Angular Momentum

Version 001 – Angular Momentum This print-out should have 47 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Airplane Momentum 001 (part 1 of 2) 10.0 points An airplane of mass 18691 kg flies level to the ground at an altitude of 14 km with a constant speed of 178 m/s relative to the Earth. What is the magnitude of the airplane’s angular momentum relative to a ground observer directly below the airplane in kg ·m2/s? Correct answer: 4.6578 × 1010 kg · m2 /s. Explanation: Since the observer is directly below the airplane, L = hmv 002 (part 2 of 2) 10.0 points Does this value change as the airplane continues its motion along a straight line? 1. Yes. L decreases as the airplane moves. 2. Yes. L increases as the airplane moves. 1 axis perpendicular to the circle and through its center? 2 ~ = 24 kg · m correct 1. kLk s 2 ~ = 12 m 2. kLk s ~ =9 N·m 3. kLk kg 2 ~ = 13.5 kg · m 4. kLk s2 ~ = 18 N · m 5. kLk kg Explanation: The angular momentum is ~ = m v r = (2 kg) (3 m/s) (4 m) kLk = 24 kg · m2 /s . Conical Pendulum 04 004 10.0 points A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. 3. Yes. L changes in a random pattern as the airplane moves. 3. 2 m 5. No. L = constant. correct Explanation: L = constant since the perpendicular distance from the line of flight to Earth’s surface doesn’t change. 003 10.0 points A 2 kg object moves in a circle of radius 4 m at a constant speed of 3 m/s. A net force of 4.5 N acts on the object. What is the magnitude of the angular mo~ of the object with respect to an mentum kLk ◦ 9.8 m/s2 29 4. Yes. L changes with certain period as the airplane moves. v 9 kg Calculate the magnitude of the angular momentum of the bob about a vertical axis through the supporting point. The acceleration of gravity is 9.8 m/s2 . Correct answer: 40.5335 kg · m2 /s. Explanation: Version 001 – Angular Momentum Let : ℓ = 3.2 m , θ = 29◦ , g = 9.8 m/s2 , m = 9 kg . and Consider the free body diagram. 2 Decelerated Grinding Wheel 005 (part 1 of 2) 10.0 points The motor driving a grinding wheel with a rotational inertia of 0.8 kg m2 is switched off when the wheel has a rotational speed of 39 rad/s. After 6.7 s, the wheel has slowed down to 31.2 rad/s. What is the absolute value of the constant torque exerted by friction to slow the wheel down? Correct answer: 0.931343 N m. θ Explanation: We have T τ ∆t = ∆L = ∆(I ω) , so that mg Newton’s second law in the vertical and horizontal projections, respectively, gives T cos θ − m g T cos θ 2 T sin θ − m ω ℓ sin θ T =0 = m g and =0 = m ω2 ℓ , where the radius of the orbit is R = ℓ sin θ. Dividing, mg m ω2 ℓ r g ω= . ℓ cos θ cos θ = ~r and ~v are perpendicular, where r = ℓ sin θ , so the angular momentum L = m ~r ×~v will be L = m ω (ℓ sin θ)2 r g ℓ3 = m sin2 θ cos θ 2 = (9 kg) sin 29◦ r (9.8 m/s2 ) (3.2 m)3 × cos 29◦ = 40.5335 kg · m2 /s . I |ω1 − ω0 | ∆t1 (0.8 kg m2 ) (39 rad/s − 31.2 rad/s) = 6.7 s = 0.931343 N m . |τ | = 006 (part 2 of 2) 10.0 points If this torque remains constant, how long after the motor is switched off will the wheel come to rest? Correct answer: 33.5 s. Explanation: When the wheel comes to rest, its angular speed is ω2 = 0; hence I (ω0 − ω2 ) τ I ω0 = |τ | (0.8 kg m2 ) (39 rad/s) = (0.931343 N m) = 33.5 s . ∆t2 = Holt SF 08Rev 66 007 (part 1 of 2) 10.0 points a) Calculate the angular momentum of Earth Version 001 – Angular Momentum that arises from its spinning motion on its
2 axis IE = 0.331ME RE . Correct answer: 5.84082 × 1033 kg · m2 /s. Explanation: Basic Concept: Given:
