fit line and find its slope. Compare the slope with the accepted value of 9.80 m/s2; give the % error. Question 3: From your graph of v vs. t find the speed of the picket fence when the first leading edge passed the photogate. Question 4: If the picket fence were dropped at an angle how would this affect your results? (The picket fence is in a vertical plane but its side is skewed from vertical, as shown.) Would your measured value of g be too large or too small? What percent error would be caused by dropping the picket fence at a 5° angle from vertical? At a 10° angle? 5 cm q The Force Table Equipment and Setup: Force table with strings and 4 hangers, Set of masses In this experiment, we will add vectors two ways: With the analytical method, we will calculate the magnitude and direction angle of the sum of two vectors from the magnitude and directions of the vectors we are adding. The force table will provide the experimental method, where we vary hanging weights to find the magnitude and direction angle of the vector sum. Vector Basics A vector is a physical quantity with both a magnitude and direction. A scalar has just a magnitude. Generally, the magnitude of a vector is a scalar that must be non-negative. A vector can be written in terms of components and the basis unit vectors, or in terms of just components using the angled bracket notation. 𝐴⃑ = 𝐴! 𝑥% + 𝐴" 𝑦% = 𝐴! 𝚤̂ + 𝐴" 𝚥̂ = 〈𝐴! , 𝐴" 〉 The x- and y-components of 𝐴⃑ are Ax and Ay. In the different notations, the basis unit vectors in the x- and ydirections are 𝑥% = 𝚤̂ = 〈1, 0〉 and 𝑦% = 𝚥̂ = 〈0, 1〉 respectively. y Ay  Ay y A =〈Ax , Ay 〉  y Ax  x  Ax x x The magnitude of this is A. In two dimensions we can specify a direction by an angle q, which we will refer to as the direction angle. This angle is unique, modulo 360°; this means that if two angles differ by a multiple of 360°, they describe the same direction. We use the polar coordinates convention for angles, where an angle is measured counterclockwise from the positive x-axis. The magnitude and direction angle, A and q, are related to the components, Ax and Ay, just as the polar coordinates, r and q, are related to the rectangular coordinates, x and y. The components can be found in terms of the magnitude and direction angle: 𝐴 and 𝜃 ⟹ 𝐴! and 𝐴" 𝐴! = 𝐴 cos 𝜃 and 𝐴" = 𝐴 sin 𝜃 (Equations 1) and the magnitude and direction angle can be found in terms of the components: 𝐴! and 𝐴" ⟹ 𝐴 and 𝜃 & 𝐴= @𝐴!# + 𝐴"# and 𝜃 = A tan$% &! " when 𝐴! > 0 & 180° + tan$% &! " when 𝐴! < 0 (Equations 2) The Force Table Page 2 of 8 The subtlety in finding the direction angle follows from there being two angles satisfying tan 𝜃 = 𝐴" ⁄𝐴! , the arctangent of 𝐴" ⁄𝐴! and 180° added to that arctangent. The arctangent always gives an answer in quadrants I and IV. If the angle is in those quadrants, you use the arctangent. However, if the angle is in quadrants II or III, then you need to take the angle opposite; that is when you add 180°. The sign of Ax determines which is needed. When 𝐴! > 0, the vector is in the first or fourth quadrants and when in the second or third quadrants, 𝐴! < 0. See the figures below. u Quadrant IV Quadrant I Quadrant II Quadrant III Quadrant IV Ay Quadrant II Ax < 0 u1 Quadrant I Ax > 0 A tan-1 u2 θ 180°+ tan-1 u2 -90° Ay -1 tan u1 90° 180° 270° 180°+ tan-1 u1 360° θ Ax Quadrant III Ax < 0 u2 Ax Quadrant IV Ax > 0 L⃑ are vectors then When we add vectors, the components of the sum are the sum of the components. If 𝐴⃑ and 𝐵 their sum, which we will call the resultant 𝑅L⃑, is L⃑ = 〈𝐵! , 𝐵" 〉 then 𝑅L⃑ = 𝐴⃑ + 𝐵 L⃑ = 〈𝐴! + 𝐵! , 𝐴" + 𝐵" 〉 = 〈𝑅! , 𝑅" 〉 𝐴⃑ = 〈𝐴! , 𝐴" 〉 and 𝐵 L⃑ by putting the tail of 𝐵 L⃑ at the tip of 𝐴⃑ and the resultant 𝑅L⃑ is the vector from the tail Graphically, we add 𝐴⃑ and 𝐵 L⃑. Alternatively, we can add the vectors using the parallelogram rule. Put the vector’s tails of 𝐴⃑ to the tip of 𝐵 together, complete the parallelogram and draw the vector to the opposite corner. B R By R y Bx A Ax Ay R B A Rx Forces and the Force Table This experiment is about vectors, however the vectors we will add are force vectors. Because the experiment is designed to precede a discussion of forces in the lecture course, we must outline the aspects of The Force Table Page 3 of 8 forces needed for the experiment. Since nothing will be moving in this experiment, we have what is known as static equilibrium; in static equilibrium all forces acting on a body must sum to zero: ∑ 𝐹⃑ = L0⃑. Note that the L⃑ = 〈0, 0〉. zero-vector is the vector with zeros as components: 0 We will hang masses on a string using a hanger with added masses. The total hanging mass is m, the sum of the added mass and the hanger’s mass. Weight W is a force; it is the force of gravity acting on an object. In the SI system of units we measure force in newtons N and we measure mass in kilograms kg. The magnitude of the weight is related to the mass by: 𝑊 = 𝑚𝑔 where 𝑔 = 9.80 m/s # For the purposes of this experiment, we will avoid multiplying all masses by 9.80 m/s2 to get the weight by using 𝑔∙kg as our units of force. It follows that the hanging mass in kg corresponds to the same magnitude force in 𝑔∙kg. For instance, if we have a total hanging mass of 200 grams, that is 0.20 kg and thus corresponds to a weight of 0.20 𝑔∙kg. Because there is no acceleration, the forces on the hanger must balance. The other important fact about forces for this experiment is that for an ideal pulley, the string’s tension F is the same on either side of the pulley. An ideal pulley is frictionless and light; this means that the friction in the pulley is negligible, and the mass of the pulley is small compared to the other masses. (In this experiment, since there is no acceleration, the assumption that the pulley is light is not needed.) The Force Table Page 4 of 8 Procedure for the Analytical Method Vector Addition For Vector Addition 1, 2 and 3, you will be given the magnitude and direction angle of the two (or three) force vectors you are summing. From these magnitudes and direction angles you find the vectors 𝐹⃑% and 𝐹⃑# . Then add these vectors to get the resultant 𝑅L⃑ = 𝐹⃑% + 𝐹⃑# . You will then find the magnitude R and direction angle qR of the resultant vector, the sum of the force vectors. An example will be helpful. Suppose you are given the magnitude and direction angles of 𝐹⃑% and 𝐹⃑# : 𝐹% = 0.155 𝑔∙kg, 𝜃% = 25°, 𝐹# = 0.275 𝑔∙kg and 𝜃# = 135° Find the components of the force vectors using (Equations 1). 𝐹⃑% = 〈𝐹%! , 𝐹%" 〉 = 〈𝐹% cos 𝜃% , 𝐹% sin 𝜃% 〉 = 〈0.14048, 0.06551〉 𝑔∙kg 𝐹⃑# = 〈𝐹#! , 𝐹#" 〉 = 〈𝐹# cos 𝜃# , 𝐹# sin 𝜃# 〉 = 〈−0.19445, 0.19445〉 𝑔∙kg Adding these give the resultant 𝑅L⃑ and its components: 𝑅L⃑ = 〈𝑅! , 𝑅" 〉 = 𝐹⃑% + 𝐹⃑# = 〈−0.05398, 0.25996〉 𝑔∙kg Using (Equations 2) we can find the magnitude and direction angle of the resultant. ' 𝑅 = @𝑅!# + 𝑅"# = 0.266 g∙kg and 𝜃' = 180° + tan$% a'! b = 101.7° " This is the result we are seeking. Note that the 180° was added to the arctangent because 𝑅! < 0. The vector diagram for the worked-out example For Vector Addition 3, you will be adding three force vectors. Follow the procedure outlined above but now include the third force: 𝑅L⃑ = 𝐹⃑% + 𝐹⃑# + 𝐹⃑( . The Force Table Page 5 of 8 Vector Resolution The last two parts of this experiment, Vector Resolution 1 and 2, involve finding the components of vectors. For each case you will be given F and q, the magnitude and direction angle of some force vector, and you want to find the components 𝐹! and 𝐹" . Calculating these components is a simple application of (Equations 1): 𝐹! = 𝐹 cos 𝜃 and 𝐹" = 𝐹 sin 𝜃 Experimental Procedure - the Force Table Setup • Obtain a force table (with four pulleys and mass hangers) and a set of masses. • Looking at the force table from the top, make sure that there is some visible gap between the pulleys and the edge of the force table. All pulleys should be able to rotate freely without rubbing against the edge of the force table. If the pulleys need adjustment, locate the horizontal screw near where the pulley meets edge of the force table. Loosen this screw and tilt the pulley outward until it no longer rubs against the edge of the table. • Looking at the force table from the side, make sure that all four pulleys are at the same vertical height, so that all the strings lie in a single horizontal plane (or as nearly so as possible). If one or more pulleys need adjustment, use the screw referred to in the previous step to tilt the pulley(s) until all pulleys lie very nearly in a single horizontal plane. • Hold the ring so that it is centered on the post and look down at the force table from the top. Make sure that all strings pull radially away from the center of the post. From time to time, the strings can slip from side to side along the circumference of the ring so that they are no longer radial. This results in some error in the angle giving the direction of the tensions acting on the ring. If one or more of your strings have slipped offcenter, slide them so that all strings pull radially away from the post. Pulley Pulley Mass hanger and mass Top View and Side View of Force Table The Force Table Page 6 of 8 Vector Addition To experimentally add two vector, we use the force table. We cannot find the resultant directly. We find the third vector, called the equilibrant, that cancels the other two and balances the ring in the force table. L⃑ 0 = 𝐹⃑% + 𝐹⃑# + 𝐸L⃑ It follows that the resultant can then be found from the equilibrant: 𝑅L⃑ = −𝐸L⃑ To set 𝐹⃑% and 𝐹⃑# set the pulleys at angles 𝜃% and 𝜃# . Then hang the weights F1 and F2 from the pulleys. If the force is 0.125 g×kg, then you will hang total mass of 0.125 kg = 125 grams. With those two strings pulling on the ring at the center, take a third string and with your hand find the direction needed to balance the other two forces. This is the direction angle of the equilibrant 𝐸L⃑ . Set the third pulley at that angle and vary the weights hanging from it until that force balances the other two. The magnitude of the hanging weight in g×kg is the magnitude of the equilibrant, E. But since 𝑅L⃑ = −𝐸L⃑ the magnitudes of 𝑅L⃑ and 𝐸L⃑ are equal, 𝑅 = 𝐸. The vectors are in opposite directions, so the direction angle for the resultant 𝜃' is related to the direction angle of the equilibrant 𝜃) by 𝜃' = 𝜃) ± 180° The magnitude of the resultant is the weight hanging. If the equilibrant is at 250° then the angle of the resultant is 70°. There are four pulleys and strings but you will only need three for Vector Addition 1 and Vector Addition 2. In those cases, make sure the fourth string is not hanging with any tension; it is best to set that fourth hanger on the force table. Vector Addition 3 will involve adding three force vectors and that will use all pulleys and hangers. To describe the experimental determination of R and q using the force table, we will consider the same example given in the analytical method: 𝐹% = 0.155 𝑔∙kg, 𝜃% = 25°, 𝐹# = 0.275 𝑔∙kg and 𝜃# = 135° Over pulleys set at 25° and 135°, hang 0.155 kg = 155 grams and 0.275 kg = 275 grams, respectively. Take a third string and pull in varying directions until you find the direction needed to balance the forces from the other two strings. Suppose this is at 283°; that then is the direction angle for the equilibrant. Set the third pulley there and start hanging masses over that third pulley until you get it to balance. The total mass hanging gives the magnitude of the equilibrant. Suppose this is 265 grams = 0.265 kg; this then gives the magnitude of 𝐸L⃑, the equilibrant, in 𝑔∙kg: 𝜃) = 283° and 𝐸 = 0.265 𝑔∙kg Since 𝑅L⃑ = −𝐸L⃑ we know that their magnitudes are equal and directions are opposite. This would give the values to record: 𝜃' = 𝜃) − 180° = 103° and 𝑅 = 𝐸 = 0.265 𝑔∙kg The Force Table Page 7 of 8 The diagram for the sample example Vector Resolution The last two parts of the experiment involve vector resolution, finding the components of a vector using the force table. With Vector Resolution 1, we have an acute angle which makes the procedure simple. To relate this to the Vector Addition part, think of the vector 𝐹⃑ , with 𝐹 = 0.125 𝑔∙kg and 𝜃 = 60°, as the resultant vector 𝑅L⃑; it is the sum of vectors in the x- and y-directions: 𝐹⃑ = 𝐹! 