Rolling Part I Solutions

Version 001 – Rolling Part I 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. ∆v = 7 m/s , ∆t = 3.4 s . and v 001 10.0 points A wheel of mass M and radius R rolls on a level surface without slipping. If the angular velocity of the wheel about its center is ω, what is its linear momentum relative to the surface? ω B Consider point B at the bottom of the wheel. It is at rest since there is no slipping. Its velocity 1. p = M ω R2 vB = v − ω R = 0 v ω= R 2. p = 0 3. p = M ω 2 R2 2 4. p = M ω 2 R 5. p = M ω R correct Explanation: First, we note that the wheel is rotating about its center at an angular velocity of ω, so the velocity difference between the center of the wheel and the lowest point (the point of the wheel that touches the surface) is ω R in magnitude in the horizontal direction. Since the wheel rolls without slipping (which means the velocity of the lowest point is 0) the velocity v of the center of the wheel is ω R. Obviously, the center of the wheel is also the center of the mass, so the linear momentum is p = M ω R. Bicycle Down a Hill 002 (part 1 of 3) 10.0 points A bicycle wheel of radius 0.27 m rolls down a hill without slipping. Its linear velocity increases constantly from 0 to 7 m/s in 3.4 s. What is its angular acceleration? Correct answer: 7.62527 rad/s2 . Explanation: Let : R = 0.27 m , ∆ω 1 ∆v = ∆t R ∆t 1 7 m/s = = 7.62527 rad/s2 . 0.27 m 3.4 s α= 003 (part 2 of 3) 10.0 points Through what angle does the wheel turn in the 3.4 s? Correct answer: 44.0741 rad. Explanation: 1 2 1 α t = (7.62527 rad/s2 ) (3.4 s)2 2 2 = 44.0741 rad . θ= 004 (part 3 of 3) 10.0 points How many revolutions does it make? Correct answer: 7.01461 rev. Explanation: 1 rev =2 π rad, so θ = 44.0741 rad · 1 rev = 7.01461 rev . 2 π rad Version 001 – Rolling Part I – smith keywords: Concept 08 14 005 10.0 points Which will have the greater acceleration rolling down an incline: a bowling ball or a volleyball, and why? 1. the volleyball; mass 2. no difference 3. the bowling ball; rotational inertia correct 4. the volleyball; rotational inertia 5. the bowling ball; mass Explanation: The bowling ball wins. A solid sphere of any mass and size beats both a solid cylinder and a hollow ball of any mass and size, because a solid sphere has less rotational inertia per mass than the other shapes. A solid sphere has the bulk of its mass nearer the rotational axis that extends through its center of mass, whereas a cylinder or hollow ball has more of its mass farther from the axis. The object with the least rotational inertia per mass is the least lazy and will win races. Concept 08 15 006 10.0 points What technique would help you to distinguish between two heavy identical-looking spheres of the same weight, one solid and the other hollow? 1. Hit them and compare their reflective sounds. 2. Plunge them into water and compare their densities. 3. Launch them and compare their trajectories. 4. Roll them down an incline and compare their speeds. correct 2 Explanation: If you roll them down an incline, the solid ball will reach the bottom first. (The hollow ball has more rotational inertia compared with its weight.) KE Ratio 01 007 10.0 points A rigid body of radius R and mass M is rolling on a level floor without slipping. Suppose Icm = κ M R2 . We can consider the point where the body toches the floor as a pivot, and calculate the rotational kinetic energy Kr,p around that pivot. We usually define a kinetic energy that consists of the rotational kinetic energy of the body around its center of mass, plus the kinetic energy of its center of mass motion; that is, K = Kr,cm + Kcm . Kr,p b K b vcm b What is the ratio of Kr,p to K, in general? 1. κ R 2. (κ + 1) by the parallel axis theorem 3. Zero 4. (κ − 1.  2 1 5. κ 6. 1.0 correct 7. κ2 1 κ κ 9. R 8. 