Explanation 1) Concept: Using the equation for position, we can find the velocity in terms of t by differentiating it. For second particle, using the equation for acceleration, we can find the equation for the velocity for second particle, by integrating this. Finally, we can equate the two equations to find the time. 2) Formula: i) L = m r × v ii) τ = r × F 3) Given: i) m = 30 k g ii) v → = - 2.0 t 3 i - m / s 4) Calculations: (a) We have, L = m r × v As the toy is moving along x axis and the velocity vector is also along the x axis, so the cross product r → × v → = 0 . Hence, the angular momentum is also zero. (b) We have, v → = - 2.0 t 3 i ^ m / s So, the acceleration vector can be calculated as, a → = d v → d t = - 6.0 t 2 i ^ m s 2 From this equation we can get that the acceleration vector is also along the x axis. So, r → × a → = 0 . Hence, r → × F → = 0 . So, the torque is zero. (c) For this case, we need to calculate position vector first. r ' → = r → - r → 0 Where, r 0 → = 2.0 i + 5.0 j Now, L = m r ' × v L = m r - r 0 × v L = m ( - r 0 × v ) L = 3.0 - 2.0 i ^ - 5.0 j ^ × 2.0 t 3 L = - 30 t 3 k ^ kg m/s 2 . (d) We have, τ → = r → × F → We have, F → = m a → So, τ → = m r ' → × a → = - m r → 0 × a → = - 3

Fundamentals Of Physics 11e Student Solutions Manual

11th Edition

ISBN: 9781119455127

Author: David Halliday

Publisher: WILEY

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Chapter 11, Problem 73P

To determine

To find:

a) Angular Momentum

b) Torque

c) Angular momentum at point (2.0, 5.0)

d) Torque at point (2.0, 5.0)

e) Angular momentum at point (2.0,-5.0)

f) Torque at point (2.0,-5.0)

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Chapter 11, Problem 72P

Chapter 11, Problem 74P

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Chapter 11 Solutions

Fundamentals Of Physics 11e Student Solutions Manual

Ch. 11 - Figure 11-23 shows three particles of the same...Ch. 11 - Figure 11-24 shows two particles A and B at xyz...Ch. 11 - Prob. 3Q Ch. 11 - The position vector r of a particle relative to a...Ch. 11 - In Fig. 11-26, three forces of the same magnitude...Ch. 11 - The angular momenta t of a particle in four...Ch. 11 - A rhinoceros beetle rides the rim of a horizontal...Ch. 11 - Figure 11-27 shows an overhead view of a...Ch. 11 - Figure 11-38 gives the angular momentum magnitude...Ch. 11 - Figure 11-29 shows a particle moving at constant...

