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Question
A body rotating at 20 rad/s is acted upon by a constant torque providing it a deceleration of 2 rad/s2. At what time will the body have kinetic energy same as the initial value if the torque continues to act?
Solution
Given
Initial angular acceleration, ω0 = 20 rad/s
Angular deceleration,
\[\alpha = 2\text{ rad/s}^2\]
\[\Rightarrow t_1 = \frac{\omega_0}{\alpha_1} = \frac{20}{2} = 10 s\]
So, in 10 s the body will come to rest.
The same torque continues to act on the body, therefore, it will produce the same angular acceleration.
The kinetic energy of the body is same as the initial value; therefore, its angular velocity should be equal to the initial value ω0.
Therefore, time required to come to that angular velocity,
\[t_2 = \frac{\omega}{\alpha} = \frac{20}{2} = 10 s\]
So, total time required,
\[t = t_1 + t_2 = 20 s\]
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