Center of the Circle in Uniform Circular Motion

Question 1 (5 points) Which of the following point towards the center of the circle in uniform circular motion? Ans: Acceleration, net force. [Review the definition of centripetal acceleration and centripetal force.] increased by a factor of 5 Question 2 (5 points) If the speed of an object in a uniform circular motion is constant and the radial distance is tripled, the magnitude of the radial acceleration decreases by what factor? Ans: 3, since aR = v2 . R Question 3 (9 points) Calculate the centripetal force exerted on a 1111 kg car that negotiates a 99.0 m radius curve at 25.0 m/s. Ans: mv 2 R = 7013.9 N FR = Question 4 (8 points) In Question 3, assuming an unbanked curve, find the minimum coefficient of static friction, between the tires and the road. Static friction is the reason that keeps the car from slipping. Ans: Static friction is the reason that keeps the car from slipping. To keep the car in the given unbanked curve, we need at least 7013.9 N of static friction force. Ffr, s = 7013.9N ¿s FN = 7013.9N 7013.9N ¿s = FN 7013.9N ¿s = mg ¿s = 0.644 Question 5 (10 points) Curves on some test tracks and racecourses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 119 m radius curve banked at 61.0¡ should be driven if the road is frictionless. Ans: Strat solving this problem by drawing a free-body diagram as shown in the figure. Now apply NewtonÕs second law separately along x- and y-axes. F = m ay 3 y FN cos » − mg = 0 mg FN = cos » F = m ax 3 x −FN sin » = − m aR mg sin » = m aR cos » g tan » = aR v2 R v 2 = Rg tan » g tan » = v= Rg tan » v = 45.9 m/s Question 6 (8 points) A 1.70 × 10 3-kg truck rounds an unbanked curve on the highway at a speed of 21.1 m/s. If the maximum frictional force between the surface of the road and all four of the tires is 7.77 × 10 3 , calculate the minimum radius of curvature for the curve to prevent the truck from skidding off the road. Ans: Static friction force is the reason that keeps the truck from skidding. So the maximum static friction should be equal to the centripetal force. Ffr,max = FR Ffr,max = R= mv 2 R mv 2 Ffr,max R = 97.4 m Question 7 (7.5 points) Riders on a Ferris wheel of diameter 19.0 m move in a circle with a centripetal acceleration of 2.22 m/s2 . What is the speed of the Ferris wheel? Ans: v2 R v 2 = RaR aR = v= RaR v = 4.59 m/s Question 8 (7.5 points) International Space station is about 400 km from the surface of Earth. If an astronaut's mass is 65 kg. What is his apparent weight (Normal Force) inside the station? Recall that true weight and apparent weight are two different quantities. Ans: AstronautÕs true weight will be 65 kg times the acceleration due to gravity at the height of International Space Station (ISS), or you can also use NewtonÕs law of universal gravitation to calculate the true weight. However, the apparent weight will be zero. This is because the only force that acts on the ISS is the gravitational force. There is no normal force at ISS to balance it. So the answer is 0 N. Question 9 (10 points) Calculate the force of Earth's gravity on a spacecraft at 2.00 times Earth radii above the Earth's surface if its mass is 2000 kg. Mass of Earth is 5.94 × 10 24 kg and the radius of the Earth is 6380 km. Hint: Do not forget to count the distance from the center of the Earth. Ans: Here, G = 6.67 × 10−11 F =G m E mS = 2163 N. d2 Nm2 kg 2 and d = 3 × RE = 3 × 6380 × 10 3 m. Question 10 (5 points) In the International Space Station which orbits Earth, astronauts experience apparent weightlessness because Ans: The astronauts and the station are in free fall towards the center of the Earth. Question 11 (5 points) Given that the acceleration of gravity at the surface of Mars is 3.70 m/s/s, and that Mars' radius is 3400 km, determine the mass of Mars. Chose the correct answer from the following. Ans: m R2 gR 2 m= G m = 6.41 × 1023 kg g=G Question 12 (10 points) A satellite orbits Earth at altitudes of 5000 km. What is its speed with respect to the center of Earth? The radius of the Earth is 6380 km and its mass is 5.94 × 10 24 kg. Ans: At the altitude of the satellite, there is only one force acting on it that is the EarthÕs gravitational force FG , which is also toward the center of the orbit. Here d = (6380+5000) km = 11380 km = 11380000 m. F = m aR 3 R v2 F G = mS d m E mS v2 G = mS 2 d d m v2 = G E d m v = G E = 5900.4 m/s d Question 13 (10 points) What is the period of a satellite orbiting the Moon 200 km above the Moon surface? Ignore the effect of Earth. The radius of the Moon is 1740 km. The mass of the moon is 7.35 × 10 22 kg. Express your answer in minutes. Ans: Using similar method of Question 12, you can determine the speed of the satellite. Make sure you substitute the mass of the moon and d = (1740 + 200) km = 1940 km = 1940000 km. v= G mM = 1589.7 m/s d Now use the usual definition of speed to determine the time period of the satellite. distance v= v= time 2Ãd for a satellite in a circular orbit T 2Ãd = 7667.9 s = 127.8 min T= v