Answer: C) \( I = 8 \, \text{kg} \cdot \text{m}^2, K.E. = 631 \, \text{J} \)
(i) Moment of Inertia:
The moment of inertia \( I \) for a body revolving in a circle at a distance \( R \) is:
\[
I = M R^2
\]
Here, \( M = 2 \, \text{kg} \) and \( R = 2 \, \text{m} \):
\[
I = 2 \times (2)^2 = 8 \, \text{kg} \cdot \text{m}^2
\]
(ii) Rotational Kinetic Energy:
The rotational kinetic energy \( K.E. \) is given by:
\[
K.E. = \frac{1}{2} I \omega^2
\]
To calculate \( \omega \) (angular velocity), we use the fact that the body completes 2 revolutions per second:
\[
\omega = 2\pi f
\]
where \( f = 2 \, \text{revolutions/second} \):
\[
\omega = 2\pi \times 2 = 4\pi \, \text{rad/s}
\]
Now substitute \( I = 8 \, \text{kg} \cdot \text{m}^2 \) and \( \omega = 4\pi \):
\[
K.E. = \frac{1}{2} \times 8 \times (4\pi)^2
\]
\[
K.E. = 4 \times 16\pi^2 = 64\pi^2
\]
Using \( \pi^2 \approx 9.85 \):
\[
K.E. = 64 \times 9.85 = 631.01 \, \text{J}
\]
Final Answers:
- Moment of Inertia \( I = 8 \, \text{kg} \cdot \text{m}^2 \)
- Rotational Kinetic Energy \( K.E. = 631.01 \, \text{J} \)
.png)
0 Comments