𝑥% + 𝐹" 𝑦% Set three pulleys at 0°, 90° and 240°. We then hang nothing at 60° but hang 0.125 kg = 125 grams opposite to that at 240°. This is −𝐹⃑ which you should view as the equilibrant. But now we vary the weights hanging at 0° and 90°. Since the angle is acute, both components are positive; what is hanging at 0° is the x-component and what is hanging at 90° is the y-component. The Force Table Page 8 of 8 The diagram for Vector Resolution 1 When the angle is not acute we can have negative components. This is the case in Vector Resolution 2. Hang 0.155 kg = 155 grams opposite to 𝜃 = 125°, then set other two pulleys along the axes as needed to balance – 𝐹⃑ . To get the x-component set the pulley at either 0° or 180° and for the y-component set the pulley at 90° or 270°. The hanging weights will give |𝐹! | and g𝐹" g. Be careful to include the appropriate signs for 𝐹! and 𝐹" . The Force Table – Worksheet Name________________ Group______ Data Table Forces LL⃑ Resultant - 𝑹 (magnitude and direction angle) Experimental Analytical 𝐹% = 0.125 𝑔∙kg and 𝜃% = 30° 𝑅= 𝑅= 𝐹# = 0.125 𝑔∙kg and 𝜃# = 120° 𝜃' = 𝜃' = 𝐹% = 0.075 𝑔∙kg and 𝜃% = 20° 𝑅= 𝑅= 𝐹# = 0.125 𝑔∙kg and 𝜃# = 120° 𝜃' = 𝜃' = 𝑅= 𝑅= 𝜃' = 𝜃' = 𝐹! = 𝐹! = 𝐹" = 𝐹" = 𝐹! = 𝐹! = 𝐹" = 𝐹" = Vector Addition 1 Vector Addition 2 𝐹% = 0.125 𝑔∙kg and 𝜃% = 30° Vector Addition 3 𝐹# = 0.175 𝑔∙kg and 𝜃# = 80° 𝐹( = 0.165 𝑔∙kg and 𝜃# = 150° Vector Resolution 1 Vector Resolution 2 𝐹 = 0.125 𝑔∙kg and 𝜃 = 60° 𝐹 = 0.155 𝑔∙kg and 𝜃 = 125° Show your work. Use the back of this sheet or a separate sheet. The Force Table – Worksheet Page 2 of 2 Questions L⃑ have magnitudes A and B, respectively. What are the largest and smallest Question 1 Two vectors 𝐴⃑ and 𝐵 L⃑, the sum of the two vectors? Give your answer in terms of A and B. values of the magnitude of 𝐴⃑ + 𝐵 Remember that the magnitude of a vector must be positive. Question 2 Give a different derivation of the analytical results of Vector Addition 1, using the parallelogram rule and simple geometry. Note that since the two vectors have the same magnitude and are perpendicular, the parallelogram is a square. Question 3 Using the values given in Vector Addition 2, find the magnitude and direction angle of the difference of the two vectors, 𝐹⃑# − 𝐹⃑% . Question 4 When experimentally determining the components of 𝐹⃑ in Vector Resolution 2, at what angles were the pulleys placed? Give all three angles used. Atwood’s Machine Experiment Equipment and Setup: Smart Pulley, Smart Pulley clamp, Mass Hangers (2), Set of masses, String, Capstone file – Atwood’s Machine.cap Theory Newton’s second law 𝐹⃑!"# = 𝑚𝑎⃑ is the basis for mechanics; it relates the vector sum of all the forces acting on a body to the mass of the body and its acceleration vector. To apply the second law to any problem, the first step is to draw a free-body diagram for each mass; these are vector diagrams showing all the forces acting on a body. When forces are acting in one dimension only, we may write the expression without the vector arrows, with the caveat that one-dimensional vectors are real numbers with the sign giving the direction. A standard example of applying the second law is Atwood’s machine; this consists of two masses 𝑚$ and 𝑚% connected by a light non-elastic string hanging over an ideal pulley. An ideal pulley is frictionless and light; this means that the rotational friction of the pulley can be neglected and m1 m2 the mass of the pulley, meaning the part of the pulley that rotates, is negligible compared to all the other masses in the problem. To the right is a drawing of the Atwood’s machine set-up combined with free-body T T diagram for each mass. For this experiment we will take 𝑚% to be the larger of the two masses. This implies that 𝑚% will accelerate downward a and 𝑚$ will accelerate upward. Also, since the string is not elastic and cannot stretch, the magnitudes of both accelerations are the same. We m1 g m 2 will choose the direction of motion for the system to be positive, so m2 g downward is positive for 𝑚% and upward is positive for 𝑚$ . Acting on m1 both masses there is the tension in the string T acting upward and its weight 𝑚$ 𝑔 or 𝑚% 𝑔 acting downward. The second law then gives the equations describing the motion of both masses. 𝐹net,$ = 𝑚$ 𝑎 ⟹ 𝑇 − 𝑚$ 𝑔 = 𝑚$ 𝑎 and 𝐹net,% = 𝑚% 𝑎 ⟹ 𝑚% 𝑔 − 𝑇 = 𝑚% 𝑎 Adding these two expressions together eliminates the tension T and allows us to solve for the acceleration a. 𝑎= (𝑚% − 𝑚$ )𝑔 𝑚$ + 𝑚% This will be referred to as the theoretical acceleration 𝑎theo . For this experiment, we will label the numerator of this expression as the net force 𝐹net and the denominator as the total mass 𝑚tot . 𝐹net 𝐹net = (𝑚% − 𝑚$ )𝑔 and 𝑚tot = 𝑚$ + 𝑚% ⟹ 𝑎theo = 𝑚tot This experiment uses a “smart pulley” connected to the Pasco interface to measure the acceleration of the system. The smart pulley is a light low-friction pulley; what makes it “smart” is that it has a photogate. Just as the black stripes of the picket fence blocked the beam of the photogate in the Free Fall experiment, the pulley’s spokes block the photogate’s beam here. The calculation of the acceleration is the same as in the Free Fall experiment, except here Δ𝑥 is the distance the string moves when the pulley rotates between the leading edges of the spokes. Since in the earlier experiment you completed the character-building exercise of calculating the velocity-time values from the raw data, we do not need to repeat that here. The Capstone programs knows the appropriate value of Δ𝑥 for the smart pulley and magically gives the graph of velocity versus time, as in the Free Fall lab; but here it is not magic because we understand the calculation. a Atwood’s Machine Page 2 of 3 There are two parts of this experiment: In part A we will vary 𝐹net while keeping 𝑚tot fixed. Part B will vary 𝑚tot keeping 𝐹net fixed. Each run will use a different pair of masses 𝑚$ and 𝑚% ; the Capstone program will give the velocity versus time graph and give the equation of best-fit line; this will tell you the acceleration. This acceleration will be called the experimental acceleration 𝑎exp . Procedure Lab Table Smart Pulley Setup 1. Set up the equipment as shown to the right. Connect two hangers to a string that passes over the pulley as shown. Choose the length of the string so that one hanger is touching the floor when the other is just below the pulley. Clamp 2. Turn on the Pasco interface at your lab station and plug the photogate into Digital Channel 1. 3. In the “Physics Folder” on your computer desktop open Atwood’s Machine.cap . Data Recording – Finding Acceleration for both Parts A and B 1. The procedure for using the Capstone program is the same as in the Free Fall experiment. Here hold 𝑚% just below the pulley and 𝑚$ just touching the floor. In this experiment 𝑚% must always be larger than 𝑚$ . 2. It is essential while choosing the values of the masses to make sure that the acceleration is not too large or too small. The acceleration is too large when 𝑚% is much larger than 𝑚$ . The problem with this situation is it will cause the string to slip on the pulley and the smart pulley then misreads the acceleration. If the acceleration is very small, then whatever friction there is in the pulley will become too significant. As a rule of thumb, try to keep the acceleration values between 0.5 m/s % and 5 m/s % . If one of the trials in parts A or B is outside of this range, it is fine, but if several are, the quality of your results will suffer. 3. When recording the masses you must include the mass of the hangers. The hanger masses contribute to 𝑚#.# and thus do not cancel. Measure the mass of the hangers. 4. In Capstone, click Record and release 𝑚% . When 𝑚% hits the floor, click the Stop button. 5. In Capstone, maximize the graph by clicking the “Scale to fit” tool above the graph (on left). 6. Click the “Highlighter” tool on top of the graph and drag the edges of the highlight box to select the “good” data. This is the linear part of the graph with a positive slope. The rest of the graph is what happened after m2 hit the ground. 7. Fit the “good” data by clicking the “Curve Fits” drop down menu and selecting the linear fit. Record the slope in as 𝑎"/0 in the appropriate table, either Table A or Table B. Atwood’s Machine Page 3 of 3 (A) 𝑚#.# is fixed while 𝐹!"# is varied. 1. To keep the total mass fixed while varying the net force, move small masses from one to the other. Start with 𝑚$ with a larger mass and four smaller masses on top of it. Begin with 𝑚% a little larger than the initial 𝑚$ . Between each trial move a small mass from 𝑚$ to 𝑚% . The accelerations will increase in this process, so you want to begin with a small acceleration. First Trial m1 Last Trial m2 m1 m2 2. Record your data in Table A and complete the table. Record the value of 𝑚#.# above the table. Complete the table by calculating and recording the net force, the theoretical acceleration and the percent error in the acceleration. 3. Follow the directions below the table to graph the data. (B) 𝐹!"# is fixed while 𝑚#.# is varied. 1. To keep the net force fixed while varying the total mass, add equal small masses to each mass between trials. Start with 𝑚$ small but not tiny, and with 𝑚% large compared to 𝑚$ . The accelerations will decrease in this process, so you want to begin with a larger acceleration. First Trial m1 Last Trial m2 m1 m2 2. Record your data in Table B and complete the table. Record the value of 𝐹!"# above the table. Complete the table by calculating and recording the total mass, the theoretical acceleration, and the percent error in the acceleration. Also, to assist you in graphing, calculate 1/𝑎"/0 . 