Question

A uniform circular disc A of radius r is made from a metal plate of thickness t and another uniform circular disc B of radius 4r is made from the same metal plate of thickness t/4. If equal torques act the discs A and B, initially both being at rest. At a later instant, the angular speeds of a point on the rim of A and another point on the rim of B are ${\omega }_A$ and ${\omega }_B$ respectively. Then we have:

• A. ${\omega }_A>{\omega }_B$
• B. ${\omega }_A={\omega }_B$
• C. ${\omega }_A<{\omega }_B$
• D. The relation depends on the actual magnitude of the torques.

Answer 1

The correct option is A ${\omega }_A>{\omega }_B$

Torque is equal in both cases. Let ${\alpha }_1$ and ${\alpha }_2$ be the angular acceleration of first and second disk and $I_1$ and $I_2$ be the moment of inertia of first and second displacement.

Thus,

$I_1{\alpha }_1=I_2{\alpha }_2$

$\left(\frac{{\alpha }_1}{{\alpha }_2}\right)=\left(\frac{I_1}{I_2}\right)⟶\left(1\right)$

Now mass of first disk $M_1=\pi P^{2}t.\rho$

and Of second disk $M_2=\pi (4r{)}^2\frac{t}{4}=4\pi r^2\rho t$

Here $P$ is the density of metal plate

So, $I_1=M_1{r}^2=\pi \rho t.r^4$

$I_2=M_2(4r{)}^2=\pi \rho t.64r^4$

so, $\frac{{\alpha }_1}{{\alpha }_2}=64$

$\Rightarrow {\alpha }_1=64{\alpha }_2$

${\alpha }_1>{\alpha }_2$

Again angular speed

$w=\alpha t$

Hence,

$\frac{{\omega }_A}{{\omega }_B}=\frac{{\alpha }_1}{{\alpha }_2}=64\Rightarrow {\omega }_A>{\omega }_B$