Suppose the rotational kinetic energy of a solid disk is 1000 J when the disk has an angular velocity of 3 rad/s. What is the disk's moment of inertia? If the mass of the disk is 50 kg, what must be its radius? moment of inertia: A. 276.5 kg-m² D. 238.5 kg-m? В. 222.2 kg-m? E. 239.2 kg-m? С. 175.9 kg-m? F. 170.3 kg-m² |--/ radius: А. 2.490 m D. 2.536 m В. 2.981 m Е. 3.341 m С. 3.212 m F. 3.195 m
Suppose the rotational kinetic energy of a solid disk is 1000 J when the disk has an angular velocity of 3 rad/s. What is the disk's moment of inertia? If the mass of the disk is 50 kg, what must be its radius? moment of inertia: A. 276.5 kg-m² D. 238.5 kg-m? В. 222.2 kg-m? E. 239.2 kg-m? С. 175.9 kg-m? F. 170.3 kg-m² |--/ radius: А. 2.490 m D. 2.536 m В. 2.981 m Е. 3.341 m С. 3.212 m F. 3.195 m
College Physics
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Author:Raymond A. Serway, Chris Vuille
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Question
![**Rotational Kinetic Energy and Moment of Inertia of a Solid Disk**
Consider a solid disk with a rotational kinetic energy of 1000 J when it has an angular velocity of 3 rad/s. We need to find the disk's moment of inertia. Further, with a disk mass of 50 kg, we will determine the disk's radius.
**Moment of Inertia Options:**
A. 276.5 kg·m²
B. 222.2 kg·m²
C. 175.9 kg·m²
D. 238.5 kg·m²
E. 239.2 kg·m²
F. 170.3 kg·m²
**Radius Options:**
A. 2.490 m
B. 2.981 m
C. 3.212 m
D. 2.536 m
E. 3.341 m
F. 3.195 m
To find the moment of inertia (\(I\)), we use the formula for rotational kinetic energy (\(K\)):
\[ K = \frac{1}{2} I \omega^2 \]
Solving for \(I\):
\[ I = \frac{2K}{\omega^2} \]
Substituting \(K = 1000\) J and \(\omega = 3\) rad/s:
\[ I = \frac{2 \times 1000}{3^2} = \frac{2000}{9} \approx 222.2 \, \text{kg·m}^2 \]
The correct choice for the moment of inertia is B. 222.2 kg·m².
For the radius (\(r\)) of the disk, use the formula for the moment of inertia of a solid disk:
\[ I = \frac{1}{2} m r^2 \]
Solving for \(r\):
\[ r = \sqrt{\frac{2I}{m}} \]
Substituting \(I = 222.2\) kg·m² and \(m = 50\) kg:
\[ r = \sqrt{\frac{2 \times 222.2}{50}} \approx 2.981 \, \text{m} \]
The correct choice for the radius is B. 2.981 m.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c7cea9f-8590-4f6c-8a87-13eb139b4e53%2Fb1e72fe4-4a60-4e9b-ac99-b782e68d7488%2Fel2h1m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Rotational Kinetic Energy and Moment of Inertia of a Solid Disk**
Consider a solid disk with a rotational kinetic energy of 1000 J when it has an angular velocity of 3 rad/s. We need to find the disk's moment of inertia. Further, with a disk mass of 50 kg, we will determine the disk's radius.
**Moment of Inertia Options:**
A. 276.5 kg·m²
B. 222.2 kg·m²
C. 175.9 kg·m²
D. 238.5 kg·m²
E. 239.2 kg·m²
F. 170.3 kg·m²
**Radius Options:**
A. 2.490 m
B. 2.981 m
C. 3.212 m
D. 2.536 m
E. 3.341 m
F. 3.195 m
To find the moment of inertia (\(I\)), we use the formula for rotational kinetic energy (\(K\)):
\[ K = \frac{1}{2} I \omega^2 \]
Solving for \(I\):
\[ I = \frac{2K}{\omega^2} \]
Substituting \(K = 1000\) J and \(\omega = 3\) rad/s:
\[ I = \frac{2 \times 1000}{3^2} = \frac{2000}{9} \approx 222.2 \, \text{kg·m}^2 \]
The correct choice for the moment of inertia is B. 222.2 kg·m².
For the radius (\(r\)) of the disk, use the formula for the moment of inertia of a solid disk:
\[ I = \frac{1}{2} m r^2 \]
Solving for \(r\):
\[ r = \sqrt{\frac{2I}{m}} \]
Substituting \(I = 222.2\) kg·m² and \(m = 50\) kg:
\[ r = \sqrt{\frac{2 \times 222.2}{50}} \approx 2.981 \, \text{m} \]
The correct choice for the radius is B. 2.981 m.
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