3. Follow the directions below the table to graph the data. Atwood’s Machine – Data Sheet Name____________ Group____ (A) Variation of Acceleration with Fnet Keeping mtot Fixed mtot = _______________ kg Table A m1 (kg) m2 (kg) aexp 2 (m/s ) Fnet atheo (N) 2 (m/s ) % error in a Graph Fnet versus aexp and include the best-fit line. Find the slope of this line and compare the slope with mtot. Give the % error. Atwood’s Machine – Data Sheet Page 2 of 2 (B) Variation of Acceleration with mtot Keeping Fnet fixed Fnet = _______________ N Table B m1 m2 aexp mtot atheo (kg) (kg) (m/s2) (kg) (m/s2) % error in a 1/aexp (s2/m) Graph mtot versus 1/aexp with the best-fit line. Find the slope of this line and compare the slope with Fnet. Give the % error. Numerical and Symbolic Methods in One Dimensional Dynamics Newton's second law is the basis for dynamics. it is the rule for evolving a dynamical system forward in time. Here we will discuss the problem of one dimensional motion. The second law in one dimension states: F(x, v, t) = m a, where F is the net force which is a function of x, the position, v, the velocity, and perhaps t, the time. If at some instant t the position and velocity are given by x and v, then an infinitesimal time step ⅆ t later the values change by the rule: t → t + ⅆt x → x + v ⅆt v→v+ F(x,v,t) m ⅆt . In a numerical simulation we approximate the infinitesimal time step with a small time step Δ t. We will label an instant by i and the rules for taking a time step are ti → ti+1 = ti + Δ t xi → xi+1 = xi + vi Δ t vi → vi+1 = vi + F(xi ,vi ,ti ) m Δt . We may numerically evolve a problem by starting with the initial conditions: t0 , x0 and v0 , and by giving the mass m and the force function F. The second law is the rule that gives the differential equation for the motion. A differential equation is an equation for a function that involves the function and its derivatives. After the third semester of calculus students take a course in differential equations to learn how to solve them. Here we will just use Mathematica to solve them. The second law gives a second order differential equation for the position as a function of time x(t): ⅆ2 ⅆ t2 x(t) = 1 m Fx(t), ⅆ ⅆt x(t), t Note that the order of a differential equations is the highest number of derivatives of the unknown function. (A) Free Fall For free fall we will take down as the positive direction and the force becomes F(x, v, t) = m g. We will take the initial conditions to be t0 = 0, x0 = 0 and v0 = 0. We will also take the mass to be 2 kg. We will choose the final time to be t = 2 s and choose our time interval at first to be the relatively large value of Δ t = 0.2 s. To evaluate Mathematica input place the cursor anywhere in an input cell and hit Shift+Enter. This will be done for all Mathematica input that follows. t[0] = 0; x[0] = 0; v[0] = 0; t[final] = 2; Δt = 0.2; m = 2; g = 9.8; F[x_, v_, t_] := m g The following is the Mathematica code for finding the numerical solution. XNum and VelNum are the functions that interpolate the discrete position and velocity points with continuous functions. 2 Dynamics in 1D.nb iMaximum = (t[final] - t[0]) / Δt; Do[ t[i + 1] = t[i] + Δt; x[i + 1] = x[i] + v[i] Δt; v[i + 1] = v[i] + Δt F[x[i], v[i], t[i]] / m;, {i, 0, iMaximum - 1} ]; XNum = Interpolation[Table[{t[i], x[i]}, {i, 0, iMaximum}]]; VelNum = Interpolation[Table[{t[i], v[i]}, {i, 0, iMaximum}]]; The following is the Mathematica code for finding the symbolic solution. The function XSym is the result of DSolve, the Mathematica function for solving a differential equation. VelSym is the symbolic velocity function, found by differentiating the position XSym. XSym = DSolve[{F[X[t], X '[t], t] ⩵ m X ''[t], X[0] ⩵ x[0], X '[0] ⩵ v[0]}, X, t]〚1, 1, 2〛; VelSym = XSym '; Print["x(t) = ", XSym[t]] Print["v(t) = ", VelSym[t]] x(t) = 4.9 t2 v(t) = 9.8 t This is the code for plotting the numerical solution, the wide yellow line, and the symbolic solution as a narrow blue line. The first plot is x vs. t and the second is v vs. t. GraphicsColumn[{Plot[{XNum[T], XSym[T]}, {T, 0, t[final]}, PlotStyle → {{RGBColor[1, 1, 0.4`], Thickness[0.02`]}, {RGBColor[0, 0, 0.6`], Thickness[0.008`]}}, AxesLabel → {"t ", "x "}, Background → GrayLevel[0.6`]], Plot[{VelNum[T], VelSym[T]}, {T, 0, t[final]}, PlotStyle → {{RGBColor[1, 1, 0.4`], Thickness[0.02`]}, {RGBColor[0, 0, 0.6`], Thickness[0.008`]}}, AxesLabel → {"t ", "v "}, Background → GrayLevel[0.6`]]}] The Mathematica function CalculateAndGraph is now defined to follow the above procedure of calculating the numerical and symbolic solutions, and then graphing those solutions. Dynamics in 1D.nb CalculateAndGraph := iMaximum = t[final] - t[0] 3 ; Δt Dot[i + 1] = t[i] + Δt; x[i + 1] = x[i] + v[i] Δt; Δt F[x[i], v[i], t[i]] v[i + 1] = v[i] + ;, {i, 0, iMaximum - 1}; m XNum = Interpolation[Table[{t[i], x[i]}, {i, 0, iMaximum}]]; VelNum = Interpolation[Table[{t[i], v[i]}, {i, 0, iMaximum}]]; XSym = Quiet[DSolve[{F[X[t], X′ [t], t] ⩵ m X′′ [t], X[0] ⩵ x[0], X′ [0] ⩵ v[0]}, X, t]〚1, 1, 2〛]; VelSym = XSym′ ; Print["x(t) = ", XSym[t]]; Print["v(t) = ", VelSym[t]]; GraphicsColumn[{Plot[{XNum[T], XSym[T]}, {T, 0, t[final]}, PlotStyle → {{RGBColor[1, 1, 0.4`], Thickness[0.02`]}, {RGBColor[0, 0, 0.6`], Thickness[0.008`]}}, AxesLabel → {"t ", "x "}, Background → GrayLevel[0.6`]], Plot[{VelNum[T], VelSym[T]}, {T, 0, t[final]}, PlotStyle → {{RGBColor[1, 1, 0.4`], Thickness[0.02`]}, {RGBColor[0, 0, 0.6`], Thickness[0.008`]}}, AxesLabel → {"t ", "v "}, Background → GrayLevel[0.6`]]}] Note that with the very crude value of Δ t = 0.2 s, where there are only 10 intervals, the symbolic and numerical values are quite different. To improve this we must choose a smaller Δ t to make a more accurate numerical approximation. Now we will try Δ t = 0.02 s and redo the numerical solutions and plots. Δt = 0.02; CalculateAndGraph (B) Motion with Viscous Damping Viscous damping is a resistive force R that is proportional to the velocity R = -b v. Take the only force to be this. The force function becomes: F(x, v, t) = -b v. Below is a repeat of the Mathematica code for free fall. Modify it to change the force function to this and change the initial conditions and other parameters to t0 = 0, x0 = 0, v0 = 3 m , s m = 2 kg and b = 5 kg . s t[0] = 0; x[0] = 0; v[0] = 0; t[final] = 2; Δt = 0.02; m = 2; g = 9.8; F[x_, v_, t_] := m g CalculateAndGraph (C) Free Fall with Viscous Damping Now combine free fall with viscous damping. The force function becomes: F(x, v, t) = m g - b v. 4 Dynamics in 1D.nb Once again, below is a repeat of the Mathematica code for free fall. Modify it to change the force function to this and change the initial conditions and other parameters to t0 = 0, x0 = 0, v0 = 0, m = 2 kg, g = 9.80 m s2 and b = 5 kg . s t[0] = 0; x[0] = 0; v[0] = 0; t[final] = 2; Δt = 0.02; m = 2; g = 9.8; F[x_, v_, t_] := m g CalculateAndGraph (D) Free Fall with Quadratic Damping For free fall at sufficiently small speeds viscous damping is an accurate model. As speeds increase and the flow of air becomes turbulent, the resistive force becomes quadratic. R = -c v2 . Combining quadratic friction with free fall we get: F(x, v, t) = m g - c v2 . Once again, below is a repeat of the Mathematica code for free fall. Modify it to change the force function to this and change the initial conditions and other parameters to t0 = 0, x0 = 0, v0 = 0, m = 2 kg, g = 9.80 m s2 and c = 0.5 kg . m t[0] = 0; x[0] = 0; v[0] = 0; t[final] = 2; Δt = 0.02; m = 2; g = 9.8; F[x_, v_, t_] := m g CalculateAndGraph (E) Oscillations with a Spring and Mass A spring produces a force that is proportional to the amount the spring is stretched. If it is stretched from its equilibrium position by x then F ∝ x. We define k, the spring constant, as the constant of proportionality and write: F = -k x. The minus sign indicates that this is the force of the spring on the mass and not the force on the spring. F(x, v, t) = -k x. Once again, below is a repeat of the Mathematica code for free fall. Modify it to change the force function to this and change the initial conditions and other parameters to t0 = 0, x0 = 2, v0 = 0, m = 2 kg and k = 90 N . m t[0] = 0; x[0] = 0; v[0] = 0; t[final] = 2; Δt = 0.02; m = 2; g = 9.8; F[x_, v_, t_] := m g CalculateAndGraph If the error is too large between the symbolic solution, the blue line, and the numerical solution, the yellow line then ou will need a smaller value of Δ t. Try Δ t = .002. To repaste the above input move the mouse curser below this text and type Ctrl-L. Dynamics in 1D.nb 5 (F) Damped Oscillations with a Spring and Mass Now combine the spring with viscous damping. Here you should see oscillations that damp out with time. F(x, v, t) = -k x - b v. Once again, below is a repeat of the Mathematica code for free fall. Modify it to change the force function to this and change the initial conditions and other parameters to t0 = 0, x0 = 2, v0 = 0, m = 2 kg, k = 60 Modify the Δ t as in the previous part. t[0] = 0; x[0] = 0; v[0] = 0; t[final] = 2; Δt = 0.02; m = 2; g = 9.8; F[x_, v_, t_] := m g CalculateAndGraph N m and b = 5 kg . s Name______________________ Group_____ Numeric and Symbolic Methods in 1-D Dynamics Open this file on the computer: Physics Folder → Mathematica → Dynamics in 1D.nb Follow the instructions in the file and you will eventually generate a graph of 𝑣(𝑡) and 𝑥(𝑡) for each part (A) → (F). You do not need to print these out, but show them to your instructor and have them verify that you have done them correctly. Instructor Initials______ In addition to generating the graphs in Mathematica, there are additional questions below for you to work out. (B) Motion with Viscous Damping !" Differential Equation: ∑ 𝐹 =− 𝑏𝑣 = 𝑚 !# Guess: 𝑣(𝑡) = 𝐴𝑒 $%# Initial Conditions: 𝑣(0) = 𝑣& Determine the unknown constants A and b in terms of the known constants m, b, and v0, then give the complete expression for time-dependent velocity. (C) Free Fall with Viscous Damping From the graphs in part (C), numerically estimate the terminal velocity. In terms of the values of the parameters given in part (C), calculate the terminal velocity and compare with your estimate. (D) Free Fall with Quadratic Damping From the graphs in part (D), numerically estimate the terminal velocity. In terms of the values of the parameters given in part (D), calculate the terminal velocity and compare with your estimate. (E) Oscillations with a Spring and Mass !!' Differential Equation: ∑ 𝐹 =− 𝑘𝑥 = 𝑚 !