L = Iω = 0.331M R2 ω R = 6.37 × 106 m M = 5.98 × 1024 kg ω = 1 rev/day 3 Merry Go Round and a Boy 009 10.0 points The sketch shows the top view of a merrygo-round which is rotating clockwise. A boy jumps on the merry-go-round in three different ways: I) from the left, II) from the top, and III) from the right. For all three cases, the boy lands on the same spot. II ω Solution: L = 0.331(5.98 × 1024 kg)   1 rev 6 2 · (6.37 × 10 m) · 1 day     2π rad 1 day 1h · 1 rev 24 h 3600 s 33 2 = 5.84082 × 10 kg · m /s 008 (part 2 of 2) 10.0 points b) Calculate the average angular momentum of Earth that arises from its orbital motion about the sun. Correct answer: 2.66465 × 1040 kg · m2 /s. Explanation: Basic Concept: Given: III I
L = Iω = M r 2 ω r = 1.496 × 1011 m ω = 1 rev/365.25 days Solution: L = (5.98 × 1024 kg)(1.496 × 1011 m)2    1 rev 2π rad · 365.25 day 1 rev    1 day 1h · 24 h 3600 s 40 = 2.66465 × 10 kg · m2 /s Compare the final angular momenta of the merry-go-round for the three cases. 1. LIII > LII > LI 2. LI = LII = LIII 3. LI = LIII > LII 4. LI > LII > LIII correct 5. LII > LI = LIII Explanation: With respect to the center of the merrygo-round, the angular momentum of the boy is: I) clockwise, II) 0, and III) counterclockwise. The angular momentum of the merrygo-round is along the clockwise direction. By adding the two angular momenta in each case, we get LI > LII > LIII . Net L of point masses 010 10.0 points Two objects are moving in the x-y plane as shown. Version 001 – Angular Momentum The magnitude of their total angular momentum (about the origin O) is 4 counter-clockwise as the positive angular direction. 1. 8 kg · m2 /s ~ = m v0 R cos θ k̂ 1. L 2. 24 kg · m2 /s ~ = m v0 R sin θ k̂ 2. L 3. 32 kg · m2 /s ~ = m v0 R k̂ 3. L 4. 0 kg · m2 /s correct ~ = m v0 4. L 5. 96 kg · m2 /s ~ = 0 correct 5. L Explanation: The angular momentum of a particle is ~ = I~ω , where we remember that given by L the ω is positive for counterclockwise rotation in the x − y plane and negative for clockwise rotation. We can see above that the two particles are rotating in opposite senses. For a point particle, the moment of inertia is given by I = mr 2 . Consequently, we have: R k̂ 2 Explanation: ~ = ~r × ~p = 0 L since ~r = 0 . 012 (part 2 of 4) 10.0 points Using the origin as the pivot, find the angular momentum when the particle is at the highest point of the trajectory. m v0 2 sin θ cos2 θ Lnet = L1 + L2 2g = I 1 ω1 − I 2 ω2 m v0 3 ~ 2. L = + sin2 θ cos θ = (m1 r12 )ω1 − (m2 r22 )ω2 2g 2 = (1 kg)(2 m)2 (4 rad/s) − (4 kg)(1 m)2 (4 rad/s) ~ = − m v0 sin2 θ cos θ 3. L 2g = 0 kg · m2 /s 2 ~ = + m v0 sin2 θ cos θ 4. L 2g Projectile Analysis 01 3 011 (part 1 of 4) 10.0 points ~ = − m v0 sin θ cos2 θ 5. L A particle of mass m moving in the gravi2g tational field of the Earth is launched with m v0 2 ~ 6. L = − sin θ cos2 θ an initial velocity vo at an angle θ with the 2g horizontal. 3 ~ = + m v0 sin θ cos2 θ 7. L C y 2g vh 3 ~ = − m v0 sin2 θ cos θ 8. L 2g h ~ is in the ̂ direction. θ 9. L A x O R ~ is in the ı̂ direction. 10. L k̂ k̂ k̂ k̂ k̂ k̂ k̂ k̂ correct vo ~ =+ 1. L vR Using the origin as the pivot, find the angular momentum when it is at the origin. Use Explanation: Denote the position vector from the origin to the highest point of the trajectory as ~rh . Version 001 – Angular Momentum At the highest point ~vh = ~vx = vo cos θ ı̂ and ~ = ~rh × ~ph = ~rh × m~vh L so that ~ = m |~vh | |~rh| sin θh . |L| The maximum height is h = |~rh | sin θh = vo2 sin2 θ 2g and since |~vh | = |~v0x | = vx , the magnitude of the angular momentum is ~ = m h vx = m |L| = vo2 sin2 θ (vo cos θ) 2g m vo3 sin2 θ cos θ . 2g The direction is −k̂ by the right hand rule of the cross product. 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 the ground. 2 m v0 3 sin2 θ cos θ g 2 ~ = − 2 m v0 sin θ cos2 θ 2. L g 2 ~ = − 2 m v0 sin2 θ cos θ 3. L g 3 ~ = − 2 m v0 sin θ cos2 θ 4. L g ~ is in the ı̂ direction. 