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Moon revolves around the earth in an orbit of radius $\mathrm{R}$ with time period of revolution $\mathrm{T}$. It also rotates about its own axis with a time period $\mathrm{T}$. If mass of the moon is $\mathrm{M}$ and its radius is ' $r$ ', the total kinetic energy of the moon is
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The correct answer is:
$\frac{2 M \pi^2 R^2}{T^2}+\frac{4 M r^2 \pi^2}{5 T^2}$
Kinetic Energy of Moon = Transational Kinetic Energy + Rotational Kinetic Energy.
$$
\mathrm{KE}=\mathrm{KE}_{\mathrm{T}}+\mathrm{KE}_{\mathrm{R}}
$$
$\begin{aligned} & =\frac{1}{2} \mathrm{Mv}^2+\frac{1}{2} \mathrm{I} \omega^2 \\ & =\frac{1}{2} \mathrm{M}\left(\mathrm{R}^2 \omega^2\right)+\frac{1}{2}\left(\frac{2}{5} \mathrm{Mr}^2\right)\left(\omega^2\right)\left[\begin{array}{c}\because \mathrm{v}=\mathrm{r} \omega \\ \omega=\frac{2 \pi}{\mathrm{T}} \\ \mathrm{MR} \times \frac{2}{2} \mathrm{Mr}^2\end{array}\right] \\ & \mathrm{KE}=\frac{2 \mathrm{M} \pi^2 \mathrm{R}^2}{\mathrm{~T}^2}+\frac{4 \mathrm{Mr}^2 \pi^2}{5 \mathrm{~T}^2} \\ & \end{aligned}$
$$
\mathrm{KE}=\mathrm{KE}_{\mathrm{T}}+\mathrm{KE}_{\mathrm{R}}
$$
$\begin{aligned} & =\frac{1}{2} \mathrm{Mv}^2+\frac{1}{2} \mathrm{I} \omega^2 \\ & =\frac{1}{2} \mathrm{M}\left(\mathrm{R}^2 \omega^2\right)+\frac{1}{2}\left(\frac{2}{5} \mathrm{Mr}^2\right)\left(\omega^2\right)\left[\begin{array}{c}\because \mathrm{v}=\mathrm{r} \omega \\ \omega=\frac{2 \pi}{\mathrm{T}} \\ \mathrm{MR} \times \frac{2}{2} \mathrm{Mr}^2\end{array}\right] \\ & \mathrm{KE}=\frac{2 \mathrm{M} \pi^2 \mathrm{R}^2}{\mathrm{~T}^2}+\frac{4 \mathrm{Mr}^2 \pi^2}{5 \mathrm{~T}^2} \\ & \end{aligned}$
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