# ! Guess: 𝑥(𝑡) = 𝐴 cos(𝜔𝑡) + 𝐵 sin(𝜔𝑡) Initial Conditions: 𝑥(0) = 𝑥& 𝑣(0) = 0 Determine the unknown constants A, B, and ω in terms of the known constants m, k, and x0, then give the complete expression for time-dependent position. Friction Equipment and Setup: Friction Trays (3 types), 2 cart masses, Set of masses, Hanger and string, Table Clamp, rod and super pulley, Pasco interface, Capstone file – Friction.cap In this experiment you will determine the coefficient of kinetic friction between a pair of surfaces. You will repeat this for several pairs of surfaces. You will also design an experiment to determine the static coefficient of friction between a pair of surfaces. Theory A mass, 𝑚! , on a horizontal surface is connected by a pulley to a hanging mass, 𝑚" . We assume the pulley has negligible mass, that the rope does not slip on the pulley, and that there is negligible friction between the pulley and its axis. The acceleration of the system can be computed from Newton’s 2nd law m1 𝐹#$% = 𝑚&'& 𝑎 where 𝑚&'& = 𝑚! + 𝑚" is the mass of the system. m2 Assuming the system is in motion, this can be written as 𝑚" 𝑔 − 𝑓( = (𝑚! + 𝑚" )𝑎. where 𝑓( = 𝜇( 𝑚! 𝑔 is the kinetic friction between 𝑚! and the horizontal surface. Procedure Setup Smart Pulley Friction Tray table clamp mass hanger Pasco interface Friction Page 2 of 3 (S.1) Use a scale to find the mass of a plastic-bottom friction tray and one other friction tray—choose either felt cork. Note: all three trays are made of black plastic, but the bottom of the plastic-bottom tray is white. (S.2) Use the table clamp to attach the rod with super pulley to the end of the table. Plug the photogate into digital input 1 on the interface box. Turn on the interface box and start the Capstone file Friction.cap in Physics/Capstone folder. If you spin the pulley, the light on the back of the photogate should blink. The photogate detects when the spokes block an LED light shining into a detector. The diameter of the pulley is already in the computer, so it can use the angular speed of the pulley to determine linear speed of the string on the outer edge of the pulley, and thus, determine the speed of the system. (S.3) Cut a string about 1 m long. Tie one end with an overhand loop and use the loop to attach the string to the plastic bottom friction tray. Attach the other end to a hanger by wrapping the string around the slot until it catches. Place the string over the pulley. Put one cart mass in the tray. Place a small mass on the hanger so that it hangs evenly, but the tray does not slide when released from rest. Adjust the pulley so that it is straight, and the string attached to the tray is horizontal. overhand loop connect friction tray connect hanger (A) Coefficient of Kinetic Friction: changing the normal force (A.1) Place mass on the hanger and release the friction tray. WARNING: Block the table clamp with you hand so that the tray does not run into it! You can also put the second cart mass in front of the table clamp for added protection. If the friction tray does not move, add more mass to the hanger. Once you have the system working: (A.2) Record the mass 𝑚! (tray + cart mass). Record mass 𝑚" (mass on hanger + 5g for hanger). (A.3) Pull the friction tray back to about 50cm away from the pulley. Stop the hanger from swinging. Block the pulley with your hand so that the tray does not hit it. Click record and release the tray. Capstone should stop recording automatically. Note: The start and stop conditions are set for 5cm and 30cm respectively. If you have trouble with them, you can adjust them or turn them off (the control is below the graph display.) (A.4) Adjust the display so you can see the entire graph. Use the graph controls above the graph to select the “good" data. Fit the graph with a linear fit and record the slope as 𝑎$)* . If the graph is not linear (generally when 𝑎$)* < 0.500 m/s " ) add a little mass to the hanger and try again. Note: Controls for adjusting the graph are located above the graph to the left. You may need to click on the graph to see them. (See image below.) Friction Page 3 of 3 Curve fit tool and drop-down menu. Auto-scale: Use this first Run selector tool: To select which run to see Highlighter: Use this to box-in “good” data and exclude extraneous points To expand the graph: Mouse over either axis until you get a double arrow. Use mouse scroll or click and drag to expand or contract the axis. To recenter the graph: Mouse over the graph until you get a hand. Click and drag to move the graph. (A.5) Record data for 5 runs with different values of hanging mass, 𝑚" . Try to keep the acceleration below 3.00 m/s " so that the cart does go flying when it is stopped. (A.6) Add the second cart mass to the friction tray and record data for 5 runs with different values of 𝑚" . Try to use the same values of hanging mass 𝑚" as previously. Although it may not be possible to use all the same values of 𝑚" , try to use at least 2 of the same values. Note: get all your data before you start analyzing. (B) Coefficient of kinetic friction: changing surfaces (B.1) Disconnect the hanging mass, detach the string from the friction tray. Connect the second friction tray to the system. Place one cart mass in the tray. Record 𝑚! (tray + added mass). Note: The system works best if the mass in the tray is distributed evenly and does not slide—hence the use of the cart mass. However, on some lab tables, this may be too much weight, especially for the cork bottom tray. You may need to use a smaller mass from you mass set in this case. (B.2) Repeat the experiment. Record data for 5 runs with different values of hanging mass 𝑚" . (C) Coefficient of Static friction Devise an experiment to determine the coefficient of static friction between two surfaces. Run this experiment three times, record data and determine the coefficient of static friction. Take notes. You will need to present a brief explanation of your experiment and present your data and results. Friction – Data Sheet Name________________ Group______ Fill in the tables. Remember to include proper units at the top of columns. Part A: Changing Normal Force write the equation used to solve for 𝑓( ______________________________ write the equation used to solve for 𝜇( _______________________________ Plastic base friction tray Table A1. 𝑚! _____________ kg 𝑚" ( ) 𝑎$)* ( ) 𝑓( ( ) computed value 𝜇( computed value Average Table A2. 𝑚" ( 𝑚! _____________ kg ) 𝑎$)* ( ) 𝑓( ( ) computed value Average 𝜇( computed value Friction - Worksheet Page 2 of 4 Table B: Coefficient surface Table B. Base of friction tray material (felt, cork) ____________________________ 𝑚! _____________ kg 𝑚" ( ) 𝑎$)* ( ) 𝑓( ( ) computed value 𝜇( computed value Average Data Part C: Create a table and record your data for part C in the space below. You will need this to answer question C-1 on the last page. Don't leave class without the data! Friction - Worksheet Page 3 of 4 Questions AB-1. What are the units of the coefficient of kinetic friction? AB-2. How would the results of the experiment differ if the mass 𝑚! were initially moving in the opposite direction (away from the pulley)? Explain. Would you still be able to find the kinetic coefficient of friction in this case? AB-3. Write the equation used to determine the force of kinetic friction. Should changing the hanging mass, 𝑚" change the force of kinetic friction on the tray? Do your experimental results support your assumption? (Be clear about HOW your results DO or DO NOT support the theory.) AB-4.Write the equation used to determine the acceleration of the system. If you changed the normal force on a tray (𝑚! 𝑔) but did not change the hanging mass (𝑚" ) should this change the force of friction on the tray? Should this change the value of coefficient of kinetic friction between the tray and the horizontal surface? Do your results from tables A1 and A2 support these claims? (Be clear about HOW your results DO or DO NOT support the theory.) Friction - Worksheet Page 4 of 4 Data recording and Question for part C. C-1. Briefly describe the procedure you used in part C to find the static coefficient of friction (one paragraph). From you data recorded in part C of the data recording section, derive the static coefficient for friction and present your results. Systems and Friction Equipment: Air track with air supply and hose, Glider, Accessory kit, Set of masses, Hanger and string, Table Clamp, rod and super pulley, Pasco interface, Capstone file – Forces and Friction.cap In this experiment we will determine the coefficient of kinetic friction between the glider and the air track. Theory: A mass, 𝑚! , on a horizontal surface is connected by a pulley to a hanging mass, 𝑚" . We assume the pulley has negligible mass, that the rope does not slip on the pulley, and that there is negligible friction between the pulley and its axis. The acceleration of the system can be computed from Newton’s 2nd law 𝐹#$% = 𝑚&'& 𝑎 where 𝑚&'& = 𝑚! + 𝑚" is the mass of the system. 𝑚! Assuming the system is in motion, this can be written as 𝑚" 𝑚" 𝑔 − 𝑓( = (𝑚! + 𝑚" )𝑎. where 𝑓( = 𝜇( 𝑚! 𝑔 is the kinetic friction between 𝑚! and the horizontal surface. Setup: bumper string loop here glider Smart Pulley table clamp Pasco interface mass hanger (not shown) 1. Place the air track on the table and connect the air hose. The other end should at the edge of the table closest to the interface box. Check that the section of track on this end is level. (Use the spirit level or, with air turned Systems and Friction Page 2 of 3 on, glider stays at rest for a little while when placed at rest on track. It is not necessary for the entire track to be level.) Place any connector in the top slot of the glider. Use a scale to find the mass of the glider. You may wish to use a rubber-band bumper on the end of the track (lower hole). 2. Use the table clamp to attach the rod with super pulley to the end of the table. Cut a string about 1 m long. Tie a loop in one end, attach the other end to a hanger. Place the loop over the connector on the top of the glider and hang the string over the pulley. Adjust the setup until everything is aligned and the string is fairly level. You may need to angle the table clamp a little to get the screw out of the way of the hanging mass. 3. Plug the photogate into digital input 1 on the interface box. Turn on the interface box and start the Capstone file called Friction.cap in Physics/Capstone folder. If you spin the pulley, the light on the back of the photogate should blink. The photogate detects when the spokes block a light (not visible) shining into a detector. The diameter of the pulley is already in the computer, so it can convert the angular motion of the pulley into linear movement of the string on the outer edge of the pulley, and thus, the movement of the system. (A) Accelerating without friction. 1. Turn the air supply on. Adjust it to a low setting. The glider should coast without slowing down when no string is attached. 2. Attach the string and put a small mass on the hanger, or leave it empty. Record the hanging mass, 𝑚" . REMEMBER to add 5g for the hanger. 