Question

An impulse $J=mv$ is applied at one end of a stationary uniform frictionless rod of mass $m$ and length $l$ which is free to rotate in a gravity-free space. The impact is elastic. Instantaneous axis of rotation of the rod will pass through:

• A. its centre of mass
• B. the centre of mass of the rod plus ball
• C. the point of impact of the ball on the rod
• D. the point which is at a distance $\frac{2l}{3}$ from the striking end

Answer 1

The correct option is D the point which is at a distance $\frac{2l}{3}$ from the striking end

$J=mv=m{v}^2$

$\Rightarrow$ Velocity of the CM of rod $=v$

Applying impulse momentum equation about the CM of rod

$J\frac{1}{2}=I_{CM}\omega \Rightarrow \frac{mvl}{2}=\left(\frac{m{l}^2}{12}\right)\omega \Rightarrow \omega =\frac{6V}{l}$

About instantaneous axis of rotation the rod is considered to have pure rotation.

Let instantaneous axis of rotation be located at a distance x from the colliding end [Fig (b)]

$\frac{\frac{\omega l}{2}-v}{1-x}=\frac{v+\frac{\omega l}{2}}{x}$ ....(i)

Substituting the value of $\omega =6V/l$ in Eq. (i), we get $x=\frac{2}{3}l$.