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 $\frac{t}{4}$. Equal torque acts on both the discs $A$ and $B$, initially both being at rest. If at a later instant, the angular speeds of disc $A$ and $B$ 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 B ${\omega }_A>{\omega }_B$

Given that torque is equal in both cases.
Let ${\alpha }_A$ and ${\alpha }_B$ be the angular acceleration of first and second disc.
Let $I_A$ and $I_B$ be the moment of inertia of first and second disc respectively.
Thus,
${\tau }_A={\tau }_B$
$\Rightarrow I_A{\alpha }_A=I_B{\alpha }_B$
$\Rightarrow \frac{{\alpha }_A}{{\alpha }_B}=\frac{I_B}{I_A}\dots \dots \left(1\right)$
Now, mass of first disc, $M_A=\pi r^{2}t\rho$
Mass of second disc, $M_B=\pi (4r{)}^2\left(\frac{t}{4}\right)\rho$
$M_B=4\pi r^2\rho t$
where $\rho \to$ density of metal plate
So, $I_A=\frac{M_A{r}_A^2}{2}=\left(\pi r^{2}t\rho \right)\frac{r^2}{2}=\frac{\pi r^{4}t\rho }{2}$

$I_B=\frac{M_B{r}_B^2}{2}=\frac{M_2(4r{)}^2}{2}=\frac{4\pi r^2\rho t\times 16r^2}{2}$
$\Rightarrow I_B=32\pi r^{4}t\rho$
So, from $\left(1\right)$, we get
$\frac{{\alpha }_A}{{\alpha }_B}=\frac{I_B}{I_A}=64$
$\Rightarrow {\alpha }_A=64{\alpha }_B$
$\therefore {\alpha }_A>{\alpha }_B$
Again, angular speed $\omega =\alpha t$
Hence, $\frac{{\omega }_A}{{\omega }_B}=\frac{{\alpha }_A}{{\alpha }_B}=64$
$\Rightarrow {\omega }_A>{\omega }_B$