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,

\[\tau = I\alpha\]

\[\Rightarrow   F \times r = \left( m r^2 + m r^2 \right)  \alpha\]

\[ \Rightarrow 5 \times 0 . 25   =   2m r^2  \times \alpha\]

\[ \Rightarrow \alpha = \frac{1 . 25}{2 \times 0 . 5 \times \left( 0 . 25 \right)^2} = 20\text{ rad/s}^2 \]

Given

\[\text{Using }\omega = \omega_0 + \alpha t\text{, we get}\]
\[\omega = 10 + 20 \times 0 . 10 = 10 + 2 = 12\text{ rad/s}\]