Explanation Given data: Moment of inertia of the skater with arms out and one leg extended is 3.2 Kg ⋅ m 2 . Moment of inertia of the skater with arms and legs in is 0.80 Kg ⋅ m 2 . Angular velocity of the skater when her arms and legs in is 5.0 rev s . Formula used: Write the expression for law of conservation of angular momentum for two positions of the skater as follows. I i ω i = I f ω f (1) Here, I i is the moment of inertia of the skater with her arms and legs are inward position, ω i is the angular velocity of the skater when her arms and legs are inward position, I f is the moment of inertia of the skater with her arms are outward and one leg is in extended position, and ω f is the angular velocity of the skater with her arms and one leg are open outward position. Explanation: Rewrite the expression in equation (1) to find the angular velocity of the skater with her arms and one leg are open outward position as follows. ω f = I i ω i I f = ( I i I f ) ω i Substitute 0

College Physics: A Strategic Approach, Books a la Carte Edition (4th Edition)

4th Edition

ISBN: 9780134700502

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 9, Problem 43P

To determine

To find: The angular speed of the skater when her arms and one leg outward.

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Chapter 9, Problem 42P

Chapter 9, Problem 45P

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Chapter 9 Solutions

College Physics: A Strategic Approach, Books a la Carte Edition (4th Edition)

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The angular speed of the skater when her arms and one leg outward.

Solution Summary: The author explains how to find the angular speed of the skater when her arms and one leg are outward.

To find: The angular speed of the skater when her arms and one leg outward.

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Chapter 9 Solutions