5. L ~ =− 1. L k̂ correct Explanation: Consider the motion to reach the maximum height: vf = 0 = vo sin θ − g t1 vo sin θ t1 = , g so the range of the projectile is R = (vo cos θ) 2 t1 = 2 vo2 sin θ0 cos θ0 , g and the angular momentum at that position is 2 m vo3 L = m vo sin θ R = sin2 θ cos θ . g Using the right hand rule, the direction should be −k̂ . 014 (part 4 of 4) 10.0 points What torque causes its angular momentum to change? 1. The torque from the initial force required to shoot the particle. k̂ k̂ k̂ 2 ~ = + 2 m v0 sin2 θ cos θ k̂ 6. L g 2 ~ = + 2 m v0 sin θ cos2 θ k̂ 7. L g 3 ~ = + 2 m v0 sin2 θ cos θ k̂ 8. L g ~ is in the ̂ direction. 9. L ~ =+ 10. L 5 2 m v0 3 sin θ cos2 θ k̂ g 2. The torque from the centripetal acceleration. 3. The torque due to the initial velocity. 4. The torque due to the particle’s impulse. 5. The torque due to gravity. correct Explanation: Gravity is the only force acting on the particle. The change in angular momentum is negative (going from zero to negative values) because the torque of the gravity force is negative (−k̂ direction) as you can see from the ~ ×F ~. right hand rule for cross product: ~τ = R Time Dependent Momentum 015 10.0 points Version 001 – Angular Momentum – smith – (3102F16B1) The position vector of a particle of mass 2 kg is given as a function of time by ~r = (1 m) ı̂ + (5 m/s) t ̂ . Determine the magnitude of the angular momentum of the particle with respect to the origin at time 3 s . Let : m = 5 kg , v = 2.5 m/s , r = 3 m. and The angular momemntum is = 37.5 kg · m2 /s . Explanation: and Basic Concepts: ~ = ~r × ~p L Solution: ∂ ~r ~v = = vy ∂t = (5 m/s) ̂ . ~ = ~r × ~p L = m ~r × ~v ∂ ~r = m ~r × ∂t = m (~x ı̂ + ~y ̂) × ~v ̂ = m x vy k̂ = (2 kg) (1 m) (5 m/s) k̂ = 10 kg m2 /s k̂ , and is constant as a function of time since ̂ × ̂ = 0 . Tipler PSE5 10 44 016 (part 1 of 3) 10.0 points A 5 kg particle moves at a constant speed of 2.5 m/s around a circle of radius 3 m. What is its angular momentum about the center of the circle? Correct answer: 37.5 kg · m2 /s. Explanation: L = m v r = (5 kg) (2.5 m/s) (3 m) Correct answer: 10 kg m2 /s. Let : ~r = x ı̂ + y ̂ , x = 1 m, y = vy t = (5 m/s) t , t = 3 s. 6 017 (part 2 of 3) 10.0 points What is its moment of inertia about an axis through the center of the circle and perpendicular to the plane of the motion? Correct answer: 45 kg · m2 . Explanation: The moment of inertia is I = m r 2 = (5 kg)(3 m)2 = 45 kg · m2 . 018 (part 3 of 3) 10.0 points What is the angular speed of the particle? Correct answer: 0.833333 rad/s. Explanation: The angular speed is L 37.5 kg · m2 /s ω= = I 45 kg · m2 = 0.833333 rad/s . Tipler PSE5 10 79 02 019 (part 1 of 2) 10.0 points A particle of mass 5 kg moves with velocity ~v = (2 m/s)ı̂ along the line z = 0 m, y = 1 m. Find the angular momentum relative to the origin when the particle is at x = 16 m, y = 1 m. 1. −(1.0 kg · m2 /s) ĵ 2. −(1.0 kg · m2 /s) k̂ Version 001 – Angular Momentum 3. −(2.0 kg · m2 /s) k̂ 4. −(2.0 kg · m2 /s) ĵ 5. −(10 kg · m2 /s)ĵ 6. (47.7 kg · m2 /s) k̂ 7. (47.7 kg · m2 /s) ĵ 8. −(10 kg · m2 /s)k̂ correct Explanation: Let : m = 5 kg , ~v = 2 m/s , and ~r = (16 m)ı̂ + (1 m) ̂ . The momentum is ~p = m~v = (5 kg) (2 m/s)ı̂ = (10 kg · m/s)ı̂ . The angular momentum is ~ = ~r × ~p L ~ = [(16 m)ı̂ + (1 m) ̂] × (10 kg · m/s)ı̂ L = (10 kg · m2 /s) (̂ × ı̂) = −(10 kg · m2 /s)k̂ . 020 (part 2 of 2) 10.0 points ~ = (−8 N)ı̂ is applied to the particle. A force F Find the torque relative to the origin due to this force. 