Question

A uniform rod of mass M and length L is pivoted at point A. It is struck by a horizontal force which believers delivers an impulse J at a distance x from point A as shown in figure, such that the impulse delivered by pivot is zero. If the angular velocity of the rod just after the impulse given is $\frac{6J}{nML}$. Find the value of n

• A. $\frac{12J}{M{L}^2\omega }$
• B. $0$
• C. $\frac{18J}{M{L}^2\omega }$
• D. $\frac{72J}{M{L}^2\omega }$

Answer 1

The correct option is A $\frac{12J}{M{L}^2\omega }$

Solution

Conserving the angular momentum about pivot

$Jx=m{l}^2\omega /3also,V_{{}_{cm}}=\omega l/2m{v}_{cm}(x-l/2)=m{l}^2\omega /12=0x=\frac{2l}{3}\frac{6J}{nML}=\frac{\omega L}{2}n=\frac{12J}{M{L}^2\omega }$