A circular disc of moment of inertia $3.5 \mathrm{~kg} \mathrm{~m}^2$ is rotating with angular speed $30…
A circular disc of moment of inertia $3.5 \mathrm{~kg} \mathrm{~m}^2$ is rotating with angular speed $30 \mathrm{rad} \mathrm{s}^{-1}$ about an axis passing through its centre and perpendicular to its plane. The torque required to stop the disc in 5 seconds is
$84 \mathrm{Nm}$
$42 \mathrm{Nm}$
$10.5 \mathrm{Nm}$
$21 \mathrm{Nm}$
Solution
Moment of inertia of circular disc, $I=3.5 \mathrm{kgm}^2$ Angular speed, $\omega=30 \mathrm{rad} / \mathrm{s}$
The torque required to stop the disc in 5 seconds
$\tau=\frac{\mathrm{I} \omega}{\mathrm{t}}=\frac{3.5 \times 30}{5}=21 \mathrm{Nm}$