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A solid sphere of mass $5 \mathrm{~kg}$ rolls on a plane surfaces. Find its kinetic energy at an instant when its centre moves with speed $4 \mathrm{~m} / \mathrm{s}$.
Options:
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Verified Answer
The correct answer is:
$56 \mathrm{~J}$
As we know, the rotational kinetic energy of a body is given by,
$$
\mathrm{KE}=\frac{1}{2} m v^2\left[1+\frac{k^2}{R^2}\right]
$$
where,$k=$ radius of gyration
Given, $m=5 \mathrm{~kg}, v=4 \mathrm{~m} / \mathrm{s}$ and $k=\sqrt{\frac{2}{5}} R$
Hence, $\mathrm{KE}=\frac{5}{2} \times 4^2\left[1+\frac{2}{5}\right] \Rightarrow \mathrm{KE}=40 \times \frac{7}{5}=56 \mathrm{~J}$
Hence, the correct option is (1).
$$
\mathrm{KE}=\frac{1}{2} m v^2\left[1+\frac{k^2}{R^2}\right]
$$
where,$k=$ radius of gyration
Given, $m=5 \mathrm{~kg}, v=4 \mathrm{~m} / \mathrm{s}$ and $k=\sqrt{\frac{2}{5}} R$
Hence, $\mathrm{KE}=\frac{5}{2} \times 4^2\left[1+\frac{2}{5}\right] \Rightarrow \mathrm{KE}=40 \times \frac{7}{5}=56 \mathrm{~J}$
Hence, the correct option is (1).
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