Question

A uniform vertical rod of mass $m$ and length $2l$ is hinged at its highest point and is initially at rest. A small pendulum of mass $m$ and string length $l$, is also suspended from the same point. Initially it is at an angle 60${}^0$ with vertical and then it is released. The pendulum strikes the rod, just after collision, the pendulum comes to rest, the angular speed of the rod becomes $\omega$ and coefficient of restitution of this collision is e, then:

• A. $\omega =\frac{2}{3}\sqrt{\frac{g}{l}}$
• B. $e=\frac{3}{4}$
• C. $e=\frac{2}{3}$
• D. $\omega =\frac{3}{4}\sqrt{\frac{g}{l}}$

Answer 1

Answer: (B), (D)

The ball will strike the rod with a velocity of
$v=\sqrt{2gR(1-\cos \theta )}\Rightarrow v=\sqrt{gl}$
Applying angular momentum conservation about the hinge point
$m\left(\sqrt{gl}\right)l+0=0+\left(\frac{m{\left(2l\right)}^2}{3}\right)\left(\omega \right)$ get $\omega =\frac{3}{4}\sqrt{\frac{g}{l}}$