Question

A small disc and a thin uniform rod of length $l$, whose mass is $\eta$ times greater than the mass of the disc, lie on a smooth horizontal plane. The disc is set in motion, in horizontal direction and perpendicular to the rod, with velocity $v$, after which it elastically collides with the end of the rod. Find the velocity of the disc and the angular velocity of the rod after the collision. At what value of $\eta$ will the velocity of the disc after the collision be equal to zero? Reverse its direction?

Answer 1

For the closed system (disc$+$rod) the angular momentum is conserved about any axis. Thus from the conservation of angular momentum of the system about the rotation axis of rod passing through its $C.M$ gives:

$mv\frac{l}{2}={mv}^{\prime }\frac{l}{2}+\frac{\eta {ml}^2}{12}w-\left(1\right)$

($v^{\prime }$ is the final velocity of the disk and $w$ is the angular velocity of the rod)

For closed system, linear momentum is also conserved.

Hence,

$mv={mv}^{\prime }+\eta {mv}_c-\left(2\right)$ $[v_c\to$velocity of $cm$ of rod$]$

From $\left(1\right)$ and $\left(2\right)$

$v_c=\frac{lw}{3}$

$v-v^{\prime }=\eta v_c$

Applying conservation of $K.E$, as the collision is elastic

$\frac{1}{2}{mv}^2=\frac{1}{2}{mv\prime }^2+\frac{1}{2}\eta {mv}_c^2+\frac{1}{2}\frac{\eta {ml}^2}{12}{w}^2-\left(3\right)$

or, $v^2-{v\prime }^2=4\eta v_c^2$ and hence $v+v\prime =4v_c$

Then,

$v\prime =\frac{4-\eta }{4+\eta }v$ and $w=\frac{12v}{(4+\eta )l}$

Vectorically nothing that we have taken $\overrightarrow{v}\parallel \overrightarrow{v\prime }$

$\overrightarrow{u}=\left(\frac{4-\eta }{4+\eta }\right)\overrightarrow{v}$

So $\overrightarrow{u^{\prime }}=0$ for $\eta =4$ and $\overrightarrow{u^{\prime }}\downarrow \uparrow \overrightarrow{v}$ for $\eta >4$