# Solve the following :

Question:

A metre stick weighing $240 \mathrm{~g}$ is pivoted at its upper end in such a way that it can freely rotate in a vertical plane through the end. A particle of mass $100 \mathrm{~g}$ is attached to the upper end of the stick through a light string of length I $\mathrm{m}$. Initially, the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick there. Find the maximum angle through which the stick will rise.

Solution:

Velocity of ball just before the collision:-

$\frac{1}{2} m u^{2}=m g h$

$\mathrm{u}=\sqrt{2 g h}$

$u=\sqrt{2 \times 10 \times 1}$

$\mathrm{u}=\sqrt{20} \mathrm{~m} / \mathrm{sec}$

Now, Torque about hinge is zero

$L_{i}=L_{f}$

mul+0 $=l \omega$

$(0.1 \times \sqrt{20} \times 1)=\left(m l^{2}+\frac{M L^{2}}{3}\right) \omega$

$\left(0.1^{\times \sqrt{20}} \times 1\right)=\left[(0.1)^{\left.(1)^{2}+\frac{(0.24)(1)^{2}}{3}\right] \omega}\right.$

$\omega=\frac{\sqrt{20}}{(10)(0.18)}$

Now, By energy conservation

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

$\frac{1}{2}\left[(0.1)(1)^{2}+\frac{(0.24)(1)^{2}}{3}\right]\left[\frac{\sqrt{20}}{(10)(0.18)}\right]^{2}=0.1 \times g \times 1(1-\cos \theta)+\frac{0.24 \times 10 \times 1}{2}[1-\cos \theta]$

$\theta=41^{\circ}$