A solid sphere of mass $1 \mathrm{~kg}$ rolls on a table with linear speed $2 \mathrm{~m} / \mathrm{s}$, find its total kinetic energy.
Q 128.2
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Data : $M=1 \mathrm{~kg}, \mathrm{v}=2 \mathrm{~m} / \mathrm{s}$
The total kinetic energy of a rolling body,
$
E=\frac{1}{2} M v^2\left(1+\frac{k^2}{R^2}\right)
$
For a solid sphere, $k^2=\frac{2}{5} R^2$
$
\begin{aligned}
\therefore E & =\frac{1}{2} M v^2\left(1+\frac{2}{5}\right)=\frac{1}{2} \times \frac{7}{5} M v^2 \\
& =\frac{7}{10} M v^2=\frac{7}{10} \times 1 \times 2^2=\frac{7 \times 4}{10}=2.8 \mathrm{~J}
\end{aligned}
$
The total kinetic energy of a rolling body,
$
E=\frac{1}{2} M v^2\left(1+\frac{k^2}{R^2}\right)
$
For a solid sphere, $k^2=\frac{2}{5} R^2$
$
\begin{aligned}
\therefore E & =\frac{1}{2} M v^2\left(1+\frac{2}{5}\right)=\frac{1}{2} \times \frac{7}{5} M v^2 \\
& =\frac{7}{10} M v^2=\frac{7}{10} \times 1 \times 2^2=\frac{7 \times 4}{10}=2.8 \mathrm{~J}
\end{aligned}
$
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