Question

In the shown figure a mass m slides down the frictionless surface from height h and collides with a uniform vertical rod of length L and mass M. After a collision, mass m sticks to the rod. The rod is free to rotate in a vertical plane about a fixed axis through O. Find the maximum angular deflection of the rod from its initial position

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is B

Just before collision, velocity of the mass m is along the horizontal and is equal to $v_0=\sqrt{2gh}$

From the FBD during the collision, we note that net torque during collision about O = 0

∴The angular momentum of the system about O is conserved. If $L_{1}and{L}_2$ are the angular momentum of the system just before and just after the collision, then

$L_1=m{v}_{0}Land{L}_2=I\omega =(\frac{M{L}^2}{3}+m{L}^2)\omega$

By law of conservation of angular momentum,

$(\frac{M}{3}+m)L^2\omega =m{v}_{0}L$

$\Rightarrow \omega =\frac{m{v}_0}{(\frac{M}{3}+m)L}$

Let the rod deflects through an angle $\theta$

Initial energy of rod and mass system = $\frac{1}{2}l{\omega }^2$, where $l=(\frac{M{L}^2}{3}+M{L}^2)$

=$mgL[1-cos\theta ]+Mg\frac{L}{2}[1-cos\theta ]$

$\because$ From law of conservation of energy,

$\frac{1}{2}l{\omega }^2=(m+\frac{M}{2})gL(1-cos\theta )$

$\Rightarrow \frac{1}{2}(\frac{M{L}^2}{3}+M{L}^2)\times \frac{m^2{v}_0^2}{{(\frac{M}{3}+m)}^2{L}^2}=(m+\frac{M}{2})gL(1-cos\theta )$

$\Rightarrow \frac{1}{2}\frac{m^2{v}_0^2}{{(\frac{M}{3}+m)}^2}=(m+\frac{M}{2})gL(1-cos\theta )$

$\therefore cos\theta =1-\frac{1}{2}\frac{m^2{v}_0^2}{[\frac{M}{3}+m][\frac{M}{2}+m]gL}$

$\Rightarrow \theta =co{s}^{-1}[1-\frac{mgh}{\left\{\right(\frac{M}{3})+3\}\left\{\right(\frac{M}{2})+m\}gL}]$

Gain in potential energy of the system