Suppose the rod with the balls A and B of the previous problem is… - Vidyadip

Question

Suppose the rod with the balls $A$ and $B$ of the previous problem is clamped at the centre in such a way that it can rotate freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. $A$ particle $P$ of the same mass $m$ is dropped from a height $h$ on the ball $B.$ The particle collides with $B$ and sticks to it:

Find the angular momentum and the angular speed of the system just after the collision.
What should be the minimum value of $h$ so that the system makes a full rotation after the collision.

✓

Answer

The system is kept rest in the horizontal position and a particle $P$ falls from a height h and collides with $B$ and sticks to it.

Therefore, the velocity of the particle $'P\ '$ before collision $=\sqrt{2\text{gh}}$
If we consider the two bodies $P$ and $B$ to be a system. Net external torque and force $= 0$
Therefore, $\text{m}\sqrt{2\text{gh}}=\text{2m}\times\text{v}$
$\Rightarrow\text{v}'=\sqrt{\frac{(2\text{gh})}{2}}$
Therefore angular momentum of the rod just after the collision
$\Rightarrow\text{2m}(\text{v}'\times\text{r})=\text{2m}\times\sqrt{\frac{(2\text{gh})}{2}}\times\frac{\text{l}}2{}$
$\Rightarrow\text{ml}\sqrt{\frac{(2\text{gh})}{2}}$
$\omega=\frac{\text{L}}{\text{l}}=\frac{\text{ml}\sqrt{2\text{gh}}}{2\big(\frac{\text{ml}^2}{4}+\frac{2\text{ml}^2}{4}\big)}$
$=\frac{2\sqrt{\text{gh}}}{\text{3l}}=\frac{\sqrt{8\text{gh}}}{\text{3l}}$

When the mass $2m$ will at the top most position and the mass $m$ at the lowest point, they will automatically rotate. In this position the total gain in potential energy $=2\text{mg}\times\Big(\frac{\text{l}}{2}\Big)-\text{mg}\Big(\frac{\text{l}}{2}\Big)=\text{mg}\Big(\frac{\text{l}}{2}\Big)$

Therefore
$\text{mg}\frac{\text{l}}{2}=\frac{\text{l}}{2}\text{l}\omega^2$
$\Rightarrow\text{mg}\frac{\text{l}}{2}=\frac{\big(\frac{1}{2}\times3\text{ml}^2\big)}{4}\times\Big(\frac{\text{8gh}}{9\text{gl}^2}\Big)$
$\Rightarrow\text{h}=\frac{3\text{l}}{2}$

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