1. (20.0 N · m) ĵ 2. (−8 N · m) k̂ 3. (10.0 N · m) ĵ 4. (8 N · m) k̂ correct 5. (−8 N · m) ĵ 6. (8 N · m) ĵ 7. (10.0 N · m) k̂ 7 8. (20.0 N · m) k̂ Explanation: ~ = (−8 N)ı̂ . Let : F The torque due to the force is ~ ~τ = ~r × F = [(16 m)ı̂ + (1 m) ̂] × (−8 N)ı̂ = −(8 N · m) (̂ × ı̂) = (8 N · m)k̂ . AP M 1998 MC 32 33 021 (part 1 of 2) 10.0 points A wheel with rotational inertia I is mounted on a fixed, frictionless axle. The angular speed ω of the wheel is increased from zero to ωf in a time interval T . What is the average net torque τ on the wheel during this time interval? I ωf correct T I ωf 2. τ = 2 T ωf 3. τ = 2 T I ωf2 4. τ = T ωf 5. τ = T Explanation: The change of angular momentum of the wheel is ∆L = I ωf , 1. τ = so the average net torque on the wheel during this time interval is τ= I ωf L = . T T 022 (part 2 of 2) 10.0 points What is the average power input to the wheel during this time interval? Version 001 – Angular Momentum I 2 ωf 1. P = 2T2 I ωf2 2. P = correct 2T I ωf 3. P = 2T I ωf2 4. P = 2T2 I 2 ωf2 5. P = 2T2 Explanation: The change in the energy of the wheel is 1 I ωf2 in the time interval T , so the average 2 power input to the wheel is I ωf2 1 2 1 I ωf = . 2 T 2T Rotating Meter Stick 023 (part 1 of 2) 10.0 points A particle of mass 0.31 kg is attached to the 100-cm mark of a meter stick of mass 0.35 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 1.1 rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 50-cm mark. 8 and the angular momentum is L1 = I1 ω = (0.106667 kg m2 )(1.1 rad/s) = 0.117333 kg m2 /s . 024 (part 2 of 2) 10.0 points Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 0-cm mark. Correct answer: 0.469333 kg m2 /s. Explanation: The moment of inertia of the stick in this 1 case is Is = ms (1 m)2 and of the particle 3 attached to the stick Ip = mp (1 m)2 . The total moment of inertia of the system is I2 = Is + Ip 1 = (0.35 kg) (1 m)2 + (0.31 kg (1 m)2 3 = 0.426667 kg m2 , and the angular momentum is L2 = I2 ω = (0.426667 kg m2 )(1.1 rad/s) = 0.469333 kg m2 /s . Correct answer: 0.117333 kg m2 /s. Explanation: L = Iω The moment of inertia of the stick is 1 ms (1 m)2 and of the particle atIs = 12  2 1 tached to the stick Ip = mp m . The 2 total moment of inertia of the system is I1 = Is + Ip 1 = (0.35 kg) (1 m)2 + (0.31 kg) 12 = 0.106667 kg m2 ,  1 m 2 2 Rotor of a Generator 025 10.0 points A net torque of magnitude 200 N m is exerted on the rotor of an electric generator for 32 s. What is the magnitude of the angular momentum of the rotor at the end of 32 s if it was initially at rest? Correct answer: 6400 kg m2 /s. Explanation: We have: ∆L = τnet ∆t Version 001 – Angular Momentum Since the rotor was initially at rest, Li = 0, Lf = (200 N m)(32 s) 9 027 (part 2 of 2) 10.0 points What is the total angular momentum of the system about the z-axis relative to point B along y axis if d2 = 25 m? = 6400 kg m2 /s Total Angular Momentum 026 (part 1 of 2) 10.0 points Two particles move in opposite directions along a straight line. Particle 1 of mass m1 = 22 kg at x1 = 16 m moves with a 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 left). Given: Counter-clockwise is the positive angular direction. Correct answer: −21050 kg m2 /s. Explanation: Since at point B r sin θ = d2 , we find B l B = d2 p = d2 (m1 v1 + m2 v2 ) = (25 m) [(22 kg)(48 m/s) + (73 kg) (−26 m/s)] = −21050 kg m2 /s . d2 v2 v1 m2 m1 x2 x1 d1 A y x What is the total angular momentum of the system about the z-axis relative to point A along y axis if d1 = 13 m? Correct answer: 10946 kg m2 /s. Explanation: The angular momentum along z axis is given by L = r p sin θ , where p is the total linear momentum. We find that p = m1 v1 + m2 v2 , where v2 is negative. Since at point A r sin θ = −d1 , we find l A = d1 p = −d1 (m1 v1 + m2 v2 ) = (−13 m) [(22 kg)(48 m/s) + (73 kg) (−26 m/s)] = 10946 kg m2 /s . Bullet Rotates a Rod 01 028 (part 1 of 2) 10.0 points A wooden block of mass M hangs from a rigid rod of length ℓ having negligible mass. The rod is pivoted at its upper end. A bullet of mass m traveling horizontally and normal to the rod with speed v hits the block and gets embedded in it. ℓ v m M What is the angular momentum L of the block-bullet system, with respect to the pivot point immediately after the collision? 