Question

A slender rod of mass $M$ and length $L$ rests on a horizontal frictionless surface. The rod is pivoted about one of its ends. The impulse of the force exerted on the rod by the pivot when the rod is struck by a blow of impulse J perpendicular to the rod at other end is :

• A. $J$
• B. $\frac{J}{2}$
• C. $\frac{J}{3}$
• D. Information is insufficient

Answer 1

The correct option is B $\frac{J}{2}$

Let J' be the impulse exerted by the pivot on the rod.
Then from impulse momentum theorem, $J+J^{\prime }=m{v}_0$, where $v_0$ is the velocity acquired by the centre of mass.
Now apply, angular impulse $=$ change in angular momentum about the pivot.
$J\times L=\frac{M{L}^2}{3}\omega$
$J=\frac{ML}{3}\omega$,
where $\omega$ is the angular velocity of the rod.
$v_0=\frac{\omega L}{2}\Rightarrow M{v}_0=\frac{M\omega L}{2}$

$J+J^{\prime }=\frac{3J}{2}\Rightarrow J^{\prime }=\frac{J}{2}$