Question
A body rotating at 20rad/s is acted upon by a constant torque providing it a deceleration of $2rad/s^2.$ At what time will the body have kinetic energy same as the initial value if the torque continues to act?
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Answer
Initial angular velocity = 20rad/s Therefore $\alpha=2\text{rad/s}^2$
$\Rightarrow\text{t}_1=\frac{\omega_2}{\alpha_1}=\frac{20}{2}=10\text{sec}$
Therefore 10sec it will come to rest. Since the same torque is continues to act on the body it will produce same angular acceleration and since the initial kinetic energy = the kinetic energy at a instant. So initial angular velocity = angular velocity at that instant Therefore time require to come to that angular velocity,
$\Rightarrow\text{t}_2=\frac{\omega}{\alpha_2}=\frac{20}{2}=10\text{sec}$
therefore time required $= t_1 + t_2 = 20$sec.
$\Rightarrow\text{t}_1=\frac{\omega_2}{\alpha_1}=\frac{20}{2}=10\text{sec}$
Therefore 10sec it will come to rest. Since the same torque is continues to act on the body it will produce same angular acceleration and since the initial kinetic energy = the kinetic energy at a instant. So initial angular velocity = angular velocity at that instant Therefore time require to come to that angular velocity,
$\Rightarrow\text{t}_2=\frac{\omega}{\alpha_2}=\frac{20}{2}=10\text{sec}$
therefore time required $= t_1 + t_2 = 20$sec.
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