A constant external torque `tau` acts for a very brief period `/_\t` on a rotating system having moment of inertia `I` then

Options

A. The angular momentum of the system will change by `tau Delta t`
B. The angular velocity of the system will change by `(tau Delta t)/I`
C. If the system was intially at rest, it will acquire rotational kinetic anergy `(tau Delta t)^(2)/(2I)`
D. The kinetic energy of the system will change by `(tau Delta t)^(2)/I`

Answer

A, B, C

Explanation

`Iomega=tauDeltat`
hence `1/2Iomega^(2)=(tauDeltat)^(2)/(2I)`