Question

A solid disc of radius $20\text{cm}$ and mass $10\text{kg}$ is rotating with an angular velocity of $600\text{rpm}$, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc to rest in $10\text{s}$ is $\pi \times {10}^{-1}\text{Nm}$.

Answer 1

Torque is defined as,

$\tau =\frac{\Delta L}{\Delta t}=\frac{I({\omega }_f-{\omega }_i)}{\Delta t}$

Here, for the given axis of rotation, $I=\frac{M{R}^2}{2}$

$\tau =\frac{\left(\frac{M{R}^2}{2}\right)\times [0-\omega ]}{\Delta t}$

Converting, $1\text{rpm}=\frac{2\pi }{60}{\text{rad s}}^{-1}$

$\Rightarrow \tau =\frac{10\times (20\times {10}^{-2}{)}^2}{2}\times \frac{600\times 2\pi }{60\times 10}$

$\Rightarrow \tau =0.4\pi =4\pi \times {10}^{-1}\text{Nm}$

Hence, $4$ is the correct answer.