Question

A uniform square plate of mass 2.0 kg and edge 10 cm rotates about one of its diagonals under the action of a constant torque of 0.10 N-m. calculate the angular momentum and kinetic energy of the plate at the end of the fifth second after start.

Answer 1

According to the question the square is rotated about diagonal axis

Then, moment of inertia about the diagonal

$I_{diag}=\frac{1}{12}M{L}^2=\frac{1}{12}\left(2\right)(10\times {10}^{-2})=\frac{1}{60}$

Also, the torque is given

$\tau =I\alpha$

$0.10=\frac{1}{60}\times \alpha$

$\alpha =6rad/s^2$

By equation of circular motion

$\omega ={\omega }_0+\alpha \left(t\right)$

$\omega =0+6\times 5$ (${\omega }_0=0$, because starts from rest)

$\omega =30rad/s$

Thus,

Angular momentum $=I\omega$

$=\frac{1}{60}\times 30=\frac{1}{2}=0.5Kg{m}^2/s$

Kinetic energy $=\frac{1}{2}I{\omega }^2$

$=\frac{1}{2}\times \frac{1}{60}\times {\left(30\right)}^2=\frac{15}{2}=7.5J$