Version001–RollingPartI–smith–(3102F16B1) 1 This print-out should have 28 questions. ∆v =7m/s , and Multiple-choice questions may continue on ∆t =3.4 s . the next column or page – find all choices before answering.

AP M 1998 MC 6 v 001 10.0 points A wheel of mass M and radius R rolls on a ω level surface without slipping. If the angular velocity of the wheel about B its center is ω, what is its linear momentum Consider point B at the bottom of the relative to the surface? wheel. It is at rest since there is no slip- ping. Its velocity 2 1. p = MωR

2. p =0 vB = v − ωR =0 v ω = Mω2 R2 R 3. p = 2 2 4. p = Mω R ∆ω 1 ∆v α = = ∆t R ∆t 5. p = MωR correct 1 7m/s 2 = = 7.62527 rad/s . Explanation: 0.27 m 3.4 s First, we note that the wheel is rotating about its center at an angular velocity of ω, 003 (part 2 of 3) 10.0 points so the velocity difference between the center Through what angle does the wheel turn in of the wheel and the lowest point (the point of the 3.4 s? the wheel that touches the surface) is ωR in magnitude in the horizontal direction. Since Correct answer: 44.0741 rad. the wheel rolls without slipping (which means Explanation: the velocity of the lowest point is 0) the ve- locity v of the center of the wheel is ωR. 1 2 1 2 2 Obviously, the center of the wheel is also the θ = αt = (7.62527 rad/s )(3.4 s) center of the mass, so the linear momentum is 2 2 p = MωR. = 44.0741 rad .

Bicycle Down a Hill 002 (part 1 of 3) 10.0 points 004 (part 3 of 3) 10.0 points A bicycle wheel of radius 0.27 m rolls down How many revolutions does it make? a hill without slipping. Its linear velocity increases constantly from 0 to 7 m/sin3.4 s. Correct answer: 7.01461 rev. What is its angular acceleration? Explanation: 2 1 rev =2 π rad, so Correct answer: 7.62527 rad/s .

