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Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
8
Rotational Motion
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8-1: Angular Quantities
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8-2 and 8-3: Constant Angular Acceleration; Rolling
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8-4: Torque
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8-5 and 8-6: Rotational Dynamics
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8-7: Rotational Kinetic Energy
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8-8: Angular Momentum
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8-9: Angular Quantities as Vectors
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Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 60
Q

What is the angular momentum of a 0.270-kg ball revolving on the end of a thin string in a circle of radius 1.35 m at an angular speed of 10.4 rad/s?

A
5.12 kg m2/s5.12 \textrm{ kg m}^2\textrm{/s}
Giancoli 7th Edition, Chapter 8, Problem 60 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Angular momentum is moment of inertia times angular velocity and for a point mass moving in a circle of radius r, the moment of inertia will be mr squared. So we have 0.270 kilograms times 1.35 meters squared times 10.4 radians per second which gives about 5.12 kilogram meter squared per second of angular momentum.

COMMENTS
By darwisyadaniea12 on Fri, 11/19/2021 - 3:40 AM

hi shouldn't we need to convert the rad/s to m/s for the angular velocity which is 10.4?

By Mr. Dychko on Wed, 11/24/2021 - 6:13 PM

Hi Darwisyadeniea, thanks for the question. The angular momentum calculation calls for the angular velocity, which has units of radians per second, so there's no need to convert to m/s.
All the best,
Shaun

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Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.