Question

Equal torques act on the disc A and B of the previous problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of A and another point on the rim of B are ${\nu }_A$ and ${\nu }_B$ respectively. We have
(a) ${\nu }_A>{\nu }_B$

(b) ${\nu }_A={\nu }_B$

(c) ${\nu }_A<{\nu }_B$

(d) the relation depends on the actual magnitude of the torques.

Answer 1

(a) ${\nu }_A>{\nu }_B$

$\tau =I\alpha \left(\mathrm{magnitude}\right)$
For equal torque, we have:
$I_A{\alpha }_A=I_B{\alpha }_B$

I_A < I_B
⇒ ${\alpha }_A>{\alpha }_B$ ...(i)

Now, $\omega =\alpha t$
Or, $\frac{v}{r}=\alpha t$
${\nu }_A>{\nu }_B$ (using (i))