1. L = m v ℓ correct   Mm 2. L = vℓ M +m Version 001 – Angular Momentum 3. L = (M − m) v ℓ 4. L = (m + M ) v ℓ 5. L = M v ℓ Explanation: X X ~ = 0. If ~τext = 0, then L The net angular momentum of the system conserves, and Li = Lf = L = m v ℓ . 029 (part 2 of 2) 10.0 points Kf What is the fraction (the final kinetic Ki energy compared to the initial kinetic energy) in the collision? Kf 2m = Ki m+M Kf M 2. = Ki M +m Kf M 3. = Ki M −m Kf m 4. = correct Ki m+M Kf m 5. = Ki M −m Explanation: By conservation of the angular momentum 1. Li = Lf = L m v ℓ = (m + M ) vf ℓ   m vf = v m+M 1 1 m v 2 and Kf = I ωf2 where 2 2 vf 2 I = (M + m) ℓ and ωf = , so ℓ 1 Kf = (M + m) vf2 and 2 Ki = 1 m2 v2 Kf m 2 M +m = = . 1 Ki M +m 2 mv 2 10 Child on a Merrygoround 030 10.0 points A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center? (neglect friction in the bearing) 1. neither mechanical energy nor angular momentum 2. only mechanical energy 3. mechanical energy and angular momentum 4. only angular momentum correct Explanation: The forces external to the system are not exerting any torque, so the angular momentum is conserved. On the other hand, the friction force acting on the child is doing work, because she is moving towards the center (which is the direction of that force). Child on a MerryGoRound 02 v1 031 10.0 points A child of mass 52.8 kg sits on the edge of a merry-go-round with radius 1.7 m and moment of inertia 134.281 kg m2 . The merrygo-round rotates with an angular velocity of 1.9 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.612 m from the center. Now what is the angular velocity of the merry-go-round? Correct answer: 3.53803 rad/s. Explanation: When the child moves inward, the moment of inertia of the system iM GR + ichild (the merrygo-round plus the child) changes. Therefore, to conserve angular momentum, the angular Version 001 – Angular Momentum velocity of the system must change. Specifically: Linit = Lf inal (ichild + iM GR ) ω = (ichild,2 + iM GR ) ω2 The moment of inertia of the child is m r 2 . Therefore m r 2 + iM GR ω2 = ω. m r2 2 + iM GR Child on Merry Go Round 032 10.0 points A playground merry-go-round has a radius of 3 m and a rotational inertia of 600 kg · m2 /s. It is initially spinning at 0.8 rad/s when a 20 kg child crawls from the center to the rim. When the child reaches the rim the angular velocity of the merry-go-round is: 033 10.0 points A thin uniform cylindrical turntable of radius 1.8 m and mass 23 kg rotates in a horizontal plane with an initial angular speed of 8.5 rad/s. The turntable bearing is frictionless. A clump of clay of mass 5.2 kg is dropped onto the turntable and sticks at a point 0.96 m from the point of rotation. Find the angular speed of the clay and turntable. Correct answer: 7.53133 rad/s. Explanation: Let : R = 1.8 m , x = 0.96 m , M = 23 kg , ω = 8.5 rad/s , m = 5.2 kg . 1. 0.89 rad/s 3. 0.8 rad/s X 4. 0.62 rad/s correct Explanation: We conserve angular momentum, realizing that the child’s moment of inertia is given by mr 2 : and Basic Concepts: 2. 1.1 rad/s 5. 0.73 rad/s 11 ~ = d ~τext L dt ∆K = Kf − Ki = Q . From conservation of the angular momentum it follows that Li = Lf or I i ω = I f ωf , Lf = L0 where I f ωf = I 0 ω0 1

