Question

A dumb-bell consists of two identical small balls of mass $\frac{1}{2}$ kg each connected to the two ends of a 50 cm long light rod. The dumb-bell is rotating about a fixed axis through 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

$\tau =Ia$

$\Rightarrow F\times r=(m{r}^2+m{r}^2)\alpha$

$\Rightarrow 5\times 0.25=2m{r}^2\times \alpha$

$\Rightarrow \alpha =\frac{1.25}{2\times 0.5\times 0.025\times 0.25}$

= $20$

${\omega }_0=10rad/s$

$t=0.10sec$

$\omega ={\omega }_0+\alpha t$

$\Rightarrow \omega =10+0.10\times 20$

$=10+2=12rad/s$