Version001–AngularMomentum–smith–(3102F16B1) 1 This print-out should have 47 questions. axis perpendicular to the circle and through Multiple-choice questions may continue on its center? the next column or page – find all choices kg m2 before answering. 1. L~ = 24 · correct k k s Airplane m2 2. L~ = 12 001 (part 1 of 2) 10.0 points k k s An airplane of 18691 kg flies level to the N m ground at an altitude of 14 km with a constant 3. L~ =9 · k k kg of 178 m/s relative to the Earth. 2 What is the magnitude of the airplane’s ~ kg m 4. L = 13.5 ·2 relative to a ground ob- k k s 2 server directly below the airplane in kg m /s? N m · 5. L~ = 18 · kg 10 2 k k Correct answer: 4.6578 10 kg m /s. × · Explanation: Explanation: The angular momentum is

Since the observer is directly below the air- L~ = mvr = (2 kg)(3 m/s)(4 m) plane, k k 2 = 24 kg m /s . L = h m v ·

002 (part 2 of 2) 10.0 points Does this value change as the airplane contin- Conical 04 ues its along a straight line? 004 10.0 points A small metallic bob is suspended from the 1. Yes. L decreases as the airplane moves. ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle 2. Yes. L increases as the airplane moves. so that the thread describes a cone.

3. Yes. L changes in a random pattern as the airplane moves.

4. Yes. L changes with certain period as the 29 airplane moves. 2 ◦ 9.8m/s 3 . 2 m 5. No. L = constant. correct Explanation: L = constant since the perpendicular dis- v tance from the line of flight to Earth’s surface doesn’t change. 9 kg Calculate the magnitude of the angular mo- AP B 1998 MC 6 mentum of the bob about a vertical axis 003 10.0 points through the supporting point. The acceler- 2 A 2 kg object moves in a circle of radius 4 m ation of is 9.8m/s . at a constant speed of 3 m/s. A net of 2 4.5 N acts on the object. Correct answer: 40.5335 kg m /s. What is the magnitude of the angular mo- · mentum L~ of the object with respect to an Explanation: k k Version001–AngularMomentum–smith–(3102F16B1) 2

Decelerated Grinding Wheel Let : ℓ =3.2m , 005 (part 1 of 2) 10.0 points ◦ The motor driving a grinding wheel with a θ = 29 , 2 2 g =9.8m/s , and rotational of 0.8 kgm is switched off when the wheel has a m =9kg . of 39 rad/s. After 6.7 s, the wheel has slowed down to 31.2 rad/s. Consider the free body diagram. What is the absolute value of the constant exerted by to slow the wheel down?

Correct answer: 0.931343 Nm. θ Explanation: T We have

τ ∆t =∆L = ∆(Iω) ,

so that mg I ω1 ω0 τ = | − | ’s law in the vertical and | | ∆t1 horizontal projections, respectively, gives (0.8 kgm2)(39 rad/s 31.2 rad/s) = − 6.7 s T cos θ mg =0 − =0.931343 Nm . T cos θ = mg and 2 T sin θ mω ℓ sin θ =0 − 2 006 (part 2 of 2) 10.0 points T = mω ℓ, If this torque remains constant, how long after where the radius of the orbit is R = ℓ sin θ. the motor is switched off will the wheel come Dividing, to rest? mg Correct answer: 33.5 s. cos θ = 2 mω ℓ Explanation: g ω = . When the wheel comes to rest, its angular rℓ cos θ speed is ω2 = 0; hence

~r and ~v are perpendicular, where r = I (ω0 ω2) ∆t2 = − ℓ sin θ, so the angular momentum L = m ~r ~v τ × will be Iω0 = 2 τ L = mω (ℓ sin θ) | | 2 3 (0.8 kgm )(39 rad/s) 2 gℓ = = m sin θ (0.931343 Nm) rcos θ 2 ◦ = 33.5 s . = (9 kg) sin 29 (9.8m/s2)(3.2 m)3 ◦ Holt SF 08Rev 66 × r cos29 2 007 (part 1 of 2) 10.0 points = 40.5335 kg m /s . · a) Calculate the angular momentum of Earth Version001–AngularMomentum–smith–(3102F16B1) 3 that arises from its spinning motion on its 2 axis IE =0.331MERE . Merry Go Round and a Boy  009 10.0 points 33 2 Correct answer: 5.84082 10 kg m /s. The sketch shows the top view of a merry- × · Explanation: go-round which is rotating clockwise. A boy Basic Concept: jumps on the merry-go-round in three differ- ent ways: I) from the left, II) from the top, 2 L = Iω = 0.331MR ω and III) from the right. For all three cases,  the boy lands on the same spot. Given: II 6 R =6.37 10 m × 24 M =5.98 10 kg I III × ω = 1 rev/day ω Solution: 24 L =0.331(5.98 10 kg) × Compare the final angular momenta of the 6 2 1 rev (6.37 10 m) merry-go-round for the three cases. · × · 1 day 2π rad 1 day 1 h 1. LIII >LII >LI ·  1 rev   24 h  3600 s 33 2 2. L = L = L =5.84082 10 kg m /s I II III × · 3. LI = LIII >LII 008 (part 2 of 2) 10.0 points b) Calculate the average angular momentum 4. LI >LII >LIII correct of Earth that arises from its orbital motion about the sun. 5. LII >LI = LIII