Explanation: 1 rev θ = 44.0741 rad · = 7.01461 rev . 2 π rad Let : R =0.27 m , Version001–RollingPartI–smith–(3102F16B1) 2 keywords: Explanation: If you roll them down an incline, the solid Concept 08 14 ball will reach the bottom first. (The hol- 005 10.0 points low ball has more rotational inertia compared Which will have the greater acceleration with its weight.) rolling down an incline: a bowling ball or a volleyball, and why? KE Ratio 01 007 10.0 points 1. the volleyball; mass A rigid body of radius R and mass M is rolling on a level floor without slipping. Suppose 2 2. no difference Icm = κMR . We can consider the point where the body toches the floor as a pivot, 3. the bowling ball; rotational inertia cor- and calculate the rotational kinetic energy rect Kr,p around that pivot. We usually define a kinetic energy that consists of the rotational 4. the volleyball; rotational inertia kinetic energy of the body around its center of mass, plus the kinetic energy of its center 5. the bowling ball; mass of mass motion; that is, K = Kr,cm + Kcm. Explanation: The bowling ball wins. A solid sphere of any Kr,p K mass and size beats both a solid cylinder and a hollow ball of any mass and size, because b b vcm a solid sphere has less rotational inertia per mass than the other shapes. A solid sphere b 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 What is the ratio of Kr,p to K, in general? its mass farther from the axis. The object with the least rotational inertia per mass is 1. κR the least lazy and will win races. 2. (κ + 1) by the parallel axis theorem Concept 08 15 006 10.0 points 3. Zero What technique would help you to distinguish between two heavy identical-looking spheres 4. (κ − 1. of the same weight, one solid and the other 2 1 hollow? 5. κ 1. Hit them and compare their reflective 6. 1.0 correct sounds. 2 7. κ 2. Plunge them into water and compare their 1 densities. 8. κ κ 3. Launch them and compare their trajecto- 9. ries. R 10. κ 4. Roll them down an incline and compare their speeds. correct Version001–RollingPartI–smith–(3102F16B1) 3 Explanation: Both K and K are standard representa- r,p b tions of the total kinetic energy of a rigid body F K c rolling with slipping, so r,p =1.0. K A B This is easy to see from the parallel axis f P 1 2 2 theorem: K = (κ + 1) MR ω . But r,p 2 Choose either A or P as the pivoting point. vcm = Rω for rolling without slipping, so Choose A: the torque about A is τ = f(c − b) 1 2 2 1 2 clockwise and the spool will rotate clockwise. K = κMR ω + M v r,p 2 2 cm Choose P : the torque about P is τ = F (c − b) 1 2 2 clockwise and spool will still rotate clockwise. = (κ + 1) MR ω . 2 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. 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 009 (part 2 of 2) 10.0 points stem in a counterclockwise manner. Starting Choose the correct set of equations that gov- from rest, the spool rolls when the string is ern the motion of the spool. Here f is the pulled horizontally along AB as shown below. magnitude of the frictional force between the The force F is applied in such way that there spool and the horizontal surface, a is the lin- is no slippage. ear 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 evalu- b ated with respect to the center of the spool. c F Be careful about the relative signs. A B 1. Fx : F + f = ma X a Determine the direction of rotation of the τ : f c − F b = I spool. X b 2. Fx : F − f = ma 1. counterclockwise X a τ : f c − F b = I correct X c 2. Cannot be determined 3. Fx : F − f = ma X 3. clockwise correct τ : f c − F b = Ia Explanation: X 4. Fx : F − f = ma X a Fnet = ma, and (1) τ : f c − F b = I a X b τnet = Iα = I , (2) r 5. Fx : F + f = ma X Under the action of force F , the spool tends τ : f c − F b = Ia to accelerate linearly to the right, and to ro- X tate counterclockwise. Therefore the contact 6. Fx : F − f = ma point (presently at P , see the picture below), X a τ : f c + F b = I will have a tendency to move to the right c against the surface. In turn, the friction force X must point to the left. Explanation: Version001–RollingPartI–smith–(3102F16B1) 4 Using equations (1) and (2), into the paper, and in case Q, there is no torque, so the spool in case P rolls to the Fnet = F − f = ma, right, while the spool in case Q does not roll, but instead slides along the table. and for clockwise torque about the center of mass Ramp Race a τnet = f c − Fb = Iα = I , 011 10.0 points c Consider four round bodies of the same mass a where we use the relation α = for rolling 9.7 kg and the same radius 0.34 m but of dif- c ferent geometries – a solid sphere, a spherical without slipping. So the correct pair of equa- shell, a solid cylinder, and a cylindrical shell. tions is All four bodies start with zero initial veloc- ity at the same distance from the bottom of Fx : F − f = ma X the incline and roll without slipping down the Ia same inclined plane. τ : f c − Fb = . Which body reaches the bottom first? X c 1. Depends on the steepness of the incline. keywords: 2. The spherical shell. Pulling a Spool 02 010 10.0 points 3. The solid sphere. correct Consider the two orientations for pulling on a string wrapped around the axle of a giant 4. All four take the same time. transparent spool, causing it to move across the table. 5. The solid cylinder. P Q 6. The cylindrical shell.

