Moment of inertia of a body about an axis is $4 \mathrm{~kg}-\mathrm{m}^2$. The body is initially at rest…

Moment of inertia of a body about an axis is $4 \mathrm{~kg}-\mathrm{m}^2$. The body is initially at rest and a torque of $8 \mathrm{~N}$-m starts acting on it along the same axis. Work done by the torque in $20 \mathrm{~s}$, in joules, is
  1. $40$
  2. $640$
  3. $2560$
  4. $3200$

Solution

Given, $I=4 \mathrm{~kg}-\mathrm{m}^2, \tau=8 \mathrm{~N}-\mathrm{m}$ and $t=20 \mathrm{~s}$ $ \begin{gathered} \tau=l \alpha \\ \alpha=\frac{\tau}{l}=\frac{8}{4}=2 \\ \theta=\frac{1}{2} \alpha t^2 \\ \Rightarrow \quad \theta=\frac{1}{2} \times 2 \times 20 \times 20=400 \\ \omega=\tau \theta=8 \times 400=3200 \mathrm{~J} \end{gathered} $

Asked in: AP EAMCET 2013

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