Question

A system consists of two identical small balls of mass $2kg$ each connected to the two ends of a $1m$ 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 $9rad/s$. An impulsive force of average magnitude $10N$ acts on one of the masses in the direction of its velocity for $0.20s$. Calculate the new angular velocity of the system.

Answer 1

Ggiven,

$Mass=2kg,d=1m,{\omega }_0=9rad/s,f=10N,t=0.20s$

We know that the MI of the ball is:

$I=m{r}^2+m{r}^2=2m{r}^2$

Since $\tau =I\alpha$

So, $f\times r==\left(2m{r}^2\right)\alpha \Rightarrow \frac{40}{2\times 2\times 4^2}=\frac{40}{64}=0.625rad/s$

Now we have,

${\omega }_0=9rad/s,t=0.20s$

$\omega ={\omega }_0+\alpha t=9+0.625\times 0.20=9.125$