JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2023
- Medium
- Numerical
A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is \(\pi: 22\) then, the value of its angular speed will be \(...........\,rad / s\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(L =\left( I _{\text {com }}\right)(\omega) \text { and } K =\frac{1}{2}\left( I _{\text {com }}\right)\left(\omega^2\right)+\frac{1}{2} MV _{\text {com }}^2\)
\(L =\frac{2}{5} M^2 \frac{V_{\text {com }}}{R} \quad K=\frac{1}{2}\left(\frac{2}{5} M R^2\right) \frac{V_{\text {com }}^2}{ R ^2}+\frac{1}{2} M V_{\text {com }}^2\)
\(L =\frac{2 MRV _{\text {com }}}{5} \quad K =\frac{7}{10} MV _{\text {com }}^2\)
\(\text { Ratio } \frac{ L }{ K }=\frac{4}{7} \frac{ R }{ V _{\text {com }}}=\frac{\pi}{22}\)
\(\Rightarrow \omega=\frac{4}{7} \times \frac{22}{22} \times 7=4\)
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