40 2 Correct answer: 2.66465 10 kg m /s. Explanation: × · Explanation: With respect to the center of the merry- Basic Concept: go-round, the angular momentum of the boy is: I) clockwise, II) 0, and III) counterclock- 2 L = Iω = Mr ω wise. The angular momentum of the merry-  go-round is along the clockwise direction. By Given: adding the two angular momenta in each case, 11 we get LI >LII >LIII. r =1.496 10 m × ω = 1 rev/365.25 days Net L of point 010 10.0 points Solution: Two objects are moving in the x-y plane as 24 11 2 L = (5.98 10 kg)(1.496 10 m) shown. × × 1 rev 2π rad · 365.25 day  1 rev  1 day 1 h ·  24 h  3600 s 40 2 =2.66465 10 kg m /s × · Version001–AngularMomentum–smith–(3102F16B1) 4 The magnitude of their total angular mo- counter-clockwise as the positive angular di- mentum (about the origin O) is rection.

2 1. 8 kg m /s 1. L~ = m v0 R cos θ kˆ · 2 2. 24 kg m /s 2. L~ = m v0 R sin θ kˆ · 2 3. 32 kg m /s 3. L~ = m v0 R kˆ · 2 R 4. 0 kg m /s correct 4. L~ = m v0 kˆ · 2 2 5. 96 kg m /s ~ · 5. L =0 correct Explanation: Explanation: The angular momentum of a particle is given by L~ = I~ω, where we remember that L~ = ~r ~p =0 the ω is positive for counterclockwise rotation × in the x y plane and negative for clockwise since ~r =0 . rotation.− We can see above that the two particles are rotating in opposite senses. For a point 012 (part 2 of 4) 10.0 points particle, the of inertia is given by Using the origin as the pivot, find the angular 2 I = mr . momentum when the particle is at the highest Consequently, we have: point of the trajectory. 2 m v0 2 1. L~ = + sin θ cos θ kˆ Lnet = L1 + L2 2 g 3 = I1ω1 I2ω2 m v0 2 −2 2 2. L~ = + sin θ cos θ kˆ =(m1r1)ω1 (m2r2)ω2 2 g −2 2 2 = (1 kg)(2 m) (4 rad/s) (4 kg)(1 m) (4 rad/s) m v0 2 2 − 3. L~ = sin θ cos θ kˆ =0kg m /s − 2 g · 2 m v0 2 4. L~ = + sin θ cos θ kˆ Projectile Analysis 01 2 g 3 011 (part 1 of 4) 10.0 points m v0 2 5. L~ = sin θ cos θ kˆ A particle of mass m moving in the gravi- − 2 g tational field of the Earth is launched with 2 m v0 2 an initial v at an θ with the 6. L~ = sin θ cos θ kˆ o − 2 g horizontal. 3 m v0 2 y C 7. L~ = + sin θ cos θ kˆ vh 2 g 3 m v0 2 8. L~ = sin θ cos θ kˆ correct h − 2 g v o ~ θ A 9. L is in theˆ direction. x O R v ~ R 10. L is in theˆı direction. Explanation: Using the origin as the pivot, find the an- Denote the vector from the origin gular momentum when it is at the origin. Use to the highest point of the trajectory as ~rh . Version001–AngularMomentum–smith–(3102F16B1) 5

At the highest point ~vh = ~vx = vo cos θ ˆı and Explanation: Consider the motion to reach the maximum L~ = ~rh ~ph = ~rh m~vh × × height: so that vf = 0 = vo sin θ gt1 ~ − L = m ~vh ~rh sin θh . vo sin θ | | | || | t1 = , The maximum height is g