2 Ii = M R 2 I + mrf ωf = I + mr02 ω0 2   1 I = (23 kg) (1.8 m)2 ωf = ω0 2 2 I + mr   2 = 37.26 kg m2 . 600 kg · m /s = 600 kg · m2 /s + (20 kg)(3 m)2 Similarly, × (0.8 rad/s) = 0.62 rad/s I f = I i + m x2 Clay on a Turntable 01 = (37.26 kg m2 ) + (5.2 kg) (0.96 m)2 = 42.0523 kg m2 . Version 001 – Angular Momentum 12 Finally, 5. II and V only Ii ωf = ω If (37.26 kg m2 ) = (8.5 rad/s) (42.0523 kg m2 ) = 7.53133 rad/s . 6. II, III and IV only 7. III, IV and V only 8. I, II and III only 9. II, III, IV and V only Clay Rotates a Rod 01a 034 10.0 points 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 rodclay system to spin. Viewed from above the scheme is as follows: v m ℓ 0 ω ω 2m (a) before (b) during (c) after After the collisions the rod and clay system has an angular velocity ω about the pivot. Which quantity/quantities I) total mechanical energy II) total linear momentum III) total angular momentum with respect to pivot point O IV) total gravitational potential energy V) total kinetic energy is/are conserved in this process? 1. All of these 2. None of these 3. III and IV only correct 4. II and III only 10. I only Explanation: V) Total KE is not conserved. One can Kf Ii verify that = < 1. The inequality Ki If here is due to the fact that Ii is the moment of inertia of the clay and If is the moment of inertia of the clay+rod. IV) Since the motion is in a horizontal plane there is no change in the potential energy. IV) and V) together imply that the total mechanical energy (KE + PE) is not conserved. II) Linear momentum is not conserved by inspection. Apparently the linear momentum is altered by the presence of the pivot. The pivot provides an external force to the rod+clay system. III) Angular momentum is conserved, since there is no external torque applied to the rod+clay system. Clay Rotates a Rod 01 035 (part 1 of 2) 10.0 points A uniform rod, supported and pivoted at its midpoint, but initially at rest, has a mass 2 m and a length ℓ. A piece of clay with mass m and velocity v0 hits one end of the rod, gets stuck and causes the clay-rod system to spin about the pivot point O at the center of the rod in a horizontal plane. Viewed from above the scheme is Version 001 – Angular Momentum plus the moment of inertia of the clay:  2 1 ℓ 2 If = Irod + Iclay = 2mℓ + m 12 2 5 = m ℓ2 . 12 v0 m ℓ 0 ω ω (a) 2 m Before (b) During (c) After With respect to the pivot point O, what is the magnitude of the initial angular momentum Li of the piece of clay and the final moment of inertia If of the clay-rod system? After the collisions the clay-rod system has an angular velocity ω about the pivot. 1. Li = m v ℓ , 2 2. Li = m v ℓ , 3. Li = m v ℓ , 4. Li = m v ℓ , ℓ , 2 ℓ 6. Li = m v , 2 ℓ 7. Li = m v , 2 5. Li = m v 8. Li = m v ℓ , 13 7 m ℓ2 12 3 If = m ℓ2 12 5 If = m ℓ2 12 8 If = m ℓ2 12 4 If = m ℓ2 12 5 If = m ℓ2 correct 12 8 If = m ℓ2 12 4 If = m ℓ2 12 If = Explanation: Since the total external torque acting on the clay-rod system is zero, the total angular momentum is a constant of motion. The total initial angular momentum Li is simply the angular momentum of the clay, since the rod is at rest initially mvℓ Li = |~r × ~p| = m r v = . 2 The final moment of inertia If of the clayrod system is the moment of inertia of the rod 036 (part 2 of 2) 10.0 points What is the final angular speed ωf of the rod-clay system¿ 3 v 5 ℓ 6 2. ωf = v 5 5 v 3. ωf = 12 ℓ 4 4. ωf = v 6 12 v 5. ωf = 7 ℓ 5 v 6. ωf = 6 ℓ 12 7. ωf = v 7 6 v 8. ωf = correct 5 ℓ 6 v 9. ωf = 2 ℓ 12 v 10. ωf = 5 ℓ Explanation: The total angular momenta are the same: 1. ωf = Lf = Li mvℓ I f ωf = 2 5 m vℓ m ℓ 2 ωf = 12 2 6 v ωf = . 5 ℓ Concept 08 43 037 10.0 points You sit at the middle of a large turntable at an amusement park as it is set spinning on nearly frictionless bearings, and then allowed to spin freely. Version 001 – Angular Momentum When you crawl toward the edge of the turntable, does the rate of the rotation increase, decrease, or remain unchanged, and why? 1. increases; conservation of momentum 2. decreases; conservation of energy 3. increases; conservation of energy 4. decreases; conservation of momentum correct Explanation: The rotational inertia of the system (you and the rotating turntable) is least when you are at the rotational axis. As you move