JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Hard
- MCQ
A thin smooth rod of length \(L\) and mass \(M\) is rotating freely with angular speed \({\omega _0}\) about an axis perpendicular to the rod and passing through its centre. Two beads of mass \(m\) and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod will be
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Answer & Solution
Correct Answer
(D) \(\frac{{M{\omega _0}}}{{M + 6m}}\)
Step-by-step Solution
Detailed explanation
Conservation of angualr momentum about rotation axes:
\({L_i} = {L_f}\)
\(\left( {\frac{{M{\ell ^2}}}{{12}}} \right){\omega _0} = \left[ {\frac{{M{\ell ^2}}}{{12}} + 2\left( {m{{\left( {\frac{{{\ell ^2}}}{2}} \right)}^2}} \right)} \right]{\omega _f}\)
\( \Rightarrow \,{\omega _f} = \left( {\frac{M}{{M + 6m}}} \right){\omega _0}\)
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