2 2 vo sin θ so the range of the projectile is h = ~rh sin θh = | | 2 g 2 2 vo sin θ0 cos θ0 and since ~vh = ~v0x = vx, the magnitude of R =(vo cos θ)2 t1 = , the angular| momentum| | | is g 2 2 vo sin θ and the angular momentum at that position L~ = m h vx = m (vo cos θ) | | 2 g is 3 m v 2 = o sin θ cos θ. 2 g 3 2 m vo 2 L = m vo sin θR = sin θ cos θ. The direction is kˆ by the right hand rule of g the cross product.− Using the right hand rule, the direction should be k.ˆ 013 (part 3 of 4) 10.0 points − Using the origin as the pivot, find the angular momentum of the particle just before it hits 014 (part 4 of 4) 10.0 points the ground. What torque causes its angular momentum to change? 3 2 m v0 2 1. L~ = sin θ cos θ kˆ correct − g 1. The torque from the initial force required 2 to shoot the particle. 2 m v0 2 2. L~ = sin θ cos θ kˆ − g 2. The torque from the centripetal accelera- 2 2 m v0 2 tion. 3. L~ = sin θ cos θ kˆ − g 3 2 m v0 2 3. The torque due to the initial velocity. 4. L~ = sin θ cos θ kˆ − g 4. The torque due to the particle’s . 5. L~ is in theˆı direction.

2 5. The torque due to gravity. correct 2 m v0 2 6. L~ = + sin θ cos θ kˆ g Explanation: 2 Gravity is the only force acting on the par- 2 m v0 2 7. L~ = + sin θ cos θ kˆ ticle. The change in angular momentum is g 3 negative (going from zero to negative values) 2 m v0 2 8. L~ = + sin θ cos θ kˆ because the torque of the gravity force is neg- g ative ( kˆ direction) as you can see from the − ~ right hand rule for cross product: ~τ = R~ F~ . 9. L is in theˆ direction. × 3 2 m v0 2 Dependent Momentum 10. L~ = + sin θ cos θ kˆ g 015 10.0 points Version001–AngularMomentum–smith–(3102F16B1) 6 The position vector of a particle of mass 2 kg Explanation: is given as a function of time by Let : m =5kg , ~r = (1m)ˆı +(5 m/s) t ˆ. v =2.5m/s , and Determine the magnitude of the angular r =3m . momentum of the particle with respect to the The angular momemntum is origin at time 3 s . L = mvr = (5 kg)(2.5m/s)(3 m) 2 Correct answer: 10 kgm /s. 2 = 37.5 kg m /s . Explanation: ·

017 (part 2 of 3) 10.0 points Let : ~r = x ˆı + y ˆ, What is its about an axis x =1m , through the center of the circle and perpen- y = vy t =(5m/s) t, and dicular to the plane of the motion? t = 3 s . 2 Correct answer: 45 kg m . · Basic Concepts: Explanation: The moment of inertia is L~ = ~r ~p × 2 2 I = mr = (5 kg)(3 m) Solution: 2 = 45 kg m . ∂ ~r · ~v = = vy ∂t =(5m/s) ˆ. 018 (part 3 of 3) 10.0 points What is the angular speed of the particle? L~ = ~r ~p × Correct answer: 0.833333 rad/s. = m ~r ~v × ∂ ~r Explanation: = m ~r × ∂t The angular speed is = m (~x ˆı + ~y ˆ) ~v ˆ × L 37.5 kg m2/s = m x vy kˆ ω = = · I 45 kg m2 = (2 kg)(1 m)(5 m/s) kˆ · = 0.833333 rad/s . 2 = 10 kgm /s kˆ , and is constant as a function of time since Tipler PSE5 10 79 02 ˆ ˆ =0 . 019 (part 1 of 2) 10.0 points × A particle of mass 5 kg moves with velocity Tipler PSE5 10 44 ~v =(2m/s)ˆı along the line z =0m,y =1m. 016 (part 1 of 3) 10.0 points Find the angular momentum relative to the A 5 kg particle moves at a constant speed of origin when the particle is at x = 16 m,y = 2.5m/s around a circle of radius 3 m. 1 m. What is its angular momentum about the 2 center of the circle? 1. (1.0 kg m /s) ˆj − · 2 2 Correct answer: 37.5 kg m /s. 2. (1.0 kg m /s) kˆ · − · Version001–AngularMomentum–smith–(3102F16B1) 7