Ch. 11 - A cannonball and a marble roll smoothly from rest...Ch. 11 - Prob. 12Q Ch. 11 - A car travels at 80 km/h on a level road in the...Ch. 11 - An automobile traveling at 80.0 km/h has tires of...Ch. 11 - Prob. 3P Ch. 11 - A uniform solid sphere rolls down an incline. a...Ch. 11 - ILW A 1000 kg car has four 10 kg wheels. When the...Ch. 11 - Figure 11-30 gives the speed v versus time t for a...Ch. 11 - ILW In Fig. 11-31, a solid cylinder of radius 10cm...Ch. 11 - Figure 11-32 shows the potential energy Ux of a...Ch. 11 - GO In Fig. 11-33, a solid ball rolls smoothly from...Ch. 11 - A hollow sphere of radius 0.15 m, with rotational...Ch. 11 - In Fig. 11-34, a constant horizontal force Fapp of...Ch. 11 - GO In Fig. 11-35, a solid brass ball of mass 0.280...Ch. 11 - GO Nonuniform ball. In Fig. 11-36, a ball of mass...Ch. 11 - In Fig. 11-37, a small, solid, uniform ball is to...Ch. 11 - GO A bowler throws a bowling ball of radius R = 11...Ch. 11 - GO Nonuniform cylindrical object. In Fig. 11-39, a...Ch. 11 - SSM A yo-yo has a rotational inertia of 950 gcm2...Ch. 11 - Prob. 18P Ch. 11 - In unit-vector notation, what is the net torque...Ch. 11 - A plum is located at coordinates 2.0 m, 0, 4.0 m....Ch. 11 - In unit-vector notation, what is the torque about...Ch. 11 - A particle moves through an xyz coordinate system...Ch. 11 - Force F=(2.0N)i(3.0N)k acts on a pebble with...Ch. 11 - In unit-vector notation, what is the torque about...Ch. 11 - SSM Force F=(8.0N)i+(6.0N)j acts on a particle...Ch. 11 - At the instant of Fig. 11-40, a 2.0 kg particle P...Ch. 11 - SSM At one instant, force F=4.0N acts on a 0.25 kg...Ch. 11 - A 2.0 kg particle-like object moves in a plant...Ch. 11 - ILW In the instant of Fig, 11-41, two particles...Ch. 11 - At the instant the displacement of a 2.00 kg...Ch. 11 - In Fig. 11-42, a 0.400 kg ball is shot directly...Ch. 11 - A particle is acted on by two torques about the...Ch. 11 - Prob. 33P Ch. 11 - A particle is to move in an xy plane, clockwise...Ch. 11 - Prob. 35P Ch. 11 - Prob. 36P Ch. 11 - GO In Fig. 11-44, three particles of mass m = 23 g...Ch. 11 - A sanding disk with rotational inertia 1.2 103...Ch. 11 - SSM The angular momentum of a flywheel having a...Ch. 11 - A disk with a rotational inertia of 7.00 kgm2...Ch. 11 - GO Figure 11-45 shows a rigid structure consisting...Ch. 11 - Figure 11-46 gives the torque that acts on an...Ch. 11 - Prob. 43P Ch. 11 - A Texas cockroach of mass 0.17 kg runs...Ch. 11 - SSM WWW A man stands on a platform that is...Ch. 11 - The rotational inertia of a collapsing spinning...Ch. 11 - SSM A track is mounted on a large wheel that is...Ch. 11 - A Texas cockroach walks from the center of a...Ch. 11 - Two disks are mounted like a merry-go-round on...Ch. 11 - The rotor of an electric motor has rotational...Ch. 11 - SSM ILW A wheel is rotating freely at angular...Ch. 11 - GO A cockroach of mass m lies on the rim of a...Ch. 11 - GO In Fig. 11-50 an overhead view, a uniform thin...Ch. 11 - GO Figure 11-51 shows an overhead view of a ring...Ch. 11 - A horizontal vinyl record of mass 0.10 kg and...Ch. 11 - In a long jump, an athlete leaves the ground with...Ch. 11 - A uniform disk of mass 10m and radius 3.0r can...Ch. 11 - A horizontal platform in the shape of a circular...Ch. 11 - Figure 11-52 is an overhead view of a thin uniform...Ch. 11 - In Fig. 11-53, a 1.0 g bullet is tired into a 0.50...Ch. 11 - The uniform rod length 0.60 m, mass 1.0 kg in Fig....Ch. 11 - GO During a jump to his partner, an aerialist is...Ch. 11 - GO In Fig. 11-56, a 30 kg child stands on the edge...Ch. 11 - A ballerina begins a tour jet Fig. 11-19a with...Ch. 11 - SSM WWW Two 2.00 kg balls are attached to the ends...Ch. 11 - Prob. 66P Ch. 11 - Prob. 67P Ch. 11 - Prob. 68P Ch. 11 - A certain gyroscope consists of a uniform disk...Ch. 11 - A uniform solid ball rolls smoothly along a floor,...Ch. 11 - SSM In Fig. 11-60, a constant horizontal force...Ch. 11 - A thin-walled pipe rolls along the floor. What is...Ch. 11 - Prob. 73P Ch. 11 - Prob. 74P Ch. 11 - Prob. 75P Ch. 11 - A uniform block of granite in the shape of a book...Ch. 11 - SSM Two particles, each of mass 2.90 104 kg and...Ch. 11 - A wheel of radius 0.250 m, moving initially al...Ch. 11 - Wheels A and B in Fig. 11-61 are connected by a...Ch. 11 - Prob. 80P Ch. 11 - SSM A uniform wheel of mass 10.0 kg and radius...Ch. 11 - A uniform rod rotates in a horizontal plane about...Ch. 11 - A solid sphere of weight 36.0 N rolls up an...Ch. 11 - Suppose that the yo-yo in Problem 17, instead of...Ch. 11 - A girl of mass M stands on the rim of a...Ch. 11 - Prob. 86P

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a) Angular Momentum b) Torque c) Angular momentum at point ( 2.0 , 5.0 ) d) Torque at point ( 2.0 , 5.0 ) e) Angular momentum at point ( 2.0 , - 5.0 ) f) Torque at point ( 2.0 , - 5.0 )

Solution Summary: The author explains how to find Angular Momentum, Torque, and Time using the equation for position and acceleration.

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To find:

a) Angular Momentum

b) Torque

c) Angular momentum at point (2.0, 5.0)

d) Torque at point (2.0, 5.0)

e) Angular momentum at point (2.0,-5.0)

f) Torque at point (2.0,-5.0)

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Chapter 11 Solutions

a) Angular Momentum b) Torque c) Angular momentum at point ( 2.0 , 5.0 ) d) Torque at point ( 2.0 , 5.0 ) e) Angular momentum at point ( 2.0 , - 5.0 ) f) Torque at point ( 2.0 , - 5.0 )

Solution Summary: The author explains how to find Angular Momentum, Torque, and Time using the equation for position and acceleration.

Concept explainers

Videos

To find: a) Angular Momentum b) Torque c) Angular momentum at point (2.0, 5.0) d) Torque at point (2.0, 5.0) e) Angular momentum at point (2.0,-5.0) f) Torque at point (2.0,-5.0)

Want to see the full answer?

Chapter 11 Solutions

To find:

a) Angular Momentum

b) Torque

c) Angular momentum at point (2.0, 5.0)

d) Torque at point (2.0, 5.0)

e) Angular momentum at point (2.0,-5.0)

f) Torque at point (2.0,-5.0)