10. κ Version 001 – Rolling Part I Explanation: Both Kr,p and K are standard representations of the total kinetic energy of a rigid body Kr,p rolling with slipping, so = 1.0. K This is easy to see from the parallel axis 1 theorem: Kr,p = (κ + 1) M R2 ω 2 . But 2 vcm = R ω for rolling without slipping, so 1 1 2 κ M R2 ω 2 + M vcm 2 2 1 2 2 = (κ + 1) M R ω . 2 Kr,p = Pulling a Spool 008 (part 1 of 2) 10.0 points Consider a spool with inner radius b and outer radius c. A string wraps around the inner stem in a counterclockwise manner. Starting from rest, the spool rolls when the string is pulled horizontally along AB as shown below. The force F is applied in such way that there is no slippage. b F c A B Determine the direction of rotation of the spool. 1. counterclockwise 2. Cannot be determined 3. clockwise correct Explanation: Fnet = m a , and (1) a τnet = I α = I , (2) r Under the action of force F , the spool tends to accelerate linearly to the right, and to rotate counterclockwise. Therefore the contact point (presently at P , see the picture below), will have a tendency to move to the right against the surface. In turn, the friction force must point to the left. 3 b F c A f B P Choose either A or P as the pivoting point. Choose A: the torque about A is τ = f (c − b) clockwise and the spool will rotate clockwise. Choose P : the torque about P is τ = F (c − b) clockwise and spool will still rotate clockwise. The agreement in the direction between these two different points of view actually tells us that our choice of the direction of f is correct. 009 (part 2 of 2) 10.0 points Choose the correct set of equations that govern the motion of the spool. Here f is the magnitude of the frictional force between the spool and the horizontal surface, a is the linear acceleration of the center of mass, and I is the moment of inertia about the center of mass of the spool. The torque will be evaluated with respect to the center of the spool. Be careful about the relative signs. X 1. Fx : F + f = ma X a τ : f c−F b= I b X 2. Fx : F − f = ma X a τ : f c − F b = I correct c X 3. Fx : F − f = ma X τ : f c−F b= Ia X 4. Fx : F − f = ma X a τ : f c−F b= I b X 5. Fx : F + f = ma X τ : f c−F b= Ia X 6. Fx : F − f = ma X a τ : f c+F b= I c Explanation: Version 001 – Rolling Part I – smith Using equations (1) and (2), Fnet = F − f = m a , and for clockwise torque about the center of mass a τnet = f c − F b = I α = I , c a where we use the relation α = for rolling c without slipping. So the correct pair of equations is X Fx : F − f = m a X τ: fc−F b= Ia . c keywords: Pulling a Spool 02 010 10.0 points Consider the two orientations for pulling on a string wrapped around the axle of a giant transparent spool, causing it to move across the table. Q P 4 into the paper, and in case Q, there is no torque, so the spool in case P rolls to the right, while the spool in case Q does not roll, but instead slides along the table. Ramp Race 011 10.0 points Consider four round bodies of the same mass 9.7 kg and the same radius 0.34 m but of different geometries – a solid sphere, a spherical shell, a solid cylinder, and a cylindrical shell. All four bodies start with zero initial velocity at the same distance from the bottom of the incline and roll without slipping down the same inclined plane. Which body reaches the bottom first? 1. Depends on the steepness of the incline. 2. The spherical shell. 3. The solid sphere. correct 4. All four take the same time. 5. The solid cylinder. 6. The cylindrical shell. What rolling motion is observed for P and for Q, respectively? Explanation: Consider a round body rolling down an inclined plane. 1. To the right; not at all, it skids without rolling. correct 2. Not at all, it skids without rolling; to the left. 3. Both skid without rolling to the right. 