3. Pull the glider back to about 50cm away from the end stop. Stop the hanger from swinging. Click record and release the glider. It should stop recording automatically. Note: The start and stop conditions are set for 5cm and 30cm respectively. If you have trouble with them, you can adjust them or turn them off (the control is below the graph display.) Note: If the glider gets moving too fast, you can stop it by hand. 4. Adjust the display so you can see the entire graph. Fit the graph with a linear fit and record the slope as 𝑎$)* . Notes: Controls for adjusting graph are located above the graph to the left. You may need to click on the graph to see them. Curve fit tool and drop-down menu. Auto-scale: Use this first Run selector tool: To select which run to see Highlighter: (generally not needed) If you have some extraneous data points, use this to box-in “good” data so you don’t fit the entire run Mouse over either axis until you get a double arrow. Use mouse scroll or click and drag to adjust axis. Mouse over graph until you get a hand. Click and drag to move graph. 5. Record a total of 4 runs with different values of hanging mass, 𝑚" . Mass increments should be small, around 5g, so that the acceleration does not get too large. Systems and Friction Page 3 of 3 (B) Coefficient of Kinetic Friction 1. Turn off the air supply. Move the glider back a bit and place mass on the hanger until the glider breaks free of static friction and starts to move. Try a trial run. If the graph does not look fairly linear, add more mass and try again. 2. Once the graph looks linear, record 𝑚" and 𝑎%+$, . Increase 𝑚" . Record 4 runs with different values of hanging mass, 𝑚" . Note: you will need to make the mass increments much larger this time, around 20g. Name_____________________________ Group _______ Systems and Friction Datasheet Fill in the tables. Remember to include proper units at the top of columns. Table A: There is no friction, so %+$,-$)* 𝑓( = 0. Percent error = %+$, × 100%. Table B: Use a spreadsheet to find average and standard deviation. (Excel: use StDev or StDevS) Percent StDev = &%.#/.0/ /$23.%3,# 42$0.5$ × 100%. Table A: Accelerating without Friction 𝑚! _____________ kg 𝑚" ( 𝑎$)* ( ) ) 𝑎%+$, ( ) percent error Table B: Coefficient of kinetic friction 𝑚! _____________ kg 𝑚" ( ) 𝑎$)* ( ) 𝑓( ( ) computed value Average Standard Deviation Percent StdDev 𝜇( computed value Question 1 Explain how the results of Part (A) would be affected if the pulley has non-negligible mass. In your case, was it a valid to assume that the mass of the pulley was negligible? Question 2 In Part (B), how would the results of the experiment differ if the mass 𝑚! were initially moving in the opposite direction (away from the pulley)? Could you still find the kinetic coefficient of friction in this case? Explain. Question 3 How is a percent standard deviation more useful than a standard deviation in determining the validity of one’s results? Question 4 How could you use this equipment to find the static coefficient of friction between glider 𝑚! and the track? Briefly describe an experiment to accomplish this. Energy and Systems Equipment: Air track, air supply with hose, Glider, Accessory kit, Photogate on stand, Set of masses, Hanger and string, Capstone file – Energy.cap In this experiment we will analyze the conversion of potential energy into kinetic energy in a system of two masses. Theory: A mass, 𝑚! , on a horizontal surface is connected by a pulley to a hanging mass, 𝑚" . We assume the pulley has negligible mass, that the rope does not slip on the pulley, and that there is negligible friction between the pulley and its axis. Let’s start with the work and energy equation 𝑊#$ = Δ𝐾 + Δ𝑈 The first term, 𝑊#$ , is the work done by non-conservative forces. If kinetic friction is negligible, then this is zero. Although gravity does work to the system and the hanging mass, 𝑚" moves up or down, gravity is a conservative force, so this work is included as part of the potential energy term, Δ𝑈. Let’s assume the system is released from rest. The weight of mass, 𝑚" accelerates the system and the potential energy of 𝑚" is converted into kinetic energy of the system. When mass 𝑚! has traveled a distance x, mass 𝑚" has descended the same distance x. So if the initial potential energy of the system is chosen to be zero, the final potential energy of the system is 𝑈% = −𝑚" 𝑔𝑥. The initial kinetic energy of the system is zero and the final kinetic energy is 𝐾% = ! (𝑚! + 𝑚" )𝑣 " . The work and energy equation becomes: " ! 𝐾 = "(𝑚! + 𝑚" )𝑣 " , ! 𝑈 = −𝑚" 𝑔𝑥 and 0 = 𝐸 = 𝐾 + 𝑈 = "(𝑚! + 𝑚" )𝑣 " − 𝑚" 𝑔𝑥 x m1 m2 x Solving for 𝑣 " gives 𝑣" = 2𝑚" 𝑔 𝑥 𝑚! + 𝑚" However, if 𝑚! is on a sloped surface, the change in potential energy of 𝑚! must be considered. Assuming the system starts from rest, the work and energy equation must now be written as: ! 𝐾 = "(𝑚! + 𝑚" )𝑣 " , 𝑈 = −𝑚" 𝑔𝑥 + 𝑚! (𝑠𝑖𝑛𝜃)𝑔𝑥 and ! 0 = 𝐸 = 𝐾 + 𝑈 = "(𝑚! + 𝑚" )𝑣 " − 𝑚" 𝑔𝑥 + 𝑚! (𝑠𝑖𝑛𝜃)𝑔𝑥 where θ is the angle that the surface makes with the horizontal. Energy and Systems Page 2 of 3 m2 m1 x θ x x sin θ θ Solving for 𝑣 " gives 𝑣" = 2:𝑚" − 𝑚! (sin 𝜃)=𝑔 𝑥 𝑚! + 𝑚" Setup: 1. Place the air track on the table and connect the air hose. Set the other end of the track at the edge of the table near the computer and Pasco interface box. Check that the track is level. Place any connector in the top slot of the glider (this will be to hold a string in place). Use a scale to find the mass of the glider. Put the pulley from your Accessory Kit into the upper hole on the end-stop, facing outward. Place a rubber-band bumper in the lower hole of the end-stop, facing inward. Pulley glider bumper mass hanger (not shown) photogate head and stand 2. Cut a string about 1.5 m long. Tie a loop in one end, attach the other end to a hanger by putting the string through the hole in the hanger and wrapping it around until it catches in the groove. Hang the string over the pulley. Place the photogate stand near the front of the track and adjust the photogate so that it can detect the smart pulley. Plug the photogate into digital input 1 on the interface box and turn the interface on. Slowly spin Energy and Systems Page 3 of 3 the pulley and ensure that the light on the back of the photogate head flashes as it detects the spokes. Start the Capstone file called Energy.cap in Physics/Capstone folder. Important! When the glider hits the bumper, the air track may be bumped forward. Keep an eye on your system to ensure that the photogate stand does NOT get knocked off the table, and that the photogate head remains aligned to properly detect the pulley. (A) Energy and systems on a Horizontal Track. 1. Turn the air supply on. Adjust it to a low setting. Place a small amount of mass on the hanger. The total hanging mass shouldn’t be more than 30g. 2. Pull the glider back to about 60cm away from the end stop. Stop the hanger from swinging. 3. Click Start Recording and then release the glider. Click to stop recording before the glider hits the bumper. If the glider is moving too quickly when it hits the bumper, you may need to stop it by hand. 4. The graph is just for monitoring purposes. If the graph does not look smooth, try again. 5. Once you have a smooth graph, look at the table to the left of the graph. The value 𝑣 lists the average speed, but the positions 𝑥 are the starting and ending positions of each block. You will need to compute position for each value of 𝑣 that you use. The average position is found by averaging the positions before and after the row that the speed is listed in. 𝑥) + 𝑥% 𝑥&'( = 2 (Note that this is an average over position, not time. However, 𝑥&'( is a better approximation for 𝑥 than using either 𝑥) or 𝑥% ) Select 10 values of speed 𝑣, compute 𝑥&'( for each and record these values in Table A. Selecting values that are well spread throughout the range of the available data will give better results. (B) Energy and systems on a Sloped Track. 1. Turn off the air supply. Move the photogate out of the way. Carefully place one support of the track on a book or block so that the end with the pulley is slightly higher than the other end. Replace the photogate and adjust the track and photogate so that the hanging mass, 𝑚" , has clearance and the photogate can properly monitor the smart pulley. 2. Turn on the air supply. Repeat the process of releasing the glider from rest. Record 𝑣 and 𝑥&'( for 10 data points. 3. You will need to be able to compute sin 𝜃 for your track. Look at the measuring tape on the air track. Use a ruler to measure the vertical height of this measuring tape at the positions 0m and 1m (100cm). These are measurements ℎ! and ℎ" . Energy and Systems – Data Sheet Name______________ Group____ Fill in the tables. Remember to include proper units at the top of columns. Tables A and B: Use a spreadsheet to plot 𝑣 " vs. 𝑥&'( . Fit the data with a linear fit and record the slope [experimental]. Compute the theoretical slope. Compare your experimental slope to the theoretical slope by computing a percent error. *+(,-(./ Percent error = *+(, × 100%. Note: Remember to include the graphs with your lab report ! 𝑚! _____________ kg 𝑚" _____________ kg Table A. Horizontal track ! 𝐾 = "(𝑚! + 𝑚" )𝑣 " and 𝑈 = −𝑚" 𝑔𝑥 set 𝑣 ( ) ( 𝑥&'( 𝑣" ) ( K ) ( U ) ( ) 𝐸 =𝐾+𝑈 ( ) 1 2 3 4 5 6 7 8 9 10 Slope from graph [exp] _____________________ Slope computed from theory [theo] ___________________ (show calculations for the slope) Percent error ___________________ Show your work. Use the back of this sheet or a separate sheet if you need more spac % diff. between K and |𝑈| Energy and Systems - Worksheet Table B. Sloped track Page 2 of 4 𝑚" _____________ kg ! 