Question

A rod of length $l$ forming an anglc $\theta$ with the horizontal strikes a frictionless floor at A with its centre of mass velocity $v_0$ and no angular velocity. Assuming that the fimpact atA is perfectly elastic. Find the angular velocity of the rod immediately after the impact

Answer 1

Let $v=$ lincar velocity of rod after impact (upwards),
$\omega =$ angular velocity of rod and $J=$ linear impulse at A during impact
Then, $J=\Delta P=P_f-P_i$
$J=mv-(-m{v}_0)$
$J=m(v+v_0).....\left(i\right)$
Angular impulse $=\Delta L$
$\therefore J\left(\frac{l}{2}\cos \theta \right)=I\omega =\frac{m{l}^2}{12}\omega$ .....(ii)
Collision is elastic $(e=1)$
$\therefore$ Relative speed of approach $=$ Relative speed of separation at point of impact
$v_0=v+\frac{l}{2}\omega \cos \theta ....\left(iii\right)$
Solving above equations, we get
$\omega =\frac{6v_0\cos \theta }{l(1+3{\cos }^2\theta )}$