A thin uniform bar of length L and mass 8 m lies on a smooth… | JEE Main 2018 (15 Apr, Shift 2)

Q67JEE Main 2018 · 15 APR · SHIFT 2Difficult

Angular Momentum & its Conservation

A thin uniform bar of length $L$ and mass $8$ m lies on a smooth horizontal table. Two point masses m and 2 m are moving in the same horizontal plane from opposite sides of the bar with speeds 2 $v$ and $v$ respectively. The masses stick to the bar after collision at a distance $\frac{L}{3}$ and $\frac{L}{6}$ respectively from the center 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:

Show answer and solution

Answer: (b)

Both masses give clockwise angular momentum about O: $L = m(2v)\frac{L}{3} + 2m(v)\frac{L}{6} = \frac{2mvL}{3} + \frac{mvL}{3} = mvL$ .
$I = \frac{1}{12}(8m)L^2 + m\left(\frac{L}{3}\right)^2 + 2m\left(\frac{L}{6}\right)^2 = \left(\frac{2}{3} + \frac{1}{9} + \frac{1}{18}\right)mL^2 = \frac{5}{6}mL^2$
$\omega = \frac{mvL}{\frac{5}{6}mL^2} = \frac{6v}{5L}$
So the answer is (b) $\frac{6v}{5L}$ .

Report a mistake

Angular Momentum & its Conservation question from JEE Main 2018 (15 Apr, Shift 2)

More questions on Angular Momentum & its Conservation