𝐾 = "(𝑚! + 𝑚" )𝑣 " and 𝑈 = −𝑚" 𝑔𝑥 + 𝑚! (𝑠𝑖𝑛𝜃)𝑔𝑥 set 𝑣 ( ) ( 𝑥&'( 𝑣" ) ( K ) ( U ) ( ) % diff. between K and |𝑈| 𝐸 =𝐾+𝑈 ( ) 1 2 3 4 5 6 7 8 9 10 Other data for part B ℎ! (m) ℎ" (m) sin 𝜃 [computed] 1.00 m ℎ= ℎ" − ℎ! θ ℎ! Slope from graph [exp] _____________________ Slope computed from theory [theo] ___________________ (show all calculations for the slope) Percent error ___________________ ℎ" Energy and Systems - Worksheet Question 1 Page 3 of 4 What are the units of the slope of the graph? Question 2 Does your data in part A support conservation of energy? Explain. Because 𝐸 = 𝐾 + 𝑈 = 0, we should have 𝐾 = |𝑈|. Were they close? Question 3 Compute the experimental acceleration for the system from your data or graphs for Part A and Part B. The equation 𝑣 " = 𝑣0" + 2𝑎Δ𝑥 may be of help. Part A: Part B: Energy and Systems - Worksheet Page 4 of 4 Question 4 Does your data in part B support conservation of energy? Explain. Because 𝐸 = 𝐾 + 𝑈 = 0, we should have 𝐾 = |𝑈|. Were they close? Question 5 How would the results of this experiment be different if the air supply were turned off and there were friction? What would happen to the values of E as the xave values progressed? Name ________________ Group ______ Conservation of Linear Momentum Equipment and Setup: Air track, air supply with hose, Photogates (2), Gliders (2), Accessory kit, Capstone file – Linear Momentum.cap In this experiment we study the conservation of linear momentum. Using the air track we can achieve nearly frictionless one-dimensional motion. Using photogate timers we can accurately measure the velocities of the gliders. At the top of each glider there is a flag with width Dx that blocks the photogate as it passes. The photogate records the total time the gate was blocked, Dt, each time the gate was blocked. We get the velocity of the glider using: ∆" 𝑣 = ± ∆# where Dx = ______________ m The reason for the sign ambiguity is that the photogate cannot determine the direction of the glider as it passes through. We must put this sign in by hand. (Recall that in one dimension the sign of the velocity gives the direction of motion.) We calculate the total momentum and the total kinetic energy using: % % 𝑝#$# = 𝑚% 𝑣% + 𝑚& 𝑣& and 𝐾#$# = & 𝑚% 𝑣%& + & 𝑚& 𝑣&& A collision is elastic when Ktot,i = Ktot,f . We can make a collision elastic, to a good approximation, using rubber band bumpers. The extreme case of inelastic collisions is called totally inelastic; it is when the masses stick together and move off as one. In this case the two final velocities are equal: vf = v1f = v2f . We achieve totally inelastic collisions using a pin adapter that sticks to a wax-filled adapter. Note that the Dt values listed refer to the corresponding velocity. For instance, Dt1f is the time needed to calculate v1f. It may be read by either photogate, depending on the case. (A) Elastic, m1 = m2 , v2i = 0 , v1f = 0 Trial Dt1i m1 = m2 = __________________ kg Dt2f Before 1 v1i v2i 0 m1 m2 Gate 1 m1 After 2 Gate 2 v1 f Trial 1 2 v1i v2f ptot,i ptot,f % diff. ptot Ktot,i m2 0 v2 f Ktot,f % diff K Conservation of Linear Momentum (B) Elastic, m1 < m2 , v2i = 0 Trial Dt1i Dt1f Page 2 of 4 m1 = __________________ kg m2 = __________________ kg Dt2f Before v1i v2i 0 m1 m2 1 Gate 1 After 2 Trial v1i v1f v2f ptot,i Gate 2 m1 m2 v1 f v2 f % diff. ptot,f Ktot,i ptot Ktot,f % diff K 1 2 (C) Elastic, m1 > m2 , v2i = 0 Trial Dt1i Dt1f m1 = __________________ kg Dt2f Before v1i v2i 0 m1 m2 1 Gate 1 1 2 Gate 2 After 2 Trial m2 = __________________ kg v1i v1f v2f ptot,i ptot,f % diff. ptot Ktot,i m1 m2 v1 f v2 f Ktot,f % diff K Conservation of Linear Momentum (D) Elastic, m1 > m2 , Both moving Trial Dt1i Dt2i Dt1f Page 3 of 4 m1 = __________________ kg Dt2f Before v1i m1 v2i m2 Gate 1 1 2 v1i v2i v1f v2f ptot,i Gate 2 m2 v2 f m1 v1 f After Trial m2 = __________________ kg % diff. ptot,f Ktot,i ptot Ktot,f % diff K 1 2 (E) Totally Inelastic, m1 = m2 , v2i = 0 Trial Dt1i m1 = m2 = __________________ kg v1i Before m1 Dtf v2i 0 m2 Gate 1 1 After 2 Trial 1 2 v1i vf ptot,i ptot,f % diff. ptot Ktot,i Ktot,f Gate 2 m1 m2 vf vf 𝐾' − 𝐾( 𝐾' 𝑚& 𝑚% + 𝑚& % diff. Conservation of Linear Momentum Page 4 of 4 Problem 1 For elastic one dimensional collisions: 𝑣%' + 𝑣%( = 𝑣&' + 𝑣&( . Verify the accuracy of this for both trials of part (D). Compare 𝑣%' + 𝑣%( . and 𝑣&' + 𝑣&( . Give their % difference. Trial 𝑣%' + 𝑣%( 𝑣&' + 𝑣&( % difference 1 2 Problem 2 Show theoretically that for a totally inelastic collision with m2 initially at rest that: )! *)" )! =+ +# $ ,+# -# (Hint: Write the kinetic energy as 𝐾 = &+ and use the fact that 𝑝' = 𝑝( .) Name __________________ Group _____ Conservation of Angular Momentum (“A” Base Platform) Equipment and Setup: Smart Pulley, “A” Base Rotational Platform with Photogate, Bar, Center Post, Capstone file – Angular Momentum - A Base.cap (A) Sliding Mass on a Bar Setup • Remove the disk and ring, and place the bar and center post on the platform. • Remove the photogate from the smart pulley and attach to the small steel bar at the base of the platform. Adjust its height so that the bar rotates freely. • Level the platform. o Keep the heavy sliding mass far from the center of the bar so that it is not balanced. o There are two screws on the base for leveling. Turning clockwise raises that side and turning counterclockwise lowers it. o Align the bar to be parallel to the leveling screws. Adjust the screws together in the same sense (both clockwise or counterclockwise) until the bar stays parallel. o The align the bar perpendicular with the screws and rotate the screws together but in the opposite sense (clockwise for one and counterclockwise for the other) until it balances. o Repeat the previous two steps with smaller adjustments until the bar is level in both orientations. Data • Measure and record the length and width of the bar, L and W. Because the heavy pivot at the center of the bar’s base contributes negligibly to its moment of inertia, the value of the mass of the bar without the heavy pivot at its center was premeasured and is given below as M. • Measure and record the mass of the sliding mass, m. There is a second identical mass, which you will not use in the experiment, but can use it to find m. • Set the stopper to a position far from the center. Push the sliding mass against it and record the position of its center, xi. It is difficult to identify the center of the sliding mass, so find its midpoint by averaging the positions of the sides. You may also move the stopper a bit to make it easier to identify the midpoint. • Pull the mass all the way toward the center until it is against the pulley. Find the position of its center as you did in the previous step; this position is xf.. • Put the sliding mass against the stopper. Pass the attached string around the pulley and up the center post, so it can be pulled. • Start the bar rotating and start recording data. While rotating, pull the sliding mass toward the center. After hitting the pulley near the center, stop recording. • Read the initial and final angular velocities off the Capstone graph and record your results in the table. • Repeat this twice for a total of three trials. Bar: 𝑀 = 0.5232 kg 𝐿 = ______________________ m Sliding Mass: 𝑚 = ______________________ kg 𝑊 = ______________________ m 𝑥! = _____________________ m 𝑥" = ______________________ m Conservation of Angular Momentum Trial 1 𝝎𝒊 (rad/s) Page 2 of 4 Trial 2 𝝎𝒇 (rad/s) 𝝎𝒊 (rad/s) Trial 3 𝝎𝒇 (rad/s) 𝝎𝒊 (rad/s) 𝝎𝒇 (rad/s) Calculations • Calculate Ibar the moment of inertia of the bar. Include the formula you use for this calculation. Treat the bar as a uniform rectangular plate. The lack of uniformity across the width of the bar will introduce a negligible error. • Treating the sliding mass as a point mass at either xi or xf from the center, find Ii and If, the moments of inertia of the bar with sliding mass before and after you drag it. These moments will be the sum of the moment of the bar and the moment of a point mass a distance of xi or xf from the center. • Using 𝐿 = 𝐼𝜔, calculate the initial and final angular momenta Li and Lf.. Theoretically these should be equal but your values will differ due to experimental error. Calculate the percent differences between your experimental values. The percent difference for two experimental values is: • | difference | & average × 100%. Calculate the initial and final kinetic energies, where 𝐾 = ' 𝐼𝜔' . Also, calculate the energy gain Δ𝐾. 𝐼bar = _______________________ kg m' Ibar: Show formulas and work. 𝐼! = _______________________ kg m' Ii and If: 𝐼" = _______________________ kg m' Show formulas and work. Trial 𝑳𝒊 𝑳𝒇 % Diff. 𝑲𝒊 𝑲𝒇 ∆𝑲 = 𝑲𝒇 − 𝑲𝒊 1 2 3 (B) Rotational Collisions Setup • Remove the bar and center post and place the disk and ring back on the platform. There is a slot on the platter to accept the ring; keep this on top. Conservation of Angular Momentum Page 3 of 4 Data • Measure and record the mass and diameter of the disk, m1 and d1. • Measure and record the mass and average diameter of the ring, m2 and d2. The average diameter is the average of its inside and outside diameters. A simple way to find this is to measure the diameter from the inside edge of one side to the outside edge of the other. • With the ring removed, start the disk rotating and start recording data. While rotating, drop the ring onto the disk so that it lands in the slot. Stop recording. • Read the initial and final angular velocities off the Capstone graph and record your results in the table. • Repeat this twice for a total of three trials. Disk: 𝑚& = ______________________ kg 𝑑& = _____________________ m Ring: 𝑚' = ______________________ kg 𝑑' = _____________________ m Trial 1 Trial 2 𝝎𝒊 (rad/s) 𝝎𝒇 (rad/s) 𝝎𝒊 (rad/s) Trial 3 𝝎𝒇 (rad/s) 𝝎𝒇 (rad/s) 𝝎𝒊 (rad/s) Calculations • Calculate I1, the moment of inertia of the disk and I2, the moment of inertia of the ring. Include the formulas you use for these calculations. Be careful to not use diameters as radii in the formulas. • Before the ring is dropped, the initial case, there is only the disk with moment I1 rotating with angular velocity wi. After the ring is dropped, the final case, both the disk with moment I1 and the ring with moment I2 are rotating together with angular velocity wf. • As in Part (A) calculate the initial and final angular momenta, Li and Lf, and calculate their percent differences. Also, calculate the initial and final kinetic energies, Ki and Kf, • The fraction of energy lost is ,! -," ,! . In the Conservation of Linear Momentum experiment we showed that for the case of a totally inelastic collision with m2 initially at rest, the fraction of energy lost is Here we have the rotational analog of that, so the fraction of energy lost here should be • ,! -," 0# . 0$ /0# .# . .$ /.# 0# 0# , where is the theoretical result. The percent error ,! 0$ /0# 0$ /0# experimental+theoretical between an experimental value and a theoretical value is: × 100%. theoretical Calculate the percent error between Disk: 𝐼1 = _______________________ kg m' Show formulas and work. and Ring: 𝐼2 = _______________________ kg m' Trial Conservation of Angular Momentum 𝑳𝒊 𝑳𝒇 % Diff. 