Question

A dumb-bell consists of two identical small balls of mass $1/2kg$ each connected to the two ends of a $50cm$ long light rod. The dumb-bell is rotating about a fixed axis through the center of the rod and perpendicular to it at an angular speed of $10rad/s$. Am the impulsive force of average magnitude $5.0N$ acts on one of the masses in the direction of its velocity for $0.10s$. Find the new angular velocity of that system.

Answer 1

Given: A dumb-bell consists of two identical small balls of mass $\frac{1}{2}$kg each connected to the two ends of a 50cm long light rod. The dumb-bell is rotating about a fixed axis through the center of the rod and perpendicular to it at an angular speed of 10rad/s. Am the impulsive force of average magnitude 5.0N acts on one of the masses in the direction of its velocity for 0.10s.

To find the new angular velocity of that system

Solution:

Moment of inertia of the dumb-bell

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

And Torque is

$\tau =I\alpha ⟹F\times r=\left(2m{r}^2\right)\times \alpha ⟹\alpha =\frac{Fr}{2m{r}^2}$

Substituting the values given mass, $m=\frac{1}{2}=0.5kg$, $r=\frac{50}{2}cm=0.25m$, we get

$⟹\alpha =\frac{5\times 0.25}{2\times 0.5\times (0.25{)}^2}⟹\alpha =20rad/s^2$

Now given,

${\omega }_0=10rad/s$ and $t=0.10s$

Using the formula,

$\omega ={\omega }_0+\alpha$ and substituting the corresponding values, we get

$\omega =10+20\times 0.10⟹\omega =10+2=12rad/s$

is the new angular velocity of that system