4. To the left; to the left. 5. To the right; to the right. 6. To the left; to the right. Explanation: Take the pivot as the point where the spool touches the table. In case P , there is a torque x mg N f The forward motion of the body is governed by Newton’s Second Law m ax = Fxnet = m g sin θ − f (1) Version 001 – Rolling Part I – smith while the rotation around the body’s center is governed by the torque equation net Icm α = τCM ax Icm =fR R ax Icm 2 = f R 5 (1 + κ/2) m v2 2 (1 + 2 κ) 5. m v2 2 Explanation: 4. (2) I = κ m R2 v ω= R Adding eqs. (1) and (2), we obtain ICM m ax + 2 ax = mg sin θ R 2 m ax R + ICM ax = m g R2 sin θ m g R2 sin θ ax = m R2 + ICM (3) All four bodies have same masses, same radii, and they roll down the same incline from the same height, but they have different moments of inertia, so by to eq. (3) the body with the smallest moment of inertia has the highest downhill acceleration. Specifically, for the spheres Isolid = 2 M R2 5 and Ihollow = 2 M R2 , 3 and for the cylinders 1 Isolid = M R2 2 and Ihollow = M R2 . The solid sphere has smallest ICM of the four bodies, so it rolls faster and is the first to get to the bottom of the incline. Rolling Wheel 012 (part 1 of 2) 10.0 points A wheel of mass m and radius R has a moment of inertia κ m R2 and rolls with a speed v . What is the total kinetic energy of the wheel? (2 + κ) 1. m v2 2 (1 + κ) 2. m v 2 correct 2 (1 + κ) 3. m v2 4 Ktotal = Ktranslational + Krotational 1 1 = m v2 + I ω2 2 2
 v 2 1 1 = m v2 + κ m R2 2 2 R 1 = (1 + κ) m v 2 2 013 (part 2 of 2) 10.0 points If, due to an exernal force acting on the center of mass of the wheel, the wheel accelerates on level ground while rolling with a constant acceleration g/10 , what is the frictional force on the wheel at its point of contact? κmg correct 10 (1 + κ) m g 2. 10 κmg 3. 20 (+κ) m g 4. 20 1. 5. 2 κ m g Explanation: g 10 a α= R τ = r Ffriction = I α a= Ffriction = = Iα R   g/10 (κ m R2 ) R κmg = 10 R Version 001 – Rolling Part I – smith StringOnSpool 014 10.0 points A big spool is on a tabletop and is pulled with a constant force T applied on a string wrapped around the inner cylinder of the spool. ~ T 6 from rest near the top of the track at a height h, where h is large compared to 47 cm. 0.4 kg 47 cm D B h A 1.4 m 54 ◦ P Which way does it roll? The string does not slip on the cylinder, and the spool does not slip on the table. 1. B 2. None of these 3. A correct What is the minimum value of h (in terms of the radius of the loop R) such that the sphere completes the loop? The acceleration due to gravity is 9.8 m/s2 . The moment of 2 inertia for a solid sphere is m r 2 . 5 Correct answer: 2.511 m. 4. D; it moves upward, losing contact with the tabletop. Explanation: 5. It does not roll because the torque is zero; it skids without rolling. Note that initially the center of mass of the sphere is a distance h + r above the bottom of the loop and as the mass reaches the top of the loop, this distance above the reference is 2 R− r. The conservation of energy requirement gives 1 1 m g (h + r) = m g (2 R − r) + m v 2 + I ω 2 . 2 2 2 For the sphere I = m r 2 and v = r ω so that 5 the expression becomes 7 2 gh +2gr = 2gR + v . (1) 10 h = hmin when the speed of the sphere at the top of the loop satisfies the condition X m v2 F = mg = R−r 2 v = g (R − r) . Explanation: If we use the right-hand rule, the torque about either the center of mass or the point where the spool touches the table is directly into the paper away from you, so the spool must roll in direction A. Note that the magnitude of the torque about the CM (τ = r⊥ T ) is just the radius of the inner cylinder times the magnitude of the string tension. Slope and a Loop 02 015 (part 1 of 2) 10.0 points A small, solid sphere of mass 0.4 kg and radius 47 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 1.4 m at the end of the slope. It starts ∆Krot + ∆Ktrans + ∆U = 0 Version 001 – Rolling Part I – smith Substituting this into Eq. 1 gives hmin = 2 (R − r) + If the spool does not slip on the tabletop, and the string does not slip on the spool, in which direction will the spool roll in the two different orientations shown? 