Question

A system consists of two identical small balls of mass $2$ kg each connected to the two ends of a $1$ m long light rod. The system is rotating about a fixed axis through the centre of the rod and perpendicular to it at angular speed of $9\,rad\,{\sec ^{ - 1}}$ . An impulsive force of average magnitude $10\,N$ acts on one of the masses in the direction of its velocity for $0.20\,s$ . Calculate the new angular velocity of the system.

Answer 1

Hint: Initially the system was in equilibrium as no external torque was acting on it. Calculate the moment of inertia of the system with respect to the axis of rotation. The impulsive force will provide some torque to the system and thus its angular acceleration will increase. Due to its angular velocity will also increase. Calculate the angular acceleration and then find the angular velocity.

Complete step by step answer:
The mass of the identical balls is given as, $m = 2\,kg$
Length of the rod, $l = 1\,kg$
Magnitude of the impulsive force, $f = 10\,N$
Time for which the force acts, $t = 0.20\,s$
The new angular velocity changes due to the change in angular acceleration provided by the torque. We know that torque $\tau $ is equal to the product of moment of inertia ${\rm I}$ and the angular acceleration $\alpha $ .
The moment of inertia of the system along the axis of rotation will only include the moment of inertia due to the identical balls as the moment of inertia of the rod will be zero along the axis of rotation which coincides with the centre of the rod.
The moment of inertia of the balls be given as:
$I = m{r^2} + m{r^2}$
$I = 2m{r^2}$
Here, $r$ is the distance of the balls from the axis of rotation, we have $r = \dfrac{d}{2} = \dfrac{1}{2}m$
As, $\tau = {\rm I}\alpha $ and $\tau = F \times r$ , we can have:
$F \times r = I\alpha $
$ \Rightarrow \alpha = \dfrac{{F \times r}}{I}$
Substituting the values, we get
$\alpha = \dfrac{{10 \times \left( {\dfrac{1}{2}} \right)}}{{2 \times 2 \times {{\left( {\dfrac{1}{2}} \right)}^2}}}$
$ \Rightarrow \alpha = 5\,rad\,{s^{ - 1}}$
The initial speed is given as: ${\omega _0} = 9\,rad\,{\sec ^{ - 1}}$ , $t = 0.20\,s$ and $\alpha = 5\,rad\,{s^{ - 1}}$
Thus, the new angular velocity $\omega $ of the system will be:
$\omega = {\omega _0} + \alpha t$
 $ \Rightarrow \omega = 9 + 5 \times \left( {0.2} \right)$
$ \Rightarrow \omega = 10\,rad\,{s^{ - 1}}$
The new angular velocity of the system is $\omega = 10\,rad\,{s^{ - 1}}$

Note:
When some unbalanced external force acts on a system performing rotational motion, torque acts on the system and thus the mass or the angular acceleration of the system changes. Torque is the cross product of force and the distance from the axis of rotation. Also, torque is given as a moment of inertia multiplied by the angular acceleration of the system.