The angular momentum of a wheel having a rotational inertia of $0.2 \mathrm{~kg} \mathrm{~m}^2$ about its…

The angular momentum of a wheel having a rotational inertia of $0.2 \mathrm{~kg} \mathrm{~m}^2$ about its symmetric axis decreases from 4 to $2 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$ in $4 \mathrm{~s}$. The average power of the wheel is
  1. $7.5 \mathrm{~W}$
  2. $15 \mathrm{~W}$
  3. $5 \mathrm{~W}$
  4. $12 \mathrm{~W}$

Solution

We have, $\mathrm{I}=0.2 \mathrm{~kg} / \mathrm{m}^2$ $\begin{aligned} & \mathrm{L}_{\mathrm{f}}=4 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \\ & \mathrm{~L}_{\mathrm{i}}=2 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\end{aligned}$ $\begin{aligned} & \mathrm{L}_{\mathrm{f}}=\mathrm{I} \omega_{\mathrm{f}} \Rightarrow \omega_{\mathrm{f}}=\frac{4}{0.2}=20 \mathrm{rad} / \mathrm{s} \\ & \mathrm{L}_{\mathrm{i}}=\mathrm{I} \omega_{\mathrm{i}} \Rightarrow \omega_{\mathrm{i}}=\frac{2}{0.2}=10 \mathrm{rad} / \mathrm{s}\end{aligned}$ So, power $=\frac{\text { work done }}{\text { time taken }}=\frac{\Delta \mathrm{K}}{\Delta \mathrm{t}}$ $\begin{aligned} & =\frac{\frac{1}{2} \times 0.2 \times 20^2-\frac{1}{2} \times 0.2 \times 10^2}{4} \\ & =7.5 \mathrm{~W}\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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