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A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be:
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Verified Answer
The correct answer is:
$\frac{K^2}{K^2+R^2}$
Total energy $=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2$
$=\frac{1}{2} m v^2\left(1+\frac{K^2}{R^2}\right)$
$\begin{aligned}
& \text {Rotational energy }=\frac{1}{2} I \omega^2 \\
& =\frac{K^2+R^2}{1+\frac{K^2}{R^2}}
\end{aligned}$
$\begin{aligned}
& \text {Required fraction }=\frac{K^2 / R^2}{1+K^2 / R^2} \\
& =\frac{K^2}{R^2+K^2}
\end{aligned}$
$=\frac{1}{2} m v^2\left(1+\frac{K^2}{R^2}\right)$
$\begin{aligned}
& \text {Rotational energy }=\frac{1}{2} I \omega^2 \\
& =\frac{K^2+R^2}{1+\frac{K^2}{R^2}}
\end{aligned}$
$\begin{aligned}
& \text {Required fraction }=\frac{K^2 / R^2}{1+K^2 / R^2} \\
& =\frac{K^2}{R^2+K^2}
\end{aligned}$
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