Question

A body rotating with $20\,rad/s$ is acted upon by a uniform torque providing it with an angular deceleration of $2\,rad/{s^2}$ . At which time will the body have kinetic energy same as the initial value if the torque continues?

Answer 1

Hint:In order to this question, to calculate the exact time, the body have kinetic energy same as the initial value if the torque continuously, we will first equate the equation in which Initial kinetic energy is equal to final kinetic energy to find the final angular velocity. And then we will apply a kinematic equation to conclude our final solution.

Complete step by step answer:
Given that: Initial angular Velocity, ${w_{initial}} = 20\,rad/s$ and Angular Acceleration, $\alpha = 2\,rad/{s^2}$. As per the question, Initial kinetic energy is equal to final kinetic energy-
$\dfrac{1}{2}I{w^2}_{initial} = \dfrac{1}{2}I{w^2}_{final} \\
\Rightarrow {w^2}_{initial} = {w^2}_{final} = {20^2} \\
\Rightarrow {w_{final}} = \pm 20 \\ $
Final velocity has two solution, but only
${w_{final}} = - 20$ is impossible.
Now, applying kinematic equation:
${w_{final}} = {w_{initial}} + \alpha t \\
\Rightarrow - 20 = 20 + 2t \\
\therefore t = 20\sec \\ $
Hence, in $20\sec $ body will achieve equal kinetic energy as its initial value.

Note: The angular velocity is a measure of how rapidly the central angle changes over time, whereas the linear velocity is a measure of how fast the arc length varies over time. The radian measure of the angle divided by the time it takes to sweep out this angle gives the point's angular velocity.