outward, the rotational inertia of the system increases. Applying conservation of angular momentum, as you move toward the outer rim, you increase the rotational inertia of the spinning system, thus decreasing the angular speed. Also, if you don’t slip as you move out, you exert a friction force on the turntable opposite to its direction of rotation, thereby also slowing it down. Concept 08 45 038 10.0 points Strictly speaking, as more and more skyscrapers are built on the surface of the Earth, does the day tend to become longer or shorter? And strictly speaking, does the falling of autumn leaves tend to lengthen or shorten the 24-hour day? What physical principle supports your answers? 1. lengthen; shorten; conservation of angular momentum correct 2. shorten; lengthen; conservation of inertia 3. lengthen; shorten; conservation of kinetic energy 4. shorten; lengthen; conservation of angular torque 14 Explanation: Applying conservation of angular momentum, as the radial distance of mass increases, the angular speed decreases. The mass of material used to construct skyscrapers is lifted, slightly increasing the radial distance from the Earth’s spin axis. This would tend to slightly decrease the earth’s rate of rotation, which in turn tends to lengthen days slightly. The opposite effect occurs for falling leaves, since their radial distance from the earth’s axis decreases. As a practical matter, these effects are entirely negligible! Conceptual 07 05 039 10.0 points Why does a helicopter have a tail rotor? 1. This keeps the helicopter from spinning out of control. correct 2. None of these 3. This makes the helicopter move faster. 4. The tail rotor is just for decoration. 5. This increases safety if any of the rotors breaks down. Explanation: Without the tail rotor, the body would spin in the opposite direction of the rotors. Conceptual Rotating KE 040 10.0 points How can the rotational kinetic energy Kr of a rigid body be written in terms of the magnitude L fo its angular momentum L and its rotational moment of inertia I? L2 correct 2I 1 2. Kr = I L 2 √ 3. Kr = 2 L I 1. Kr = 4. Kr = 1 I L2 2 Version 001 – Angular Momentum L I Explanation: For a symmetric rigid body, we know that ~ =I~ L ω . Putting this relation into the definition of angular rotational kinetic energy, 5. Kr = 1 I ω2 2  2 1 L = I 2 I 2 L = 2I 15 So, KEf = KEi L2f 2 If L2i 2 Ii = Ii Ii 4 = = If 0.75 Ii 3 Kr = Figure Skater 041 (part 1 of 2) 10.0 points A figure skater rotating on one spot with both arms and one leg extended has moment of inertia Ii . She then pulls in her arms and the extended leg, reducing her moment of inertia to 0.75 Ii. What is the ratio of her final to initial kinetic energy? 042 (part 2 of 2) 10.0 points Consider the following statements for the figure skater: I. Angular momentum was conserved. II. Mechanical energy was conserved. III. The kinetic energy changed because of energy dissipation due to friction. IV. Her rotation rate changed in response to a torque exerted by pulling in her arms and leg. Which is the correct combination of statements? 1. I, II, IV 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 6. 16/9 7. 8/3 8. 3/4 9. 2 Explanation: The angular momentum was conserved, so Li = Lf 1 I ω2, 2 L2 KEr = 2I Since L = I ω and KEr = Explanation: (I) The angular momentum was conserved since there was no external torque acting on the skater. (II) From part 1, we found that the kinetic energy increased. and the potential energy didn’t change, so the total mechanical energy increased. (II) is wrong. (III) We found that the total kinetic energy increased, and therefore, we can infer that there was no energy dissipation. (IV) The force pulling her arms in would be perpendicular to her rotation, and therefore, exert no torque. Ergo, the only correct statement is (I). Figure Skater Spin Version 001 – Angular Momentum 043 (part 1 of 2) 10.0 points A figure skater on ice spins on one foot. She pulls in her arms and her rotational speed increases. Choose the best statement below: 1. Her angular speed increases because air friction is reduced as her arms come in. 2. Her angular speed increases because her angular momentum increases. 