Question
A body rotating at $20 \ \text{rad/s}$ is acted upon by a constant torque providing it a deceleration of $2\ce{rad/s}^2$. At what time will the body have kinetic energy same as the initial value if the torque continues to act?
✓
Answer
Initial angular velocity $= 20 \ce{ rad/s}$
Therefore $\alpha=2\text{ rad/s}^2$
$\Rightarrow\text{t}_1=\frac{\omega_2}{\alpha_1}=\frac{20}{2}=10\text{ sec}$
Therefore $10\sec$ 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 $= t_1 + t_2 = 20\ \sec$.
Therefore $\alpha=2\text{ rad/s}^2$
$\Rightarrow\text{t}_1=\frac{\omega_2}{\alpha_1}=\frac{20}{2}=10\text{ sec}$
Therefore $10\sec$ 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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