2 3. (2.0 kg m /s) kˆ 8. (20.0N m) kˆ − · · 2 Explanation: 4. (2.0 kg m /s) ˆj − · 2 ~ 5. (10 kg m /s)ˆj Let : F =( 8 N)ˆı. − · − 2 The torque due to the force is 6. (47.7 kg m /s) kˆ · 2 ~τ = ~r F~ 7. (47.7 kg m /s) ˆj × · = [(16 m)ˆı +(1 m)ˆ] ( 8 N)ˆı 2 × − 8. (10 kg m /s)kˆ correct = (8 N m) (ˆ ˆı) − · − · × Explanation: = (8 N m)kˆ . · Let : m =5kg , ~v =2m/s , and AP M 1998 MC 32 33 ~r = (16 m)ˆı +(1 m)ˆ. 021 (part 1 of 2) 10.0 points A wheel with rotational inertia I is mounted The momentum is on a fixed, frictionless axle. The angular ~p = m~v = (5 kg)(2 m/s)ˆı = (10 kg m/s)ˆı. speed ω of the wheel is increased from zero to · ωf in a time interval T . The angular momentum is What is the average net torque τ on the wheel during this time interval? L~ = ~r ~p × Iωf 1. τ = correct T L~ = [(16 m)ˆı +(1 m)ˆ] (10 kg m/s)ˆı 2 × · Iωf = (10 kg m /s) (ˆ ˆı) 2. τ = · × T 2 2 ˆ ωf = (10 kg m /s)k . 3. τ = − · T 2 2 Iωf 020 (part 2 of 2) 10.0 points 4. τ = T A force F~ =( 8N)ˆı is applied to the particle. ωf Find the torque− relative to the origin due 5. τ = T to this force. Explanation: 1. (20.0N m) ˆj The change of angular momentum of the · wheel is 2. ( 8N m) kˆ ∆L = Iωf , − · 3. (10.0N m) ˆj so the average net torque on the wheel during · this time interval is 4. (8 N m) kˆ correct L Iωf · τ = = . T T 5. ( 8N m) ˆj − · 6. (8 N m) ˆj 022 (part 2 of 2) 10.0 points · What is the average power input to the wheel 7. (10.0N m) kˆ during this time interval? · Version001–AngularMomentum–smith–(3102F16B1) 8

2 and the angular momentum is I ωf 1. P = 2 2 T L1 = I1 ω 2 2 Iωf = (0.106667 kgm )(1.1 rad/s) 2. P = correct 2 2 T =0.117333 kgm /s . Iωf 3. P = 2 T 2 Iωf 024 (part 2 of 2) 10.0 points 4. P = 2 Calculate the angular momentum of the sys- 2 T 2 2 tem when the stick is pivoted about an axis I ωf 5. P = perpendicular to the table through the 0-cm 2 T 2 mark. Explanation: 2 The change in the of the wheel is Correct answer: 0.469333 kgm /s. 1 2 Iω in the time interval T , so the average 2 f Explanation: power input to the wheel is The moment of inertia of the stick in this 1 2 2 case is Is = ms (1 m) and of the particle 1 2 1 Iωf 3 Iω = . 2 2 f T 2 T attached to the stick Ip = mp (1 m) . The total moment of inertia of the system is Rotating Meter Stick 023 (part 1 of 2) 10.0 points I2 = Is + Ip A particle of mass 0.31 kg is attached to 1 2 2 = (0.35 kg)(1 m) + (0.31 kg(1 m) the 100-cm mark of a meter stick of mass 3 2 0.35 kg. The meter stick rotates on a horizon- =0.426667 kgm , tal, frictionless table with an angular speed of 1.1 rad/s. and the angular momentum is Calculate the angular momentum of the system when the stick is pivoted about an L2 = I2 ω axis perpendicular to the table through the 2 50-cm mark. = (0.426667 kgm )(1.1 rad/s) 2 =0.469333 kgm /s . 2 Correct answer: 0.117333 kgm /s. Explanation: Rotor of a Generator L = Iω 025 10.0 points A net torque of magnitude 200 Nm is exerted The moment of inertia of the stick is 1 2 on the rotor of an electric generator for 32 s. Is = ms (1 m) and of the particle at- What is the magnitude of the angular mo- 12 2 1 mentum of the rotor at the end of 32 s if it tached to the stick Ip = mp m . The was initially at rest? 2  total moment of inertia of the system is 2 Correct answer: 6400 kgm /s. I1 = Is + Ip 2 Explanation: 1 2 1 = (0.35 kg)(1 m) + (0.31 kg) m We have: 12 2  2 =0.106667 kgm , ∆L = τnet ∆t Version001–AngularMomentum–smith–(3102F16B1) 9

