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 the kinetic energy of the plate at the end of the fifth second after the start.

Answer 1

Given:
Torque acting on the plate = $\tau =0.10\mathrm{N}-\mathrm{m}$
Mass of the plate = $m=2\mathrm{kg}$
$$\begin{gathered} \mathrm{On}\mathrm{applying}\tau =I\alpha ,\mathrm{we}\mathrm{get}: \\ \frac{m{r}^2}{12}\times \alpha =0.10\mathrm{N}-\mathrm{m} \\ \Rightarrow \alpha =60\mathrm{rad}/\mathrm{s} \end{gathered}$$

Let ω be the angular velocity after time t (t = 5 s).
$$\begin{gathered} \mathrm{Therefore},\mathrm{we}\mathrm{have}: \\ \omega ={\omega }_0+at \\ \Rightarrow \omega =60\times 5=300\mathrm{rad}/\mathrm{s} \\ \mathrm{Angular}\mathrm{momentum}, \\ I\omega =\left(\frac{0.10}{60}\right)\times 300 \\ =0.50\mathrm{kg}-m^2/s \\ \mathrm{Kinetic}\mathrm{energy}, \\ \frac{1}{2}I{\omega }^2=\frac{1}{2}\times \left(\frac{0.10}{60}\right)\times {300}^2 \\ =75\mathrm{joule} \end{gathered}$$