What rolling motion is observed for P and Explanation: for Q, respectively? Consider a round body rolling down an in- clined 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. mg x 4. To the left; to the left. N f 5. To the right; to the right. The forward motion of the body is governed by Newton’s Second Law 6. To the left; to the right. Explanation: Take the pivot as the point where the spool net touches the table. In case P , there is a torque max = Fx = mg sin θ − f (1) Version001–RollingPartI–smith–(3102F16B1) 5 while the rotation around the body’s center is (1 + κ/2) 2 governed by the torque equation 4. m v 2 net Icm α = τCM (1+2 κ) 2 5. m v ax 2 Icm = fR R Explanation: a I x = f (2) cm 2 2 R I = κmR v Adding eqs. (1) and (2), we obtain ω = R Ktotal = Ktranslational + Krotational ICM ma + a = mg sin θ 1 2 1 2 x R2 x = m v + Iω 2 2 2 2 2 max R + ICM ax = mgR sin θ 1 2 1 2 v 2 = m v + κmR mgR sin θ 2 2 R ax = 2 (3)
  mR + ICM 1 2 = (1 + κ) m v 2 All four bodies have same masses, same radii, and they roll down the same incline 013 (part 2 of 2) 10.0 points from the same height, but they have different If, due to an exernal force acting on the center moments of inertia, so by to eq. (3) the body of mass of the wheel, the wheel accelerates with the smallest moment of inertia has the on level ground while rolling with a constant highest downhill acceleration. Specifically, for acceleration g/10 , what is the frictional force the spheres on the wheel at its point of contact? 2 2 2 2 κmg Isolid = MR and Ihollow = MR , 1. correct 5 3 10 (1 + κ) mg and for the cylinders 2. 10 1 2 2 κmg Isolid = MR and Ihollow = MR . 3. 2 20 (+κ) mg The solid sphere has smallest ICM of the 4. four bodies, so it rolls faster and is the first to 20 get to the bottom of the incline. 5. 2 κmg Explanation: Rolling Wheel g 012 (part 1 of 2) 10.0 points a = A wheel of mass m and radius R has a moment 10 2 a of inertia κmR and rolls with a speed v. α = What is the total kinetic energy of the R wheel? τ = r Ffriction = Iα Iα (2 + κ) 2 Ffriction = 1. m v R 2 10 (κmR2) g/ (1 + κ) 2 R 2. m v correct =   2 R (1 + κ) 2 κmg 3. m v = 4 10 Version001–RollingPartI–smith–(3102F16B1) 6 from rest near the top of the track at a height StringOnSpool h, where h is large compared to 47 cm. 014 10.0 points 0.4 kg 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 47 cm spool. T~ D

h B A ◦ 1.4m 54 P

Which way does it roll? The string does not slip on the cylinder, and the spool does not slip on the table. What is the minimum value of h (in terms of the radius of the loop R) such that the 1. B sphere completes the loop? The acceleration 2 due to gravity is 9.8 m/s . The moment of 2. None of these 2 2 inertia for a solid sphere is mr . 5 3. A correct Correct answer: 2.511 m. 4. D; it moves upward, losing contact with Explanation: the tabletop. ∆Krot +∆Ktrans +∆U =0 5. It does not roll because the torque is zero; Note that initially the center of mass of the it skids without rolling. sphere is a distance h + r above the bottom of the loop and as the mass reaches the top of the Explanation: loop, this distance above the reference is 2 R− If we use the right-hand rule, the torque r. The conservation of energy requirement about either the center of mass or the point gives where the spool touches the table is directly 1 2 1 2 mg (h + r) = mg (2 R − r) + m v + Iω . into the paper away from you, so the spool 2 2 must roll in direction A. Note that the magni- 2 2 For the sphere I = mr and v = rω so that tude of the torque about the CM (τ = r⊥ T ) 5 is just the radius of the inner cylinder times the expression becomes the magnitude of the string tension. 7 2 g h +2 gr =2 gR + v . (1) 10 Slope and a Loop 02 h = hmin when the speed of the sphere at 015 (part 1 of 2) 10.0 points the top of the loop satisfies the condition A small, solid sphere of mass 0.4 kg and radius m v2 47 cm rolls without slipping along the track F = mg = consisting of slope and loop-the-loop with ra- R − r X 2 dius 1.4 m at the end of the slope. It starts v = g (R − r) . Version001–RollingPartI–smith–(3102F16B1) 7 Substituting this into Eq. 1 gives If the spool does not slip on the tabletop, and the string does not slip on the spool, in 7 hmin =2(R − r) + (R − r) which direction will the spool roll in the two 10 different orientations shown? =2.7(R − r) =2.7(1.4m − 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 3. To the right; to the right correct What are the force component in the hori- zontal direction on the sphere at the point P, 4. To the left; to the right which has coordinates (−R, 0) if we take the center of the loop as origin, and if h =3 R ? 5. To the left; to the left Explanation: Correct answer: 19.6903 N. Pick the spot at which each spool touches Explanation: the floor. The only torque around this spot is When the sphere is initially at h =3 R and the torque exerted by the force T, which is in finally at point P, the conservation of energy the same direction (into the paper and away gives from you) on both spools. Therefore, both spools will accelerate in rotation in the same 1 2 1 2 mg (3 R + r) = mgR + m v + m v sense, which is the direction of T, implying 2 5 rightward translational motion. 2 10 v = (2 R + r) g, 7 Torque on Yo-Yo 018 10.0 points so at point P, Fy = −mg and A yo-yo, arranged as shown, rests on a fric- m v2 tionless surface. F = N = + x R − r 10 2 R + r = + mg 7 R − r 10 2(1.4m)+0.47 m = + 7 1.4m − 0.47 m 2 × (0.4 kg)(9.8m/s ) = 19.6903 N . When a force F~ is applied to the string as shown, the yo-yo:

Spool Tug 1. moves to the right and does not rotate 017 10.0 points A large spool is pulled across a level table- 2. moves to the left and rotates clockwise top by a constant force applied to a string wrapped tightly around the inner part of the 3. moves to the right and rotates clockwise spool. 4. moves to the left and rotates counter- T clockwise T 5. moves to the right and rotates counter- clockwise correct Version001–RollingPartI–smith–(3102F16B1) 8 Explanation: In which direction is the net torque vector By assumption there is no friction on the ~τ due to the force applied to the door handle? yo-yo. Thus, the only force acting on the yo- The force is toward you, away from the door. yo is the applied force, which exerts a coun- terclockwise torque about the center of mass 1. D of the yo-yo. Also, it is the only horizontal force on the 2. A correct yo-yo, so the net force is in its direction, to the right. 3. C

Component of Net Torque 4. B 019 10.0 points A force F~ = (2 N)ˆı + (3.5 N)ˆ is applied 5. Insufficient information is given. to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The Explanation: force is applied at the point R~ = (3.5 m)ˆı + ~τ (4.7 m)ˆ. r ~τ = r × F Find the z-component of the net torque. F ~r points from the hinge to the spot where Correct answer: 2.85 Nm. F~ acts, so by the right hand rule, ~r × F~ points Explanation: in the direction A, along the hinge line. Basic Concept: ~τ = ~r × F~ Maximize Torque 021 10.0 points Solution: From the definition of torque A force is to be applied to a wheel. The torque can be maximized by: ~τ = R~ × F~ =[Rx ˆı + Ry ˆ] × [Fx ˆı + Fy ˆ] 1. applying the force near the rim, radially = [(3.5 m)ˆı + (4.7 m)ˆ] × [(2 N)ˆı + (3.5 N)ˆ] outward from the axle = [(3.5 m)(3.5 N) − (4.7 m)(2 N)]kˆ ˆ 2. applying the force near the axle, radially = (2.85 Nm) k outward from the axle In this case, the net torque only has a z- component. 3. applying the force near the axle, parallel to the tangent to the wheel

Door Handle Torque ◦ 020 10.0 points 4. applying the force at the rim, at 45 to A door is opened in the usual way, in the the tangent direction indicated. 5. applying the force at the rim, tangent to the rim correct A Explanation: D C Let’s look at how torque is defined via cross product, where we take F~ to be the magnitude B 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: Version001–RollingPartI–smith–(3102F16B1) 9 023 (part 1 of 2) 10.0 points A particle is located at the vector position ~τ = ~r × F~ |~τ| = |~r| F~ sin θ ~r = (1.5 m)ˆı + (4.3 m)ˆ

Where θ is the angle between the vectors ~r and the force acting on it is and F~ . F~ = (2.4 N)ˆı + (1.8 N)ˆ. So, to maximize the force, we want F~ to be applied as far from the center of the wheel as What is the magnitude of the torque about possible, and we want it to be perpendicular the origin? to the wheel. Thus, it should be near the rim, and tangent to the wheel. Correct answer: 7.62 Nm.