Since the rotor was initially at rest, Li = 0, 027 (part 2 of 2) 10.0 points Lf = (200 Nm)(32 s) 2 What is the total angular momentum of the = 6400 kgm /s system about the z-axis relative to point B along y axis if d2 = 25 m? Total Angular Momentum 2 026 (part 1 of 2) 10.0 points Correct answer: 21050 kgm /s. Two particles move in opposite directions − along a straight line. Particle 1 of mass Explanation: m1 = 22 kg at x1 = 16 m moves with a Since at point B speed v1 = 48 m/s (to the right), while the particle 2 of mass m2 = 73 kg at x2 = 24 m − moves with a speed v2 = 26 m/s (to the r sin θ = d2 , left). − Given: Counter-clockwise is the positive we find angular direction. B d lB = d2 p v 2 v 2 1 = d2 (m1 v1 + m2 v2) m m 2 1 = (25 m)[(22 kg)(48 m/s) x x2 1 +(73 kg)( 26 m/s)] −2 = 21050 kgm /s . d1 y −

x A Bullet Rotates a Rod 01 What is the total angular momentum of 028 (part 1 of 2) 10.0 points the system about the z-axis relative to point A wooden block of mass M hangs from a rigid A along y axis if d1 = 13 m? rod of length ℓ having negligible mass. The rod is pivoted at its upper end. A bullet of 2 Correct answer: 10946 kgm /s. mass m traveling horizontally and normal to Explanation: the rod with speed v hits the block and gets The angular momentum along z axis is embedded in it. given by L = r p sin θ, where p is the total linear momentum. We ℓ find that p = m1 v1 + m2 v2 , v M where v2 is negative. Since at point A m r sin θ = d1 , − What is the angular momentum L of the we find block-bullet system, with respect to the pivot lA = d1 p point immediately after the collision? = d1 (m1 v1 + m2 v2) − =( 13 m)[(22 kg)(48 m/s) 1. L = mvℓ correct − +(73 kg)( 26 m/s)] M m 2− 2. L = vℓ = 10946 kgm /s . M + m Version001–AngularMomentum–smith–(3102F16B1) 10

3. L =(M m) vℓ Child on a Merrygoround − 030 10.0 points 4. L =(m + M) vℓ A child is standing on the edge of a merry-go- round that is rotating with f. The 5. L = Mvℓ child then walks towards the center of the Explanation: merry-go-round. For the system consisting of the child plus If ~τ = 0, then L~ =0 . ext the merry-go-round, what remains constant X X The net angular momentum of the system as the child walks towards the center? (ne- conserves, and glect friction in the bearing) Li = Lf = L = mvℓ. 1. neither nor angular momentum 029 (part 2 of 2) 10.0 points Kf What is the fraction (the final kinetic 2. only mechanical energy Ki energy compared to the initial ) 3. mechanical energy and angular momen- in the collision? tum

Kf 2 m 1. = 4. only angular momentum correct Ki m + M Kf M 2. = Ki M + m Explanation: Kf M The external to the system are not 3. = Ki M m exerting any torque, so the angular momen- − tum is conserved. On the other hand, the fric- Kf m 4. = correct tion force acting on the child is doing , be- K m + M i cause she is moving towards the center (which Kf m 5. = is the direction of that force). Ki M m − Explanation: Child on a MerryGoRound 02 v1 By conservation of the angular momentum 031 10.0 points A child of mass 52.8 kg sits on the edge of Li = Lf = L a merry-go-round with radius 1.7 m and mo- 2 mvℓ =(m + M) vf ℓ ment of inertia 134.281 kgm . The merry- m go-round rotates with an of vf = v m + M  1.9 rad/s. The child then walks towards the center of the merry-go-round and stops at a 1 2 1 2 Ki = m v and Kf = Iω where 0.612 m from the center. Now what 2 2 f 2 vf is the angular velocity of the merry-go-round? I =(M + m) ℓ and ω = , so f ℓ 1 2 Correct answer: 3.53803 rad/s. K = (M + m) v and f 2 f

2 Explanation: 1 m 2 v When the child moves inward, the moment of Kf 2 M + m m = = . inertia of the system iMGR +ichild (the merry- Ki 1 2 M + m go-round plus the child) changes. Therefore, m v 2 to conserve angular momentum, the angular Version001–AngularMomentum–smith–(3102F16B1) 11 velocity of the system must change. Specifi- 033 10.0 points cally: A thin uniform cylindrical turntable of ra- dius 1.8 m and mass 23 kg rotates in a hori- Linit = Lfinal zontal plane with an initial angular speed of (ichild + iMGR) ω =(ichild,2 + iMGR) ω2 8.5 rad/s. The turntable bearing is friction- 2 less. A clump of clay of mass 5.2 kg is dropped The moment of inertia of the child is mr . onto the turntable and sticks at a point 0.96 m Therefore from the point of rotation. 2 Find the angular speed of the clay and mr + iMGR ω2 = 2 ω . turntable. mr2 + iMGR Correct answer: 7.53133 rad/s. Child on Merry Go Round 032 10.0 points Explanation: A playground merry-go-round has a radius of 2 3 m and a rotational inertia of 600 kg m /s. It is initially spinning at 0.8 rad/s when· a Let : R =1.8m , 20 kg child crawls from the center to the rim. x =0.96 m , When the child reaches the rim the angular M = 23 kg , velocity of the merry-go-round is: ω =8.5 rad/s , and 1. 0.89 rad/s m =5.2 kg .