𝑲𝒊 Page 4 of 4 𝑲𝒇 𝑲𝒊 − 𝑲𝒇 𝑲𝒊 𝑰𝟐 𝑰𝟏 + 𝑰𝟐 % Error 1 2 3 Questions A-1 Does your data in Part (A) support conservation of angular momentum? A-2 Kinetic energy should not be conserved in Part (A) and the energy gain Δ𝐾 should be positive. Where does the energy come from? A-3 Suppose in calculating the moment of inertia you used the formula for a thin rod of length L, and not that for an 𝐿 × 𝑊 rectangular plate. What percent error would be introduced in the calculated value of the moment of the bar, taking the rectangular plate as the accepted or theoretical value? Would the moment be larger or smaller than the accepted value? B-1 Does your data in Part (B) support conservation of angular momentum? B-2 Kinetic energy in part (B) should not be conserved. Where does the energy lost go? Name __________________ Group _____ Conservation of Angular Momentum Equipment and Setup: Smart Pulley, Rotational Apparatus, Capstone file – Angular Momentum.cap Moments of Inertia (Measurements and Calculations) and Angular Velocity Data I1 is the moment of inertia of the lower platter that rotates before the dropped object falls on it. • I2 is the moment of inertia of the dropped object. • Find the I2 values, the moments of inertia of the three objects to be dropped and calculate their moments. Show your work. • Note that the formulas for the moments typically involve the radius while the diameters are measured here. Be careful! • To find I1, the moment of the lower platter, with the sprocket, use the moment of the platter without the sprocket. The platters are identical except for the sprocket; the additional mass of the sprocket contributes negligibly to the moment since it is so close to the axis of rotation. • The angular velocity data is found from Capstone. Take three trials for each of the three dropped objects. Data for Moments Formulas for Moments Calculated Moment I2 (kg×m2) Trial • Platter 1 Mass = ____________ kg 2 Diam. = ___________ m Bar 3 Mass = ____________ kg 4 Length = __________ m 5 Width = ___________ m 6 Ring 7 Mass = ____________ kg 8 Diam. = ___________ m 9 I1 = I2,platter = ____________________ kg×m2 wi (rad/s) wf (rad/s) Conservation of Angular Momentum Page 2 of 3 Calculations of Angular Momentum and Kinetic Energy • Before the object is dropped, the initial case, there is only the base platter with moment I1 rotating with angular velocity wi. After the object is dropped, the final case, both the base platter with moment I1 and the dropped object with moment I2 are rotating together with angular velocity wf. • Conservation of angular momentum implies that theoretically Li = Lf. The experimental values will differ due to experimental error. Calculate the percent differences between your experimental values of Li and Lf. |difference| The percent difference for two experimental values is: average × 100% • Energy (kinetic energy in this case) is not conserved. The fraction of energy lost is +! ,+" +! . In the Conservation of Linear Momentum experiment we showed that for the case of a totally inelastic collision -# with m2 initially at rest the fraction of energy lost is . Here we have the rotational analog of that, so -$ .-# /# the fraction of energy lost here should be . /$ ./# +! ,+" /# /# , where is the theoretical result. The percent error +! /$ ./# /$ ./# experimental"theoretical between an experimental value and a theoretical value is: × 100%. theoretical • Tria l Calculate the percent error between Platter 1 2 3 Bar 4 5 6 Ring 7 8 9 Li Lf % Diff. and Ki Kf 𝑲𝒊 − 𝑲𝒇 𝑲𝒊 𝑰𝟐 𝑰𝟏 + 𝑰𝟐 % Error Conservation of Angular Momentum Page 3 of 3 Question 1 Does your data support conservation of angular momentum? Question 2 Kinetic energy should not be conserved. Where does the energy lost go? Question 3 If the ring is placed off center, how will this affect your results? Will the experimental moment of inertia be too large or too small? Suppose the ring is off center by 1 cm, what percent error would this introduce into the moment? Orbits and Kepler’s Laws Equipment and Setup: Mathematica file – Kepler.nb Theory Kepler’s Laws Kepler found three laws describing planetary orbits about the sun. He did his work after Copernicus and before Newton. The three laws are: 1. Planets move in elliptical orbits about the sun, which is at one of the focal points of the ellipse. a is the semi-major axis. b is the semi-minor axis. focus focus 2b c = a2 -b2 is the distance from the center to each focus. 2a 2c 2. A radial line from the sun to a planet sweeps out equal areas in equal time intervals. Sun 3. The orbital period squared is proportional to the semi-major axis cubed. 𝑇 ! ∝ 𝑎" General Orbital Dynamics The gravitational force on mass m due to a second mass M is 𝐺𝑀𝑚 𝐹⃑ = − ! 𝑟̂ 𝑟 where M is at the origin, 𝑟⃑ is the vector from M to m and 𝑟̂ = 𝑟⃑/𝑟 is the unit vector in the direction of 𝑟⃑. Orbits and Kepler’s Laws Page 2 of 2 We will assume that M is much larger than m (M >> m) so that M stays fixed at the origin. The inverse square law shown above is not the only possibility. A central force is one that points toward the sun. More generally, we can imagine some central force that is attractive and decreases with distance, going to zero at infinity. Here, we will consider only central forces that are inverse power laws, meaning that they drop off as the inverse of r to a power p, 𝐹 ∝ 1/𝑟 #  GMm Fp = − p r̂ = −Fp r̂ . r Since we are interested in orbits about the sun, we take M to be the mass of the sun. m is the planet’s mass which, since 𝑎⃑ = 𝐹⃑ /𝑚 does not involve the kinematics of the orbit. A very useful choice of units will simplify our discussion; measure all lengths in astronomical units, AU, where 1 AU is the radius of the earth's orbit and measure time in years. In these units, the speed of a circular orbit of radius 1 AU with period 1 year is r = 1 AU and T = 1 year Þ 𝑣 = !$% & = 2𝜋 AU year or 𝑣 = 2𝜋 (in Solar System Units) To simplify the expressions, we will omit the units and just refer to this as Solar System Units. For circular orbits, since 𝐹# /𝑚 = 𝑎- , we have 𝑣! 𝐺𝑀 4𝜋 ! ! ⎧ ⟹ 𝑣 = = 𝐺𝑀 𝑟 𝑟 #./ 𝑟 #./ = 𝑎 = (in Solar System Units) ! 𝑟# 4𝜋 ! #0/ ⎨ 2𝜋 ! #0/ =𝑟 ⎩8 𝑇 9 𝑟 ⟹ 𝑇 = 𝐺𝑀 𝑟 To simplify the two expressions above we used that in Solar System units, the earth’s circular orbit has 𝑣 = 2𝜋 and 𝑟 = 1. This gives 𝐺𝑀 = 4𝜋 ! ; we inserted that into the expressions above at the right. Newtonian Gravity Newton was able to show that for central forces, the inverse square law 𝑝 = 2 uniquely gives Kepler’s laws. To show this mathematically is beyond the scope of this course but we may demonstrate this using a computer simulation. Orbits – Worksheet Name________________ Group______ Computer Setup • This orbit simulation has a planet or satellite initially at a distance r0 from the sun with an initial velocity of magnitude v0 perpendicular to the radial direction. • The interactive panel has two tabs at the left. The “Other Power Laws” tab gives a central force proportional to the inverse of the pth power of r, 𝐹 ∝ 1/𝑟 # ; under this tab one can choose any value of p. • The “Newtonian Gravity” tab gives the physically correct case of 𝑝 = 2, with the additional controls and orbit information. There are three buttons listed as “Display Swept Area?”. The “Static with value” button shows a box to choose a value of Δ𝑡; above the graphic it gives a slider to vary the value of t, the starting time of the interval, and lists the value of the area. The “Animated” button shows the area in an animation. Still under the “Newtonian Gravity” tab, to the right of these buttons is orbit data. • To the right of the tab area are three checkboxes. “Animate” opens animation controls above the graphic. (Note that the “Animate” checkbox cannot be selected when the “Display Swept Area?” button is chosen under the “Newtonian Gravity” tab.) “Show r0 and v0” displays the initial position and the initial velocity, which is perpendicular to the radial line form the sun. “Show Earth Orbit” shows the earth’s orbit as a reference. • There are three input fields at the right of the controls to vary the initial values of v0 and r0, and to vary the stopping time tmax. Section A: Other Power Laws Use the “Other Power Law” tab for this section. 1. The simplest power law is where gravity is inversely proportional to the distance from the sun. This is the 𝑝 = 1 case. Set the value of r0 to 1AU. Use 𝑣1 = 1 × 2𝜋 AU/yr and describe the orbit. 2. Keeping 𝑝 = 1 and 𝑟1 = 1AU set 𝑣1 = 0.5 × 2𝜋 AU/yr. Describe the results. Is the orbit closed? 3. Keeping 𝑝 = 1 and 𝑟1 = 1AU set 𝑣1 = 1.5 × 2𝜋 AU/yr. Choose a larger value of tmax if needed. Describe the results. Is the orbit closed? Orbits - Worksheet Page 2 of 3 4. Now consider the 𝑝 = 3 case, an inverse cube law. Set the value of r0 to 1AU. Use 𝑣1 = 1 × 2𝜋 AU/yr and describe the orbit. 5. Keeping 𝑝 = 3 and 𝑟1 = 1AU set the speed just under that for a circular orbit. Use 𝑣1 = 0.99 × 2𝜋 AU/yr. Describe the Orbit. 6. Keeping 𝑝 = 3 and 𝑟1 = 1AU set the speed just over that for a circular orbit: 𝑣1 = 1.01 × 2𝜋 AU/yr. Describe the results. 7. Now look at the 𝑝 = 2 case, the inverse square law. Set the value of r0 to 1AU and use 𝑣1 = 0.5 × 2𝜋 AU/yr. This gives Kepler’s closed elliptical orbit. With slight deviations from the inverse square law we would expect elliptical orbits that do not quite close. Keeping the same values of v0 and r0 try 𝑝 = 1.98 and 𝑝 = 2.02. Describe the results. Section B: Kepler’s Laws Use the “Newtonian Gravity” tab for this section. 1. Use 𝑟1 = 1AU, 𝑣1 = 0.5 × 2𝜋 AU/yr and 𝑡234 = 1yr. Record rmin, rmax and T. rmin = ______________ AU, rmax = ______________ AU, T = ______________ yr, Calculate the semi-major axis a, then find a 3 and T 2 . With these units Kepler’s third law implies the values of a 3 and T 2 should be equal. Compare them by finding their percent difference. a = ____________ AU, a 3 = _____________ AU3, T 2= _____________ yr2, % diff. = ____________ Orbits - Worksheet Page 3 of 3 2. Use 𝑟1 = 1AU, 𝑣1 = 0.5 × 2𝜋 AU/yr and 𝑡max = 1 yr. In the “Display Swept Area?” buttons choose “Animated”. Set Δ𝑡 = 0.05yrs and click the play button above the graphic. Describe what you observe. 3. Keeping the same values of r0 and v0 choose𝑡max = 0.5 yr. In the “Display Swept Area?” buttons choose “Static with value”. Using the same value Δt = 0.05yrs record the value of Area at 𝑡 = 0 yr . With the slider adjust t until the swept area is nearest the sun and record t and Area. t = 0 yrs Þ Area = ______________ AU2, t = ______________ yr Þ Area = ______________ AU2 Discuss why the two area values are equal. 4. Halley’s comet has a aphelion distance of 𝑟max = 35AU and a perihelion distance of 𝑟min = 0.59AU. Begin with 𝑟0 = 35AU ; this will become rmax. For “Display Swept Area?” choose “None”. Choose different small values of v0 until you get the correct value of rmin. Set a long tmax to guarantee a full orbit; 𝑡max = 100 yr will suffice. Values of rmin between 0.55 and 0.63 AU will be close enough. (You must be systematic about this; if the orbit is unbound or if rmin is too large then choose a smaller v0 value and if rmin is to small choose a larger v0.) Record rmin, v0 and the period. rmin = ____________ AU, rmax = 35 AU, v0 = ____________× 2𝜋 AU/yr, T = ____________ yr, 5. The escape speed at some distance from the sun is 2 times that for a circular orbit. Choose 𝑟1 = 1𝐴𝑈 and 𝑣1 = √2 × 2𝜋 AU/yr. (Input Sqrt[2] into the field.) Take a large value of tmax: 𝑡max = 30 yr. What is the shape of the orbit at this critical value? 6. Now consider a speed larger than this critical escape speed. Choose 𝑟1 = 1𝐴𝑈, 𝑣1 = 1.5 × 2𝜋 AU/yr and 𝑡max = 30 yr . What is the shape of this orbit? Name __________________ Group _____ The Period of a Simple Pendulum Equipment and Setup: Support rod and base, Pendulum Clamp, Pendulum bob set, string, Meter stick, Protractor, Photogate, Capstone file – Pendulum.cap A simple pendulum consists of a small bob of mass m at the end of a string of length L. The size of the bob must be small compared to the length of the string and take L to be the distance to the center of the bob. The angle the bob makes from vertical is θ. If the bob is lifted to a maximum angle θ0 (the amplitude) and is released from rest then the motion will be planar. Applying the rotational form of Newton’s second law: τnet = I α to the pendulum gives −𝐿𝑚𝑔 sin 𝜃 = 𝑚𝐿! 