Question

A constant external torque $\tau$ acts for a very brief period $\Delta t$ on a rotation system having moment of inertia $I$. Then :

• A. The angular momentum of the system will change by $\tau \Delta t$
• B. The angular velocity of the system will change by $\frac{\tau \Delta t}{I}$
• C. If the system was initially at rest, it will acquire rotational kinetic energy $\frac{(\tau \Delta t{)}^2}{2I}$
• D. The kinetic energy of the system will change by $\frac{(\tau \Delta t{)}^2}{2I}$

Answer 1

Answer: (A), (B), (C)

The angular momentum of the system will change by $\tau \Delta t$
The angular velocity of the system will change by $\frac{\tau \Delta t}{I}$
If the system was initially at rest, it will acquire rotational kinetic energy $\frac{(\tau \Delta t{)}^2}{2I}$

We know that torque in terms of angular momentum is given as $\tau =\frac{dL}{dt}$. Thus for time period $\Delta t$ the angular momentum would be given as $\tau \Delta t$. Thus option A is correct. Also from the relation $\tau =\frac{I\omega }{\Delta t}$ we get $\omega$ as $\frac{\tau \Delta t}{I}$. Thus option B is correct. Also the rotational kinetic energy is given as $\frac{1}{2}I{\omega }^2$ which gives $\frac{(\tau \Delta t{)}^2}{2I}$.Thus option C is correct. The change in kinetic energy may also have translational component, thus option D is wrong.