Physics - Rotation Question with Solution | TestHub

Initial angular velocity ${\omega }_0=300\text{rpm}=\frac{300\times 2\pi }{60}=10\pi \text{rad/s}$

Final angular velocity $\omega =0$

Time $t=20\text{s}$

Using the equation of rotational motion $\omega ={\omega }_0+\alpha t$:

$0=10\pi +\alpha (20)$

$\alpha =-0.5\pi {\text{rad/s}}^2$

Moment of inertia of the solid disc about its central axis is $I=\frac{1}{2}M{R}^2$

$I=\frac{1}{2}\times 20\times (0.4{)}^2=1.6\text{kg}{\text{m}}^2$

Magnitude of retarding torque $\tau =I|\alpha |$

$\tau =1.6\times 0.5\pi =0.8\pi \text{Nm}$

Comparing with $\tau =x\pi \text{Nm}$, we get $x=0.8$.

Answer: $0.8$