𝛼 Using that α is the second time derivative of θ gives the second order ordinary differential equation: 𝑑! 𝑔 𝜃 = −𝜔! sin 𝜃 where 𝜔 = 3 4𝐿 𝑑𝑡 ! For sufficiently small angles we may approximate the sine of an angle with the angle in radians: sin 𝜃 ≅ 𝜃. Using this small angle approximation, we get a differential equation: 𝑑! 𝜃 = −𝜔! 𝜃 𝑑𝑡 ! This is of the form of the standard differential equation for Simple Harmonic Motion. We can conclude that for sufficiently small angles we have simple harmonic motion. Since the period T of simple harmonic motion is related to the angular frequency w by 𝑇 = 2𝜋⁄𝜔, we can find an expression for the period of a pendulum for small angle oscillations 𝑇 = 2𝜋3𝐿4𝑔 (small angle formula) We will refer to the amplitude for simple harmonic motion as θ0. The above expression for T applies for small θ0, but there is no expression for the period of a pendulum for large amplitudes (large θ0) that can be written in terms of everyday functions. It is possible to derive a series expansion for the period that converges well after a couple of terms. The expansion is: 12 θ0 12 32 θ0 12 32 52 θ0 𝑇 = 2𝜋3𝐿4𝑔 C1+ 2 sin2 + 2 2 sin4 + 2 2 2 sin6 + ⋯L 2 2 2 4 2 2 4 6 2 Truncating this series after three terms gives a good approximation that works up to relatively large angle. We will refer to this as the approximate large angle formula. 12 θ0 12 32 θ0 𝑇 = 2𝜋3𝐿4𝑔 C1+ 2 sin2 + 2 2 sin4 L 2 2 2 4 2 (approximate large angle formula) The period of a pendulum is independent of the mass of the bob. We have neglected air resistance (friction) in this analysis. The force of air resistance on a bob depends on the size, shape and texture of the bob. For a given size bob friction has a larger effect on lighter bobs. To minimize the effect of air resistance we will use the densest bob. The Period of the Simple Pendulum Page 2 of 4 (A) The Pendulum at Large Angles We will study the pendulum at large angles and test the approximate large angle formula. Use the lead bob to minimize the effect of friction. Keep the length of the pendulum constant and as long as is practical with your setup. For theoretical calculations use the approximate large angle formula. Give the percent error between the experimental period and the theoretical period. Also give the percent error between the two theoretical period formulas, the large angle and small angle formulas. Mass: m = _____________ Amp. θ0 Length: L = _____________ Period Experimental: Texp Theoretical: Ttheo Small Angle Period: T0 = _____________ % Error (between Texp and Ttheo) % Error (between T0 and Ttheo) 5° 10° 20° 30° 45° 60° (B) The Effect of the Mass of the Bob Neglecting friction, the period of a pendulum is independent of mass. The only dependence is due to friction effects. You should notice that the very light bobs are slowed down somewhat. Use four different bobs: lead, aluminum, wood and cork. It is not necessary to measure the angle but keep it small. It is essential to keep this length constant as you swap out the different bobs. Use a long string as in part (A). Use the small angle formula for the theoretical calculations. Length: L = ____________________ Mass: m Use a small angle. Period Experimental: Texp Theoretical: Ttheo % Error The Period of the Simple Pendulum Page 3 of 4 (C) The Effect of Length and a Calculation of g We will now study the effect of length on period for small angles. Measure the period for five different lengths. Use a long string as in part (A) for the longest case. Make the shortest length approximately 1/5 the longest and use approximately equal differences between lengths. Use the small angle formula, T0, for the theoretical calculations of period. Mass: m = ____________________ Length: L Use a small angle. Period Experimental: Texp Theoretical: Ttheo % Error Question 1 Using the data from Part (C), graph (Texp)2 versus L. and include the best-fit line. Find the slope of the best-fit line. slope = ___________________ Question 2 Starting with 𝑇 = 2𝜋 P𝐿⁄𝑔, find the theoretical expression for the slope of a 𝑇 ! versus L graph. The Period of the Simple Pendulum Page 4 of 4 Question 3 Using the expression found in Question 2 solve for gexp, the experimental value of g, from your experimental value of the slope in Question 1. Compare the experimental value of g to the accepted value of 9.80 m/s2. gexp = ___________________ % Error in g = ___________________ Question 4 On the moon gravity is approximately 1/6 as strong as on earth. For pendulums of the same length find the ratio of the period on the moon to the period on the earth, Tmoon/ Tearth ? Question 5 If the force of air resistance on each bob is approximately the same on all the bobs used, then explain why the lighter bobs are affected by it more. Name __________________ Group _____ Boyle’s Law Equipment and Setup: Syringe, Pressure sensor, Temperature sensor, Capstone file – Boyle's Law.cap Part I. Theory When temperature is not very low and pressure is not very high, gases follow ideal gas law: 𝑝𝑉 = 𝑁𝑘𝑇, 𝑜𝑟 𝑝𝑉 = 𝑛𝑅𝑇 When temperature T is kept constant, the product of pressure p and volume V for a certain amount of gas is constant, which is called Boyle’s law, discovered by British scientist Robert Boyle in 1662 by experiment. That is: 𝑝! 𝑉! = 𝑝" 𝑉" = 𝑝# 𝑉# = ⋯. The pV diagram is like figure 1. Figure 1. The pV diagram at constant temperature. The purpose of this experiment is to verify this law. Boyle’s Law Page 2 of 4 Part II. Procedure 1. Set the plunger for a volume of 50 cc (mL). Connect the syringe to pressure sensor unit and the temperature sensor unit; then connect the pressure sensor unit to port A and the temperature sensor unit to port B on Capstone box, as shown in Fig. 2. Fig. 2. (a) Pressure sensor and temperature sensor connected to syringe. (b) Sensors connected to Capstone box. 2. Double click on Boyle’s Law in Capstone in Physics folder. 3. Click Constant Temp on the menu. 4. Choose Data Summary if not chosen yet; the screen should be like fig. 3. Fig. 3. Screenshot of Boyles’ law lab Boyle’s Law Page 3 of 4 5. Click Preview to start recording data. Type “50” in the box labeled Volume in the first row of Table 1 (It means that the volume in the syringe is 50 cubic centimeters, or 50 mL). Click Keep Sample while the plunger is still at 50 cc. 6. Now push the plunger of the syringe to 45 mL and type “45” in the box of Volume in the 2nd row of table. Wait for 30 seconds (this is about the time needed for temperature to get back to room temperature again). 7. Click Keep Sample while the plunger is still at 45 cc. Compress the plunger to 40 cc and hold it at this position for 30 seconds. 8. Continue this process to get data for volumes of 35 cc, 30 cc, 25 cc, and 20 cc. You should see a graph like Fig. 1 in this manual. 9. Record the data to Table I and find pV values. (Note: pay attention to the units of pressure and volume to get correct unit of Joule for their product) 10. Disconnect, YES, DISCONNECT, the syringe to pressure sensor unit first, and change the volume to 20 cc. Do 2nd trial to fill the Table 2, starting with 20 cc this time. Part III. Data Tables V (mL) Table 1 (Trial 1) P (kPa) pV (J) V (mL) 50 20 45 25 40 30 35 35 30 40 25 45 20 50 Average pV = J Table 2 (Trial 2) P (kPa) Average pV = pV (J) J Boyle’s Law Page 4 of 4 Part IV. Problems 1. Plot the pV diagrams (graphs of p vs. V) for both trials, which should be turned in with the lab report. 2. Assume room temperature is 23 °C = 296 K, find the number of moles and number of molecules of air in the syringe by using the average values of pV for each of the two trials. Show your work and both answers below. 3. Suppose that at room temperature and around 100 kPa (which is about the temperature and pressure when you start the trial 1), the air mass density is 1.2754 kg/m3. Find the mass in kg and weight in Newtons of air inside the syringe for trial 1. Graphing with Excel University Physics I and II • All graphs must be created on a computer. Directions for plotting using Excel will follow. If you use other software, then your graphs must have all the same components described here. • When plotting y vs. x, the horizontal axis is x and the vertical is y. Always remember you are plotting the dependent variable versus the independent variable, or y vs. x. • The graph must be properly labeled. o At the top of each graph there should be a label that gives the variables that are plotted, for example a distance versus time graph could be labeled as “distance vs. time” or just as “x vs. t”. o Each axis must be labeled and the units should be included in brackets. For example, if an axis labels a distance x, then use the label: “x (m)”. • When plotting y vs. x, the horizontal axis is x. All graphs must have a uniform scale along both axes. Always have the axes cross at the point (0,0). • For any graph you must plot the data points, the best-fit curve and the equation of the best-fit curve. • When using Excel always use the scatter plot format. Other formats will likely cause problems with maintaining a uniform scale. o Put the data in columns; if it is a graph of y vs. x, then the x column is on the left and y column is on the right. Highlight the data and then click the Insert tab. Then Chart > Scatter then choose the option “Scatter with only markers”. o Select the Layout tab to add the labels. To label Axes click “Axes Label” and appropriately label each axis. Labeling the graph is done with the “Chart Title” choice. o If the axis origin is not through (0,0) you can force it. Under the Layout tab click Axes, then choose which axis and then “More Primary Axis Options”, or more simply right-click the axis and choose “Format Axis”. Forcing the origin is done by choosing “Axis Options” > Minimum > Fixed > (0,0). o Now you must add the best-fit line or trendline. Right-click on a data point and choose “Add Trendline”. Choose “Trendline Options” and then Linear for the best-fit line and Polynomial > Order > 2 for a parabola. Always check “Display Equation on chart.” 2 • o The legend can be removed by right-clicking it and choosing delete. o The preceding directions are for Windows. On a Mac you have menus instead of the tabs. You choose a graph with Insert > Chart > “XY Scatter”. For the trendline right-clicking a data point behaves the same as Windows. Adding the labels requires using the Formatting Palette. (View > “Formatting Palette”). “Chart Options” > Titles will allow the labeling. Example - Here is a sample quadratic plot for the position vs. time data shown below. The data points are in the (t,x) format.