Question

A dumb-bell consists of two identical small balls of mass 1/2 kg each connected to the two ends of a 50 cm long light rod. The dumb-bell is rotating about a fixed axis thorough the centre of the rod and perpendicular to it at an angular speed of 10 rad/s. An impulsive force of average magnitude 5⋅0 N acts on one of the masses in the direction of its velocity for 0⋅10 s. Find the new angular velocity of the system.

Answer 1

Moment of inertia of the dumb-bell,
$I=m{r}^2+m{r}^2=2m{r}^2$
Torque, $\tau =I\alpha$
$$\begin{gathered} \Rightarrow \mathit{}F\times r=\left(m{r}^2+m{r}^2\right)\alpha \\ \Rightarrow 5\times 0.25=2m{r}^2\times \alpha \\ \Rightarrow \alpha =\frac{1.25}{2\times 0.5\times {\left(0.25\right)}^2}=20\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Given:
${\omega }_0=10\mathrm{rad}/\mathrm{s}\mathrm{and}t=0.10\mathrm{s}$
$$\begin{gathered} \mathrm{Using}\omega ={\omega }_0+\alpha t,\mathrm{we}\mathrm{get}: \\ \omega =10+20\times 0.10=10+2=12\mathrm{rad}/\mathrm{s} \end{gathered}$$