Answer: A) \( \omega = 52.36 \, \text{rad/s}, I = 125 \, \text{kg} \cdot \text{m}^2, K.E. = 171347.5 \, \text{J} \)
(i) Angular Velocity \( \omega \):
The flywheel makes 500 revolutions per minute. To convert this to angular velocity in radians per second:
\[
\omega = 2\pi \times \text{frequency} = 2\pi \times \left( \frac{\text{rpm}}{60} \right)
\]
Substitute \( \text{rpm} = 500 \):
\[
\omega = 2\pi \times \left( \frac{500}{60} \right) = 52.36 \, \text{rad/s}
\]
(ii) Moment of Inertia \( I \):
Assuming the mass is concentrated along the rim of the flywheel, the moment of inertia is given by:
\[
I = M R^2
\]
Here, \( M = 500 \, \text{kg} \) and the radius \( R = \frac{\text{diameter}}{2} = \frac{1}{2} = 0.5 \, \text{m} \):
\[
I = 500 \times (0.5)^2 = 500 \times 0.25 = 125 \, \text{kg} \cdot \text{m}^2
\]
(iii) Rotational Kinetic Energy \( K.E. \):
The rotational kinetic energy is given by:
\[
K.E. = \frac{1}{2} I \omega^2
\]
Substitute \( I = 125 \, \text{kg} \cdot \text{m}^2 \) and \( \omega = 52.36 \, \text{rad/s} \):
\[
K.E. = \frac{1}{2} \times 125 \times (52.36)^2
\]
First, calculate \( (52.36)^2 \):
\[
(52.36)^2 = 2741.56
\]
Now calculate \( K.E. \):
\[
K.E. = \frac{1}{2} \times 125 \times 2741.56 = 62.5 \times 2741.56 = 171347.5 \, \text{J}
\]
Final Answers:
- Angular Velocity \( \omega = 52.36 \, \text{rad/s} \)
- Moment of Inertia \( I = 125 \, \text{kg} \cdot \text{m}^2 \)
- Rotational Kinetic Energy \( K.E. = 171347.5 \, \text{J} \)
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