7 (R − r) 10 = 2.7 (R − r) = 2.7 (1.4 m − 0.47 m) 1. Insufficient information is given. = 2.511 m . 2. To the right; to the left 016 (part 2 of 2) 10.0 points What are the force component in the horizontal direction on the sphere at the point P, which has coordinates (−R, 0) if we take the center of the loop as origin, and if h = 3 R ? Correct answer: 19.6903 N. Explanation: When the sphere is initially at h = 3 R and finally at point P, the conservation of energy gives m g (3 R + r) = m g R + v2 = 7 1 1 m v2 + m v2 2 5 10 (2 R + r) g , 7 3. To the right; to the right correct 4. To the left; to the right 5. To the left; to the left Explanation: Pick the spot at which each spool touches the floor. The only torque around this spot is the torque exerted by the force T, which is in the same direction (into the paper and away from you) on both spools. Therefore, both spools will accelerate in rotation in the same sense, which is the direction of T, implying rightward translational motion. Torque on Yo-Yo 018 10.0 points A yo-yo, arranged as shown, rests on a frictionless surface. so at point P, Fy = −m g and m v2 R−r 10 2 R + r =+ mg 7 R−r 10 2(1.4 m) + 0.47 m =+ 7 1.4 m − 0.47 m × (0.4 kg)(9.8 m/s2 ) Fx = N = + ~ is applied to the string as When a force F shown, the yo-yo: = 19.6903 N . Spool Tug 017 10.0 points A large spool is pulled across a level tabletop by a constant force applied to a string wrapped tightly around the inner part of the spool. 1. moves to the right and does not rotate 2. moves to the left and rotates clockwise 3. moves to the right and rotates clockwise 4. moves to the left and rotates counterclockwise T T 5. moves to the right and rotates counterclockwise correct Version 001 – Rolling Part I – smith Explanation: By assumption there is no friction on the yo-yo. Thus, the only force acting on the yoyo is the applied force, which exerts a counterclockwise torque about the center of mass of the yo-yo. Also, it is the only horizontal force on the yo-yo, so the net force is in its direction, to the right. Component of Net Torque 019 10.0 points ~ = (2 N)ı̂ + (3.5 N) ̂ is applied A force F to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The ~ = (3.5 m)ı̂ + force is applied at the point R (4.7 m) ̂. Find the z-component of the net torque. Correct answer: 2.85 N m. Explanation: Basic Concept: ~ ~τ = ~r × F Solution: From the definition of torque ~ ×F ~ ~τ = R = [Rx ı̂ + Ry ̂] × [Fx ı̂ + Fy ̂] = [(3.5 m)ı̂ + (4.7 m) ̂] × [(2 N)ı̂ + (3.5 N) ̂] = [(3.5 m) (3.5 N) − (4.7 m) (2 N)]k̂ = (2.85 N m) k̂ In this case, the net torque only has a zcomponent. Door Handle Torque 020 10.0 points A door is opened in the usual way, in the direction indicated. A D C B 8 In which direction is the net torque vector ~τ due to the force applied to the door handle? The force is toward you, away from the door. 1. D 2. A correct 3. C 4. B 5. Insufficient information is given. Explanation: ~τ = r × F r ~τ F ~r points from the hinge to the spot where ~ acts, so by the right hand rule, ~r × F ~ points F in the direction A, along the hinge line. Maximize Torque 021 10.0 points A force is to be applied to a wheel. The torque can be maximized by: 1. applying the force near the rim, radially outward from the axle 2. applying the force near the axle, radially outward from the axle 3. applying the force near the axle, parallel to the tangent to the wheel 4. applying the force