Question:
A circular ring of mass 10 kg and radius 1 m is rotating at 210 revolutions in a minute. It is brought to stop in 2 s. The required average power is:
A circular ring of mass 10 kg and radius 1 m is rotating at 210 revolutions in a minute. It is brought to stop in 2 s. The required average power is:
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Average power required to stop a rotating object = rotational kinetic energy divided by stopping time. Convert rpm to rad/s before calculation.
Updated On: Jul 18, 2026
- 980 W
- 1210 W
- 1340 W
- 1580 W
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The Correct Option is B
Solution and Explanation
Step 1: Convert angular speed to SI units.
The given rotational speed is \(210\) revolutions per minute. Convert to angular velocity in rad/s: \[ \omega = 210 \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 7 \pi \, \text{rad/s} \approx 21.99 \, \text{rad/s} \]
Step 2: Moment of inertia of the ring.
For a circular ring of mass \(M = 10 \, \text{kg}\) and radius \(R = 1 \, \text{m}\): \[ I = MR^2 = 10 \times 1^2 = 10 \, \text{kg m}^2 \]
Step 3: Rotational kinetic energy before stopping.
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 10 \times (21.99)^2 \approx 5 \times 483.6 \approx 2418 \, \text{J} \]
Step 4: Average power to bring the ring to rest.
Power is work done per unit time. Here work done equals initial kinetic energy (since it is brought to rest): \[ P_{\text{avg}} = \frac{KE_{\text{rot}}}{t} = \frac{2418}{2} \approx 1209 \, \text{W} \]
Step 5: Round to nearest option.
The nearest option is \(1210 \, \text{W}\).
Step 6: Final conclusion.
\[ \boxed{1210 \, \text{W}} \]
The given rotational speed is \(210\) revolutions per minute. Convert to angular velocity in rad/s: \[ \omega = 210 \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 7 \pi \, \text{rad/s} \approx 21.99 \, \text{rad/s} \]
Step 2: Moment of inertia of the ring.
For a circular ring of mass \(M = 10 \, \text{kg}\) and radius \(R = 1 \, \text{m}\): \[ I = MR^2 = 10 \times 1^2 = 10 \, \text{kg m}^2 \]
Step 3: Rotational kinetic energy before stopping.
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 10 \times (21.99)^2 \approx 5 \times 483.6 \approx 2418 \, \text{J} \]
Step 4: Average power to bring the ring to rest.
Power is work done per unit time. Here work done equals initial kinetic energy (since it is brought to rest): \[ P_{\text{avg}} = \frac{KE_{\text{rot}}}{t} = \frac{2418}{2} \approx 1209 \, \text{W} \]
Step 5: Round to nearest option.
The nearest option is \(1210 \, \text{W}\).
Step 6: Final conclusion.
\[ \boxed{1210 \, \text{W}} \]
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