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

#### Solution

(a) Angular momentum \[= m\nu r\]

Let the particle P collides the ball B with a speed *u* and system moves with speed *v *just after the collision.

Applying the law of conservation of linear momentum, we get

\[mu = 2m\nu - m\nu = m\nu\]

\[\therefore u = \nu\]

\[\text{Velocity, }u = \sqrt{2gh}\]

\[\text{and }r = \frac{L}{2}\]

Initial angular momentum of system about COM of rod,

\[mur = m \times \sqrt{2gh} \times \frac{L}{2}\]

\[= \frac{mL\sqrt{gh}}{\sqrt{2}}\]

Angular momentum of system about COM of the rod just after the collision \[= l\omega\]

\[I = \frac{2m L^2}{4} + \frac{m L^2}{4} = \frac{3m L^2}{4}\]

Applying the law of conservation of angular momentum and obtaining the value of *ω*, we get

\[\omega = \frac{mL\frac{\sqrt{gh}}{\sqrt{2}}}{\frac{3m L^2}{4}} = \frac{\sqrt{8gh}}{3L}\]

(b) When the mass 2*m* and *m* are at the top most position and at the lowest point, respectively, they will automatically rotate. In this position, we have

Total gain in potential energy

\[= 2mg\left( \frac{L}{2} \right) - mg\left( \frac{L}{2} \right)\]

Kinetic energy = \[\frac{1}{2}I \omega^2\]

Therefore, by the law of conservation of energy, we have

\[mg\frac{L}{2} = \frac{1}{2} \times \frac{3}{4}m L^2 \times \left( \frac{8gh}{9g L^2} \right)\]

\[ \Rightarrow h = \frac{3L}{2}\]