MHT CET · Physics · Rotational Motion
A solid sphere of mass 'M' and radius 'R' is rotating about its diameter. A disc of same mass and radius is also rotating about an axis passing through its centre and perpendicular to the plane but angular speed is twice that of the sphere. The ratio of kinetic energy of disc to that of sphere is
- A \(5: 1\)
- B \(6: 1\)
- C \(4: 1\)
- D \(3: 1\)
Answer & Solution
Correct Answer
(A) \(5: 1\)
Step-by-step Solution
Detailed explanation
Kinetic energy is given by :
For sphere \(\mathrm{k}_{\mathrm{s}}=\frac{1}{2} \mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}\)
For \(\operatorname{disc} \mathrm{k}_{\mathrm{d}}=\frac{1}{2} \mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}\)
\(\frac{\mathrm{k}_{\mathrm{d}}}{\mathrm{k}_{\mathrm{s}}}=\frac{\mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}}{\mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}}=\frac{\frac{1}{2} \mathrm{MR}^{2}(2 \omega)^{2}}{\frac{2}{5} \mathrm{MR}^{2} \omega^{2}}=5\)
For sphere \(\mathrm{k}_{\mathrm{s}}=\frac{1}{2} \mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}\)
For \(\operatorname{disc} \mathrm{k}_{\mathrm{d}}=\frac{1}{2} \mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}\)
\(\frac{\mathrm{k}_{\mathrm{d}}}{\mathrm{k}_{\mathrm{s}}}=\frac{\mathrm{I}_{\mathrm{d}} \omega_{\mathrm{d}}^{2}}{\mathrm{I}_{\mathrm{s}} \omega_{\mathrm{s}}^{2}}=\frac{\frac{1}{2} \mathrm{MR}^{2}(2 \omega)^{2}}{\frac{2}{5} \mathrm{MR}^{2} \omega^{2}}=5\)
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