A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be _______ ms$^{-1}$.
Answer (integer)
40
Solution
$\frac{1}{2} m v_{\mathrm{cm}}^{2}+\frac{1}{2} \times \frac{2}{5} m R^{2} \times \frac{v_{\mathrm{cm}}^{2}}{R^{2}}=2240 \mathrm{~J}$
<br/><br/>
$$
\begin{aligned}
& \frac{7}{10} m v_{\mathrm{cm}}^{2}=2240 \\\\
& v_{\mathrm{cm}}=\sqrt{\frac{2240 \times 10}{7 \times 2}}=40 \mathrm{~m} / \mathrm{sec}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
This question is part of PrepWiser's free JEE Main question bank. 158 more solved questions on Rotational Motion are available — start with the harder ones if your accuracy is >70%.