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. (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.

Answer 1

(a) Angular momentum = mvr Conservation of linear momentum, mu = 2mv - mv = mv

$\therefore$ u = v = mvr

Velocity = $\sqrt{2gh}$

and r = $\frac{L}{2}$

Angular momentum

=m. $\sqrt{2gh}$.$\frac{L}{2}$

= $\frac{mL\sqrt{gh}}{\sqrt{2}}$

Angular momentum = lw

$\therefore$ $\omega$ = angular velocity = $\frac{L}{I}$

or I = $\frac{2m{L}^2}{4}+\frac{m{L}^2}{4}=\frac{3m{L}^2}{4}$

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

(b) When the mass 2m will be at the top most position and the mass m be at the lowest point. They will automaticaly rotate. In this position the total gain in potential energy

$=2mg\times \left(\frac{L}{2}\right)-mg\left(\frac{L}{2}\right)$

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

$\Rightarrow h=\frac{3L}{2}$