A figure skater wants to spin faster. With her arms and leg outstretched, she has a moment of inertia1 = 3.6 kgm2 and a spin rate of 2.0 rev/s. If she can squeeze her arms and leg closer to her rotation axis, she can spin at a rate of 24 rev/s. What is her new moment of inertia for the faster spin? O 1.9 kgm2 O 0.30 kgm2 O 7.2 kgm2 O 45 kgm2
A figure skater wants to spin faster. With her arms and leg outstretched, she has a moment of inertia1 = 3.6 kgm2 and a spin rate of 2.0 rev/s. If she can squeeze her arms and leg closer to her rotation axis, she can spin at a rate of 24 rev/s. What is her new moment of inertia for the faster spin? O 1.9 kgm2 O 0.30 kgm2 O 7.2 kgm2 O 45 kgm2
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Understanding Moments of Inertia and Angular Velocity
#### Problem Statement:
A figure skater wants to spin faster. With her arms and legs outstretched, she has a moment of inertia \( I \) = 3.6 kgm\(^2\) and a spin rate of 2.0 rev/s. If she can squeeze her arms and legs closer to her rotation axis, she can spin at a rate of 24 rev/s. What is her new moment of inertia for the faster spin?
#### Options:
- 1.9 kgm\(^2\)
- 0.30 kgm\(^2\)
- 7.2 kgm\(^2\)
- 45 kgm\(^2\)
#### Explanation:
The relationship between the moment of inertia \( I \) and angular velocity \( \omega \) for a rotating body is governed by the conservation of angular momentum. Angular momentum \( L \) is given by:
\[ L = I \omega \]
Since angular momentum is conserved, the initial angular momentum \( L_1 \) must equal the final angular momentum \( L_2 \):
\[ L_1 = L_2 \]
Therefore:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Given:
- \( I_1 = 3.6 \) kgm\(^2\)
- \( \omega_1 = 2.0 \) rev/s
- \( \omega_2 = 24 \) rev/s
We need to find \( I_2 \):
\[ I_2 = \frac{I_1 \omega_1}{\omega_2} \]
Substitute the values:
\[ I_2 = \frac{3.6 \, \text{kgm}^2 \times 2.0 \, \text{rev/s}}{24 \, \text{rev/s}} \]
\[ I_2 = \frac{7.2 \, \text{kgm}^2 \text{ rev/s}}{24 \, \text{rev/s}} \]
\[ I_2 = 0.30 \, \text{kgm}^2 \]
Hence, the correct answer is:
\[ \boxed{0.30 \, \text{kgm}^2} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24666b77-75da-49bd-bf84-8c4674dedd2c%2Fdac9e494-15b2-4c49-85c7-84ba28e54d6f%2Fmqstav9.png&w=3840&q=75)
Transcribed Image Text:### Understanding Moments of Inertia and Angular Velocity
#### Problem Statement:
A figure skater wants to spin faster. With her arms and legs outstretched, she has a moment of inertia \( I \) = 3.6 kgm\(^2\) and a spin rate of 2.0 rev/s. If she can squeeze her arms and legs closer to her rotation axis, she can spin at a rate of 24 rev/s. What is her new moment of inertia for the faster spin?
#### Options:
- 1.9 kgm\(^2\)
- 0.30 kgm\(^2\)
- 7.2 kgm\(^2\)
- 45 kgm\(^2\)
#### Explanation:
The relationship between the moment of inertia \( I \) and angular velocity \( \omega \) for a rotating body is governed by the conservation of angular momentum. Angular momentum \( L \) is given by:
\[ L = I \omega \]
Since angular momentum is conserved, the initial angular momentum \( L_1 \) must equal the final angular momentum \( L_2 \):
\[ L_1 = L_2 \]
Therefore:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Given:
- \( I_1 = 3.6 \) kgm\(^2\)
- \( \omega_1 = 2.0 \) rev/s
- \( \omega_2 = 24 \) rev/s
We need to find \( I_2 \):
\[ I_2 = \frac{I_1 \omega_1}{\omega_2} \]
Substitute the values:
\[ I_2 = \frac{3.6 \, \text{kgm}^2 \times 2.0 \, \text{rev/s}}{24 \, \text{rev/s}} \]
\[ I_2 = \frac{7.2 \, \text{kgm}^2 \text{ rev/s}}{24 \, \text{rev/s}} \]
\[ I_2 = 0.30 \, \text{kgm}^2 \]
Hence, the correct answer is:
\[ \boxed{0.30 \, \text{kgm}^2} \]
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