2. 1.1 rad/s Basic Concepts: d~τ 3. 0.8 rad/s L~ = ext X dt 4. 0.62 rad/s correct ∆K = Kf Ki = Q. − 5. 0.73 rad/s From conservation of the angular momentum Explanation: it follows that We conserve angular momentum, realizing Li = Lf that the child’s moment of inertia is given by 2 mr : or Ii ω = If ωf ,

Lf = L0 where If ωf = I0ω0 2 2 1 2 Ii = MR I + mrf ωf = I + mr0 ω0 2  I  1 2 ω = ω0 = (23 kg)(1.8 m) f I + mr2  2 2 600 kg m2/s = 37.26 kgm . = · 600 kg m2/s+(20 kg)(3 m)2  · Similarly, (0.8 rad/s) × =0.62 rad/s 2 If = Ii + mx 2 2 = (37.26 kgm )+(5.2 kg)(0.96 m) 2 Clay on a Turntable 01 = 42.0523 kgm . Version001–AngularMomentum–smith–(3102F16B1) 12 Finally, 5. II and V only Ii ωf = ω If 6. II, III and IV only (37.26 kgm2) = (8.5 rad/s) 7. III, IV and V only (42.0523 kgm2) = 7.53133 rad/s . 8. I, II and III only

9. II, III, IV and V only Clay Rotates a Rod 01a 034 10.0 points 10. I only

A uniform rod has a mass 2 m and a length ℓ, and it can spin freely in a horizontal plane about a pivot point O at the center of the rod. A piece of clay with mass m and velocity v hits one end of the rod, and causes the rod- Explanation: clay system to spin. Viewed from above the V) Total KE is not conserved. One can scheme is as follows: Kf Ii v verify that = < 1. The inequality Ki If m here is due to the fact that Ii is the moment of inertia of the clay and If is the moment of ω inertia of the clay+rod. ℓ 0 IV) Since the motion is in a horizontal plane there is no change in the . ω IV) and V) together imply that the total mechanical energy (KE + PE) is not con- 2 m served. (a) before (b) during (c) after II) Linear momentum is not conserved by inspection. Apparently the linear momen- After the collisions the rod and clay system tum is altered by the presence of the pivot. has an angular velocity ω about the pivot. The pivot provides an external force to the Which quantity/quantities rod+clay system. I) total mechanical energy III) Angular momentum is conserved, since II) total linear momentum there is no external torque applied to the III) total angular momentum with respect to rod+clay system. pivot point O IV) total gravitational potential energy Clay Rotates a Rod 01 V) total kinetic energy 035 (part 1 of 2) 10.0 points is/are conserved in this process? A uniform rod, supported and pivoted at 1. All of these its midpoint, but initially at rest, has a mass 2 m and a length ℓ. A piece of clay with mass 2. None of these m and velocity v0 hits one end of the rod, gets stuck and causes the clay-rod system to spin 3. III and IV only correct about the pivot point O at the center of the rod in a horizontal plane. Viewed from above 4. II and III only the scheme is Version001–AngularMomentum–smith–(3102F16B1) 13

v0 plus the moment of inertia of the clay: 2 1 2 ℓ m I = I + I = 2 mℓ + m f rod clay 12 2 5 2 ω = mℓ . ℓ 0 12

