A ring is rolling without slipping on a plane surface as shown in figure. If speed of A is 4 m/s, speed of B is

If $\omega$ be its angular velocity and since O is the instantaneous centre of rotation, then ${\text{V}}_{\text{A}}\text{=ω×OA}$ and 

1. Rolling Without Slipping: In rolling motion without slipping, the velocity of the point of contact with the surface (let's call it $P$) is zero relative to the surface.
2. Velocity at Different Points on the Ring:
   • The center of mass (COM) of the ring has a translational velocity $v$.
   • The topmost point of the ring (point $B$) has a velocity equal to $v + R\omega$, where $\omega$ is the angular velocity of the ring and $R$ is its radius.
   • The bottommost point of the ring (point $A$) has a velocity of $v - R\omega$.
3. Rolling Condition: The rolling without slipping condition implies $v = R\omega$.

Given Data:

• Speed of point $A$: $4 \, \text{m/s}$.

For point $A$, the velocity relative to the ground is given as $v - R\omega = 4 \, \text{m/s}$. Using the rolling condition $v = R\omega$, we get:

v−v=4m/s,

which implies:

v=4m/s.

Speed of Point $B$$B$:

At the topmost point $B$, the velocity is:

vB​=v+Rω.

Substituting $R\omega = v$:

vB​=v+v=2v.

With $v = 4 \, \text{m/s}$:

vB​=2×4=8m/s.${\text{V}}_{\text{B}}\text{=ω×OB}$

$\therefore \frac{{\mathrm{V}}_{\mathrm{B}}}{{\mathrm{V}}_{\mathrm{A}}}=\frac{\mathrm{\omega }\times \mathrm{OB}}{\mathrm{\omega }\times \mathrm{OA}}=\frac{\sqrt{2}\mathrm{R}}{2\mathrm{R}}\Rightarrow \frac{{\mathrm{V}}_{\mathrm{B}}}{\mathrm{V}}=\frac{1}{\sqrt{2}}\Rightarrow {\mathrm{V}}_{\mathrm{B}}=2\sqrt{2}\mathrm{m}/\mathrm{s}$