Rigid Body Torque Explanation: 022 10.0 points Basic Concept: ˆ A force F = F0 ˆı +ˆ + k acts on a rigid ~τ = ~r × F~   body at a point r = r0 (ˆı − ˆ) away from the axis of rotation. Solution: Since neither position of the par- What is the resulting torque on the body? ticle, nor the force acting on the particle have the z-components, the torque acting on the 1. r0 F0 2 kˆ − ˆı − ˆ . correct particle has only z-component:   ~τ =[x F − y F ] kˆ 2. r0 F0 kˆ − ˆı − ˆ . y x   = [(1.5 m)(1.8 N) − (4.3 m)(2.4 N)] kˆ ˆ 3. r0 F0 2ˆ − ˆı − k . =[−7.62 Nm] k.ˆ   4. r0 F0 kˆ − 2ˆı − 2ˆ .   024 (part 2 of 2) 10.0 points 5. r0 F0 kˆ − 2ˆı − ˆ . What is the magnitude of the torque   about the point having coordinates [a,b] = 6. Zero. [(3 m), (6 m)]?

7. r0 F0 ˆı +ˆ + kˆ . Correct answer: 1.38 Nm.   Explanation: 8. r0 F0 2ˆı − kˆ − ˆ .   Reasoning similarly as we did in the pre- vious section, but with the difference that 9. r0 F0 kˆ − ˆı − 2ˆ .   relative to the point [(3 m), (6 m)] the y- Explanation: component of the particle is now [y − (6 m)], we have

~τ = r × F ~τ = {[x − a] Fy − [y − b] Fx} kˆ = F0 r0 ˆı × ˆ +ˆı × kˆ − ˆ × ˆı − ˆ × kˆ = {[(1.5 m) − (3 m)][1.8 N]   − [(4.3 m) − (6 m)][2.4 N]} kˆ = F0 r0 kˆ − ˆ + kˆ − ˆı   = {1.38 Nm} k.ˆ = F0 r0 2 kˆ − ˆ − ˆı .   Torque in 3D Torque About the Origin 025 (part 1 of 3) 10.0 points Version001–RollingPartI–smith–(3102F16B1) 10

A force F~ = Fx ˆi + Fy ˆj + Fz kˆ acts on F a particle located at X~ = (x,y,z). Given 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, rod calculate the three components of the torque vector ~τ = τx ˆi + τy ˆj + τz k.ˆ F First, calculate the τx component. wrench wrench Correct answer: 350.11 Nm.

nut nut Explanation: 1 2 The torque vector is defined as the cross product ~τ = X~ × F~ . In components, this rod F means F wrench wrench

nut nut τx = y Fz − z Fy = 350.11 Nm, 3 4 τy = z Fx − x Fz = 257.135 Nm, Which of the following correctly lists the τz = x Fy − y Fx = −543.125 Nm. 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 Nm. 4. τ2 ≥ τ1 ≥ τ4 ≥ τ3 correct

5. τ2 ≥ τ1 ≥ τ3 ≥ τ4 Explanation: See above. 6. τ2 ≥ τ4 ≥ τ3 ≥ τ1

7. τ1 ≥ τ3 ≥ τ2 ≥ τ4 027 (part 3 of 3) 10.0 points Finally, calculate the τz component. 8. τ2 ≥ τ4 ≥ τ1 ≥ τ3 Correct answer: −543.125 Nm. 9. τ3 ≥ τ1 ≥ τ2 ≥ τ4

10. τ3 ≥ τ4 ≥ τ2 ≥ τ1 Explanation: See above. Explanation: τ ≡ ~r × F~ Wrench Orientations 028 10.0 points The force as shown is in the x direction, so You are using a wrench to tighten a nut. the perpendicular radius arm is always in the Version001–RollingPartI–smith–(3102F16B1) 11 y direction. For a given force, a longer radius arm defines the greater torque τ, so

τ2 ≥ τ1 ≥ τ4 ≥ τ3 .