ω 036 (part 2 of 2) 10.0 points What is the final angular speed ωf of the (a)2 m (b) (c) rod-clay system¿ During Before After 3 v 1. ωf = With respect to the pivot point O, what 5 ℓ is the magnitude of the initial angular mo- 6 2. ωf = v mentum Li of the piece of clay and the final 5 moment of inertia If of the clay-rod system? 5 v 3. ω = After the collisions the clay-rod system has an f 12 ℓ angular velocity ω about the pivot. 4 4. ωf = v ℓ 7 2 6 1. Li = m v , I = mℓ 2 f 12 12 v 5. ωf = 3 2 7 ℓ 2. Li = mvℓ, I = mℓ f 12 5 v 6. ωf = 5 2 6 ℓ 3. Li = mvℓ, I = mℓ f 12 12 7. ωf = v 8 2 7 4. Li = mvℓ, I = mℓ f 12 6 v 8. ωf = correct ℓ 4 2 5 ℓ 5. Li = m v , I = mℓ 2 f 12 6 v 9. ωf = ℓ 5 2 2 ℓ 6. Li = m v , I = mℓ correct 2 f 12 12 v 10. ωf = ℓ 8 2 5 ℓ 7. Li = m v , I = mℓ 2 f 12 Explanation: The total angular momenta are the same: 4 2 8. Li = mvℓ, I = mℓ f 12 Lf = Li Explanation: mvℓ Since the total external torque acting on I ω = f f 2 the clay-rod system is zero, the total angular 5 2 mvℓ momentum is a constant of motion. The total mℓ ωf = initial angular momentum L is simply the 12 2 i 6 v angular momentum of the clay, since the rod ωf = . is at rest initially 5 ℓ Concept 08 43 mvℓ 037 10.0 points Li = ~r ~p = m r v = . | × | 2 You sit at the middle of a large turntable at an amusement park as it is set spinning The final moment of inertia If of the clay- on nearly frictionless bearings, and then rod system is the moment of inertia of the rod allowed to spin freely. Version001–AngularMomentum–smith–(3102F16B1) 14 When you crawl toward the edge of the Explanation: turntable, does the of the rotation in- Applying conservation of angular momen- crease, decrease, or remain unchanged, and tum, as the radial distance of mass increases, why? the angular speed decreases. The mass of ma- terial used to construct skyscrapers is lifted, 1. increases; conservation of momentum slightly increasing the radial distance from the Earth’s spin axis. This would tend to slightly 2. decreases; conservation of energy decrease the earth’s rate of rotation, which in turn tends to lengthen days slightly. The 3. increases; conservation of energy opposite effect occurs for falling leaves, since their radial distance from the earth’s axis de- 4. decreases; conservation of momentum creases. As a practical matter, these effects correct are entirely negligible!

Conceptual 07 05 Explanation: 039 10.0 points The rotational inertia of the system (you Why does a helicopter have a tail rotor? and the rotating turntable) is least when you are at the rotational axis. As you move out- 1. This keeps the helicopter from spinning ward, the rotational inertia of the system in- out of control. correct creases. Applying conservation of angular momen- tum, as you move toward the outer rim, you 2. None of these increase the rotational inertia of the spinning 3. This makes the helicopter move faster. system, thus decreasing the angular speed. Also, if you don’t slip as you move out, you exert a friction force on the turntable oppo- 4. The tail rotor is just for decoration. site to its direction of rotation, thereby also 5. This increases safety if any of the rotors slowing it down. breaks down. Concept 08 45 Explanation: 038 10.0 points Without the tail rotor, the body would spin Strictly speaking, as more and more skyscrap- in the opposite direction of the rotors. ers are built on the surface of the Earth, does the day tend to become longer or shorter? Conceptual Rotating KE And strictly speaking, does the falling of au- 040 10.0 points tumn leaves tend to lengthen or shorten the How can the rotational kinetic energy Kr of 24-hour day? What physical principle sup- a be written in terms of the mag- ports your answers? nitude L fo its angular momentum L and its rotational moment of inertia I? 1. lengthen; shorten; conservation of angular momentum correct L2 1. Kr = correct 2 I 2. shorten; lengthen; conservation of inertia 1 2. Kr = IL 2 3. lengthen; shorten; conservation of kinetic energy 3. Kr = √2 LI

1 2 4. shorten; lengthen; conservation of angular 4. Kr = IL torque 2 Version001–AngularMomentum–smith–(3102F16B1) 15 L So, 5. Kr = I Explanation: 2 For a symmetric rigid body, we know that Lf 2 L~ = I ~ω. Putting this relation into the defini- KEf If Ii Ii 4 = 2 = = = L tion of angular rotational kinetic energy, KEi i If 0.75 Ii 3 2 Ii

1 2 Kr = Iω 2 2 1 L 042 (part 2 of 2) 10.0 points = I 2  I  Consider the following statements for the fig- ure skater: L2 = I. Angular momentum was conserved. 2 I II. Mechanical energy was conserved. III. The kinetic energy changed because of Figure Skater energy due to friction. 041 (part 1 of 2) 10.0 points IV. Her rotation rate changed in response A figure skater rotating on one spot with both to a torque exerted by pulling in her arms and arms and one leg extended has moment of leg. inertia I . She then pulls in her arms and the i Which is the correct combination of state- extended leg, reducing her moment of inertia ments? to 0.75 Ii. What is the ratio of her final to initial ki- 1. I, II, IV netic energy? 2. I and II 1. 4/3 correct 3. II 2. 1 4. I correct 3. 9/16 5. I, II, III 4. 3/8

