Question

A uniform ring of radius $R$ is given a back spin of angular velocity $V_0/2R$ and thrown on a horizontal rough surface with velocity of center to be $V_0$. The velocity of the centre of the ring when it starts pure rolling will be

• A. $V_0/2$
• B. $V_0/4$
• C. $3V_0/4$
• D. $0$

Answer 1

The correct option is B $V_0/4$

Given: A uniform ring of radius R is given a back spin of angular velocity $\frac{V_0}{2R}$ and thrown on a horizontal rough surface with velocity of center to be $V_0$.

To find the velocity of the centre of the ring when it starts pure rolling

Solution:

Let the surface be enough rough to provide rolling after certain time.

Conserve the angular momentum,

Let the mass of the ring be M and pure rolling with $\omega$ after certain time.

Then,

$M{R}^2\times \frac{V_0}{2R}-MR{V}_0=M{R}^2\omega +MR\left(R\omega \right)⟹\frac{V_0}{2R}-\frac{V_0}{R}=2\omega ⟹\omega =-\frac{V_0}{4R}$

Here negative sign indicates that the rotation would be clockwise and the ring will move forward.

Hence $v=R\omega =\frac{V_0}{4}$

is the velocity of the centre of the ring when it starts pure rolling