MHT CET · Physics · Rotational Motion
A solid sphere of mass ' \(M\) ' and radius ' \(R\) ' is rotating about its diameter. A solid cylinder of same mass and radius is also rotating about its geometrical \(1: 3\) axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ( \(\mathrm{K}_{\text {sphere }}\) to \(\mathrm{K}_{\text {cylinder }}\) ) will be
- A \(1: 8\)
- B \(1: 6\)
- C \(1: 3\)
- D \(1: 5\)
Answer & Solution
Correct Answer
(D) \(1: 5\)
Step-by-step Solution
Detailed explanation
Kinetic energy of sphere is given by:
\(E_{\text {sphere }}=\frac{1}{2} I_{\text {sphere }} \omega_{\text {sphere }}^2\)
\(E_{\text {sphere }}=\frac{1}{2} \times \frac{2}{5} m R^2 \omega_{\text {sphere }}^2\)
Kinetic energy of cylinder is given by:
\(\begin{aligned} & E_{\text {cylinder }}=\frac{1}{2} I_{\text {cylinder }} \omega_{\text {cylinder }}^2 \\ & E_{\text {cylinder }}=\frac{1}{2} \times \frac{1}{2} m R^2 \omega_{\text {cylinder }}^2 \\ & \text { Given, } \omega_{\text {cylinder }}=2 \omega_{\text {sphere }} \\ & \frac{E_{\text {sphere }}}{E_{\text {cylinder }}}=\frac{1}{5}\end{aligned}\)
\(E_{\text {sphere }}=\frac{1}{2} I_{\text {sphere }} \omega_{\text {sphere }}^2\)
\(E_{\text {sphere }}=\frac{1}{2} \times \frac{2}{5} m R^2 \omega_{\text {sphere }}^2\)
Kinetic energy of cylinder is given by:
\(\begin{aligned} & E_{\text {cylinder }}=\frac{1}{2} I_{\text {cylinder }} \omega_{\text {cylinder }}^2 \\ & E_{\text {cylinder }}=\frac{1}{2} \times \frac{1}{2} m R^2 \omega_{\text {cylinder }}^2 \\ & \text { Given, } \omega_{\text {cylinder }}=2 \omega_{\text {sphere }} \\ & \frac{E_{\text {sphere }}}{E_{\text {cylinder }}}=\frac{1}{5}\end{aligned}\)
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