3. Her angular speed increases because her potential energy increases as her arms come in. 4. Her angular speed increases because her angular momentum is the same but her moment of inertia decreases. correct 5. Her angular speed increases because by pulling in her arms she creates a net torque in the direction of rotation. 6. Her angular speed increases due to a net torque exerted by her surroundings. Explanation: The initial angular momentum of the figure skater is Ii ωi . After she pulls in her arms, the angular momentum of her is If ωf . Note that If < Ii because her arms now rotate closer to the rotation axis and reduce the moment of inertia. Since the net external torque is zero, angular momentum remains unchanged, and so Ii ωi = If ωf = L. Therefore, ωf > ωi . 044 (part 2 of 2) 10.0 points And again, choose the best statement below: 1. When she pulls in her arms, her rotational potential energy increases as her arms approach the center. 2. When she pulls in her arms, her moment of inertia is conserved. 3. When she pulls in her arms, her rotational kinetic energy must decrease because of the decrease in her moment of inertia. 16 4. When she pulls in her arms, the work she performs on them turns into increased rotational kinetic energy. correct 5. When she pulls in her arms, her rotational kinetic energy is conserved and therefore stays the same. 6. When she pulls in her arms, her angular momentum decreases so as to conserve energy. Explanation: The kinetic energy of the figure skater, E= 1 1 I ω2 = L ω . 2 2 Since ω increases after she pulls in her arms as mentioned above, the total kinetic energy increases. This additional energy comes from the figure skater, namely she has to perform some work to achieve this. Ice Skater pulls arms in 045 10.0 points An ice skater with rotational inertia I0 is spinning with angular velocity ω0 . She pulls her arms in, decreasing her rotational inertia to I0 /3. Her angular velocity becomes: 1. ω0 /3 √ 2. ω0 / 3 3. ω0 4. 3ω0 correct √ 5. 3ω0 Explanation: Angular momentum is given by L = Iω, and is conserved on ice, so: I0 ω0 = Iω   1 = I0 ω 3 ω = 3ω0 Version 001 – Angular Momentum Mimic a Spinning Skater 01 046 (part 1 of 2) 10.0 points A student sits on a rotating stool holding two 2 kg objects. When his arms are extended horizontally, the objects are 1 m from the axis of rotation, and he rotates with angular speed of 0.61 rad/sec. The moment of inertia of the student plus the stool is 6 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.27 m from the rotation axis. ωi ωf 17 From conservation of the angular momentum it follows that I i ωi = I f ωf Ii ωf = ωi If = (0.61 rad/sec)  10 kg m2 6.2916 kg m2  = 0.969547 rad/s . 047 (part 2 of 2) 10.0 points Calculate the change in kinetic energy of the system. Correct answer: 1.09662 J. Explanation: (a) (b) Calculate the final angular speed of the student. Correct answer: 0.969547 rad/s. Explanation: Let : m = 2 kg , R = 1 m, r = 0.27 m , ω = 0.61 rad/sec , Is = 6 kg m2 . X = 1.09662 J . and ~ = Constant. L 1 I ω2 2 The initial moment of inertia of the system is Krot = Ii = Is + 2 m R 2 = (6 kg m2 ) + 2 (2 kg) (1 m)2 = 10 kg m2 . The final moment of inertia of the system is If = Is + 2 m r 2
= 6 kg m2 + 2 (2 kg) (0.27 m)2 = 6.2916 kg m2 . ∆K = Kf − Ki 1 1 = If ωf2 − Ii ωi2 2 2
1 = 6.2916 kg m2 (0.969547 rad/s)2 2
1 − 10 kg m2 (0.61 rad/sec)2 2 = (2.95712 J) − (1.8605 J)