Question

A uniform rod of mass m and length l is struck at an end by a force F perpendicular to the rod for a short time interval t. Calculate (a) the speed of the centre of mass, (b) the angular speed of the rod about the centre of mass, (c) the kinetic energy of the rod and (d) the angular momentum of the rod about the centre of mass after the force has stopped to act. Assume that t is so small that the rod does not appreciably change its direction while the force acts.

Answer 1

Given

Mass of the rod = m

Length of the rod = l

(a) For the centre of mass,

Acceleration,

\[a = \frac{F}{m}\]

\[\text{Velocity, }v = at\] 

\[ \Rightarrow v = \frac{Ft}{m}\]

(b) Let the angular speed about the centre of mass be ω.

Moment of inertia of rod about centre of mass = \[I = \frac{m l^2}{12}\]

\[I\omega = mvr\]

\[\Rightarrow \frac{m l^2}{12} \times \omega = m\nu\frac{l}{2}\]

\[ \Rightarrow \frac{m l^2}{12} \times \omega = m \times \frac{Ft}{m} \times \frac{l}{2}\]

\[ \Rightarrow \omega = 6\frac{Ft}{ml}\]

(c) Kinetic energy,

\[K . E .  = \frac{1}{2}m \nu^2  + \frac{1}{2}I \omega^2 \]

\[= \frac{1}{2} \times m \left( \frac{Ft}{m} \right)^2  + \frac{1}{2} \times I \omega^2 \]

\[= \frac{1}{2} \times m\left( \frac{F^2 t^2}{m^2} \right) + \frac{1}{2} \times \frac{m l^2}{12} \times \left( \frac{36 F^2 t^2}{m^2 l^2} \right)\]

\[= \frac{2 F^2 t^2}{m}\]

(d) Angular momentum about the centre of mass,

\[L = m\nu r\]

\[     = m \times \frac{Ft}{m} \times \left( \frac{l}{2} \right) = \frac{Flt}{2}\]