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 Momentum m2 2. L~ = 12 001 (part 1 of 2) 10.0 points k k s An airplane of mass 18691 kg flies level to the N m ground at an altitude of 14 km with a constant 3. L~ =9 · k k kg speed of 178 m/s relative to the Earth. 2 What is the magnitude of the airplane’s ~ kg m 4. L = 13.5 ·2 angular momentum 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 Pendulum 04 ues its motion 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 . 2m 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 gravity is 9.8m/s . at a constant speed of 3 m/s. A net force 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 inertia of 0.8 kgm is switched off when the wheel has a rotational speed 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 torque exerted by friction 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 = | − | Newton’s second 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 masses 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 moment 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 velocity v at an angle 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 position 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.