A thin uniform bar of length \(L\) and mass \(8\,m\) lies on a smooth horizontal table. Two point masses \(m\) and \(2\,m\) moving in the same horizontal plane from opposite sides of the bar with speeds \(2v\) and \(v\) respectively. The masses stick to the bar after collision at a distance \(\frac{L}{3}\) and \(\frac{L}{6}\) respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be

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Answer & Solution

Correct Answer

(A) \(\frac{v}{{6L}}\)

Detailed explanation

Linear momentum is conserved

\(2 m v-2 m v=(2 m+m+M) V_{c m}\)

\(\Rightarrow V_{c m}=0\)

Angular momentum is conserve

\(2 m v \frac{L}{6}+2 m v \times \frac{L}{3}+0=I w\) \(...(i)\)

\(I=\mathrm{M.I.}\) of rod \(+\mathrm{M.I.}\) of mass \(2 \mathrm{m}+\mathrm{M.I}\) of mass \(\mathrm{m}\)

\(=\frac{8 m L^{2}}{12}+2 m\left(\frac{L}{6}\right)^{2}+m\left(\frac{L}{3}\right)^{2}\)

\(=m L^{2}\left(\frac{2}{3}+\frac{1}{18}+\frac{1}{9}\right)=\frac{5}{6} m L^{2}\)

eqn \((1)\) becomes

\(\left(\frac{m v L}{3}+\frac{2}{3} m v L\right)=\frac{5}{6} m L^{2} w\)

\(\frac{m L}{3}(V+2 v)=\frac{5}{6} m L^{2} w\)

\(\frac{6 V}{5 L}=w\)

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