A solid cylinder of mass $M$ and radius $R$ is rotating about its geometrical axis. A solid sphere of the…

A solid cylinder of mass $M$ and radius $R$ is rotating about its geometrical axis. A solid sphere of the same mass and same radius is also rotating about its diameter with an angular speed half that of the cylinder. The ratio of the kinetic energy of rotation of the sphere to that of the cylinder will be
  1. $1: 4$
  2. $1: 5$
  3. $2: 3$
  4. $3: 2$

Solution

$I_{\text {sphere }}=I_S=\frac{2}{5} m R^2 \text { and } I_{\text {cylinder }}=I_c=\frac{1}{2} m R^2$
Let $\omega_{\mathrm{S}}=$ Angular speed of sphere, $\omega_c=$ Angular speed of cylinder. It is given that, $\omega_{\mathrm{s}}=\frac{\omega_{\mathrm{c}}}{2}$ $\begin{aligned} \therefore \quad & \frac{K \cdot E_{\text {sphere }}}{K \cdot E_{\text {cylinder }}}=\frac{I_s \omega_s^2}{I_c^2 \omega_c^2}=\frac{\frac{2}{5} m R^2 \times\left(\frac{\omega_c}{2}\right)^2}{\frac{1}{2} m R^2 \times \omega_c{ }^2} \\ & \frac{K \cdot E_{\text {sphere }}}{K \cdot E_{\text {cylinder }}}=\frac{1}{5} \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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