JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A spherical shell of \(1 \,kg\) mass and radius \(R\) is rolling with angular speed \(\omega\) on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin \(O\) is \(\frac{a}{3} R^{2} \omega\). The value of a will be ..............
- A \(2\)
- B \(3\)
- C \(5\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(L _{0}=\) angular momentum of shell about \(O\). As shell is rolling \(\text { so } V _{ cm }=\omega R\) \(L _{0}= mV _{ cm } R + I \omega\) \(=1 \times \omega R \times R +\frac{2}{3} R ^{2} \omega\) \(=\frac{5}{3} R ^{2} \omega\) \(\text { so } a =5\)
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