JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 19)
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$$^{-1}$$.
คำตอบ
40
คำอธิบาย
$\frac{1}{2} m v_{\mathrm{cm}}^{2}+\frac{1}{2} \times \frac{2}{5} m R^{2} \times \frac{v_{\mathrm{cm}}^{2}}{R^{2}}=2240 \mathrm{~J}$
$$ \begin{aligned} & \frac{7}{10} m v_{\mathrm{cm}}^{2}=2240 \\\\ & v_{\mathrm{cm}}=\sqrt{\frac{2240 \times 10}{7 \times 2}}=40 \mathrm{~m} / \mathrm{sec} \end{aligned} $$
$$ \begin{aligned} & \frac{7}{10} m v_{\mathrm{cm}}^{2}=2240 \\\\ & v_{\mathrm{cm}}=\sqrt{\frac{2240 \times 10}{7 \times 2}}=40 \mathrm{~m} / \mathrm{sec} \end{aligned} $$