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Question: A ring of mass 100 kg and diameter 2m is rotating at the rate of \(\left( \frac{300}{\pi} \right)\)r...

A ring of mass 100 kg and diameter 2m is rotating at the rate of (300π)\left( \frac{300}{\pi} \right)rpm. Then-

A

Moment of inertia is 100 kg – m2

B

Kinetic energy is 5 kJ

C

If a retarding torque of 200 N-m starts acting then it will come at rest after 5 sec.

D

All of these

Answer

All of these

Explanation

Solution

Moment of inertia = MR2

k.E of rotation = 12\frac{1}{2}Iw2

Torque = I µ where a = ω0t\frac{\omega_{0}}{t}