Question

As shown in figure, a ring of radius $R$ is rolling without slipping. What will be the magnitude of velocity of point $A$ on the ring ?

• A. $V_{cm}$
• B. $\sqrt{2}{V}_{cm}$
• C. $\frac{V_{cm}}{2}$
• D. $2V_{cm}$

Answer 1

The correct option is B $\sqrt{2}{V}_{cm}$

Condition for pure rolling (rolling without slipping),
$V_{CM}=\omega R...\left(i\right)$
$V_{CM}=$ speed of COM of the ring
Velocity of point $A$ will be,
$\overrightarrow{V_A}={\overrightarrow{V}}_{\text{CM}}+{\overrightarrow{V}}_{A,CM}...\left(ii\right)$
where ${\overrightarrow{V}}_{A,CM}$, represents velocity of $A$ w.r.t centre of ring.
$|{\overrightarrow{V}}_{A,CM}|=\omega R$


From the figure shown,
$|\overrightarrow{V_A}|=\sqrt{V_{\text{CM}}^2+{\left(\omega R\right)}^2}$
Substituting from Eq $\left(i\right)$
$\therefore |\overrightarrow{V_A}|=\sqrt{V_{\text{CM}}^2+V_{\text{CM}}^2}=\sqrt{2}{V}_{\text{CM}}$