Question

A uniform bar of mass M and length L collides with a horizontal surface. Before collision, velocity of center of mass was $v_0$ and no angular velocity. Just after collision, velocity of center of mass of bar becomes v in upward direction as shown. Angular velocity $\omega$ of the bar just after impact is :

• A. $\frac{6(v_0+v)cos\theta }{L}$
• B. $\frac{6(v_0-v)cos\theta }{L}$
• C. $\frac{(v_0+v)cos\theta }{L}$
• D. $\frac{(v_0-v)cos\theta }{6L}$

Answer 1

Answer: (A)

$\frac{6(v_0+v)cos\theta }{L}$

Angular momentum will conserve about A just before and
just after collision.
$m{v}_0.\frac{L}{2}cos\theta =\frac{m{L}^2}{12}.\omega -mv.\frac{L}{2}cos\theta$
$\omega =\frac{6(v_0+v)cos\theta }{L}.$