Question

Equal torques act on the discs $A$ 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 $v_A$ and $v_B$ respectively. We have

• A. $v_A>v_B$
• B. $v_A=v_B$
• C. $v_A<v_B$
• D. The relation depends on the actual magnitude of the torques

Answer 1

Answer: (A)

$v_A>v_B$

Given,

${\tau }_A={\tau }_B$

$I_A{\omega }_A=I_B{\omega }_B$

if, ${\omega }_A=\omega$

then,

${\omega }_B=\frac{\omega }{64}$

so,

$I_A\omega =64I{\omega }_B$

now,

linear velocity: $v_A=\omega r$ and

$v_B=\frac{\omega \left(4r\right)}{64}=\frac{\omega r}{16}=\frac{v_A}{16}$

hence,

$v_A>v_B$