at the rim, at 45◦ to the tangent 5. applying the force at the rim, tangent to the rim correct Explanation: Let’s look at how torque is defined via cross ~ to be the magnitude product, where we take F of the force causing the torque, and we take ~r to be the vector pointing from the center of ~ rotation of the object to the point at which F is applied: Version 001 – Rolling Part I – smith 9 023 (part 1 of 2) 10.0 points A particle is located at the vector position ~ ~τ = ~r × F ~ sin θ |~τ | = |~r| F ~r = (1.5 m)ı̂ + (4.3 m) ̂ Where θ is the angle between the vectors ~r ~. and F ~ to be So, to maximize the force, we want F applied as far from the center of the wheel as possible, and we want it to be perpendicular to the wheel. Thus, it should be near the rim, and tangent to the wheel. and the force acting on it is Rigid Body Torque 022  10.0 points  A force F = F0 ı̂ + ̂ + k̂ acts on a rigid body at a point r = r0 (ı̂ − ̂) away from the axis of rotation. What is the resulting torque on the body?   1. r0 F0 2 k̂ − ı̂ − ̂ . correct   2. r0 F0 k̂ − ı̂ − ̂ .   3. r0 F0 2 ̂ − ı̂ − k̂ .   4. r0 F0 k̂ − 2ı̂ − 2 ̂ .   5. r0 F0 k̂ − 2ı̂ − ̂ . Explanation: Basic Concept: 6. Zero.   7. r0 F0 ı̂ + ̂ + k̂ .   8. r0 F0 2ı̂ − k̂ − ̂ .   9. r0 F0 k̂ − ı̂ − 2 ̂ . = F0 r0 ı̂ × ̂ + ı̂ × k̂ − ̂ × ı̂ − ̂ × k̂   = F0 r0 k̂ − ̂ + k̂ − ı̂   = F0 r0 2 k̂ − ̂ − ı̂ . Torque About the Origin What is the magnitude of the torque about the origin? Correct answer: 7.62 N m. ~ ~τ = ~r × F Solution: Since neither position of the particle, nor the force acting on the particle have the z-components, the torque acting on the particle has only z-component: ~τ = [x Fy − y Fx ] k̂ = [(1.5 m) (1.8 N) − (4.3 m) (2.4 N)] k̂ = [−7.62 N m] k̂ . 024 (part 2 of 2) 10.0 points What is the magnitude of the torque about the point having coordinates [a, b] = [(3 m), (6 m)]? Correct answer: 1.38 N m. Explanation: Reasoning similarly as we did in the previous section, but with the difference that relative to the point [(3 m), (6 m)] the ycomponent of the particle is now [y − (6 m)], we have Explanation: ~τ = r × F ~ = (2.4 N)ı̂ + (1.8 N) ̂ . F  ~τ = {[x − a] Fy − [y − b] Fx} k̂ = {[(1.5 m) − (3 m)] [1.8 N] − [(4.3 m) − (6 m)] [2.4 N]} k̂ = {1.38 N m} k̂ . Torque in 3D 025 (part 1 of 3) 10.0 points Version 001 – Rolling Part I – smith 10 ~ = Fx î + Fy ĵ + Fz k̂ acts on A force F ~ = (x, y, z). Given a particle located at X Fx = 74.9 N, Fy = 76.5 N, Fz = 84.5 N, x = −0.765 m, y = 6.47 m and z = 2.57 m, calculate the three components of the torque vector ~τ = τx î + τy ĵ + τz k̂ . First, calculate the τx component. F rod F wrench Correct answer: 350.11 N m. wrench nut 1 Explanation: The torque vector is defined as the cross ~ ×F ~ . In components, this product ~τ = X means rod wrench τx = y Fz − z Fy = 350.11 N m, τy = z Fx − x Fz = 257.135 N m, τz = x Fy − y Fx = −543.125 N m. nut 2 F F wrench nut 4 nut 3 Which of the following correctly lists the arrangements (above) in descending order of torque magnitude? 1. τ4 ≥ τ2 ≥ τ3 ≥ τ1 2. τ1 ≥ τ2 ≥ τ3 ≥ τ4 026 (part 2 of 3) 10.0 points Second, calculate the τy component. 3. τ3 ≥ τ2 ≥ τ4 ≥ τ1 Correct answer: 257.135 N m. 4. τ2 ≥ τ1 ≥ τ4 ≥ τ3 correct 5. τ2 ≥ τ1 ≥ τ3 ≥ τ4 Explanation: See above. 027 (part 3 of 3) 10.0 points Finally, calculate the τz component. Correct answer: −543.125 N m. Explanation: See above. Wrench Orientations 028 10.0 points You are using a wrench to tighten a nut. 6. τ2 ≥ τ4 ≥ τ3 ≥ τ1 7. τ1 ≥ τ3 ≥ τ2 ≥ τ4 8. τ2 ≥ τ4 ≥ τ1 ≥ τ3 9. τ3 ≥ τ1 ≥ τ2 ≥ τ4 10. τ3 ≥ τ4 ≥ τ2 ≥ τ1 Explanation: ~ τ ≡ ~r × F The force as shown is in the x direction, so the perpendicular radius arm is always in the Version 001 – Rolling Part I – smith y direction. For a given force, a longer radius arm defines the greater torque τ , so τ2 ≥ τ1 ≥ τ4 ≥ τ3 . 11