5. 1/2 Explanation: (I) The angular momentum was conserved 6. 16/9 since there was no external torque acting on the skater. 7. 8/3 (II) From part 1, we found that the kinetic energy increased. and the potential energy 8. 3/4 didn’t change, so the total mechanical energy increased. (II) is wrong. 9. 2 (III) We found that the total kinetic energy Explanation: increased, and therefore, we can infer that The angular momentum was conserved, so there was no energy dissipation. (IV) The force pulling her arms in would be Li = Lf perpendicular to her rotation, and therefore, 1 2 exert no torque. Since L = Iω and KEr = Iω , 2 Ergo, the only correct statement is (I). L2 KEr = 2 I Figure Skater Spin Version001–AngularMomentum–smith–(3102F16B1) 16 043 (part 1 of 2) 10.0 points A figure skater on ice spins on one foot. She 4. When she pulls in her arms, the work pulls in her arms and her rotational speed she performs on them turns into increased increases. rotational kinetic energy. correct Choose the best statement below: 5. When she pulls in her arms, her rotational 1. Her angular speed increases because air kinetic energy is conserved and therefore stays friction is reduced as her arms come in. the same.

2. Her angular speed increases because her 6. When she pulls in her arms, her angu- angular momentum increases. lar momentum decreases so as to conserve energy. 3. Her angular speed increases because her Explanation: potential energy increases as her arms come The kinetic energy of the figure skater, in. 1 2 1 E = Iω = Lω. 4. Her angular speed increases because her 2 2 angular momentum is the same but her mo- ment of inertia decreases. correct Since ω increases after she pulls in her arms as mentioned above, the total kinetic energy 5. Her angular speed increases because by increases. This additional energy comes from pulling in her arms she creates a net torque in the figure skater, namely she has to perform the direction of rotation. some work to achieve this.

6. Her angular speed increases due to a net Ice Skater pulls arms in torque exerted by her surroundings. 045 10.0 points An ice skater with rotational inertia I0 is spin- Explanation: ning with angular velocity ω0. She pulls her The initial angular momentum of the figure arms in, decreasing her rotational inertia to skater is I ω . After she pulls in her arms, the i i I0/3. Her angular velocity becomes: angular momentum of her is If ωf . Note that I ω . i i f f f i 3. ω0

044 (part 2 of 2) 10.0 points 4. 3ω0 correct And again, choose the best statement below: 5. √3ω0 1. When she pulls in her arms, her rota- tional potential energy increases as her arms Explanation: approach the center. Angular momentum is given by L = Iω, and is conserved on ice, so: 2. When she pulls in her arms, her moment of inertia is conserved. I0ω0 = Iω 1 3. When she pulls in her arms, her rotational = I0 ω kinetic energy must decrease because of the 3  decrease in her moment of inertia. ω =3ω0 Version001–AngularMomentum–smith–(3102F16B1) 17 From conservation of the angular momentum Mimic a Spinning Skater 01 it follows that 046 (part 1 of 2) 10.0 points A student sits on a rotating stool holding two Ii ωi = If ωf 2 kg objects. When his arms are extended Ii ωf = ωi horizontally, the objects are 1 m from the axis If of rotation, and he rotates with angular speed 2 10 kgm of 0.61 rad/sec. The moment of inertia of the = (0.61 rad/sec) 2 2 student plus the stool is 6 kg m and is as- 6.2916 kgm  sumed to be constant. The student then pulls = 0.969547 rad/s . the objects horizontally to a radius 0.27 m from the rotation axis. ω ω i f 047 (part 2 of 2) 10.0 points Calculate the change in kinetic energy of the system.

Correct answer: 1.09662 J. Explanation:

∆K = Kf Ki (a) (b) − Calculate the final angular speed of the 1 2 1 2 = If ωf Ii ωi student. 2 − 2 1 2 2 = 6.2916 kgm (0.969547 rad/s) 2 Correct answer: 0.969547 rad/s.  1 2 2 Explanation: 10 kgm (0.61 rad/sec) −2 = (2.95712 J) (1.8605 J) Let : m =2kg , − = 1.09662 J . R =1m , r =0.27 m , ω =0.61 rad/sec , and 2 Is =6kgm .

L~ = Constant. X 1 2 Krot = Iω 2 The initial moment of inertia of the system is 2 Ii = Is +2 mR 2 2 = (6 kgm )+2(2 kg)(1 m) 2 = 10 kgm . The final moment of inertia of the system is 2 If = Is +2 mr 2 2 = 6 kgm +2(2 kg)(0.27 m) 2 =6.2916 kgm .