Question

A uniform ring and a uniform disc of same mass and radius are rotating with same angular speed. If a constant force is applied tangentially on the ring in opposite direction, it stops after n revolution. If the same force is applied on disc in opposing sense tangentially then the number of revolution made by it before it stops, is:

• A. $\frac{n}{2}$
• B. $n$
• C. $2n$
• D. $4n$

Answer 1

The correct option is A $\frac{n}{2}$

According to the concept of torque and by kinematics equation
${\tau }_{ring}={\tau }_{disc}$ $I_{Ring}{\alpha }_{ring}=I_{disc}{\alpha }_{disc}$ $m{r}^2{\alpha }_{ring}=\frac{1}{2}m\frac{r^2}{2}r{\alpha }_{disc}{\alpha }_{disc}$$=2_{\alpha }ring$.

Hence, for ring it will be ${\omega }^2={\omega }_0^2+2\alpha \theta$ ${\omega }^2=0+2\alpha \left(2\pi n\right)$ $\alpha ring=\frac{{\omega }_0^2}{4\pi n}$

But according to the relation of
${\omega }^2=0+2.2\alpha \left(2\pi n\right)$
$n_{disc}=\frac{n_{ring}}{2}$