Question

A ring of radius 1 m is rolling without over a rough horizontal surface with a velocity $v_0=\sqrt{2}m/s$. Two points are located at A and B on the rim of the ring what is the speed of A w.r.t. B in m/s

• A. $\sqrt{2}m/s$
• B. 1 m/s
• C. 2 m/s
• D. $2\sqrt{2}m/s$

Answer 1

The correct option is C 2 m/s

$\begin{array}{cc}\text{Given}-& \ast \text{Radius}=\mathrm{lm}\left(r\right)\\ & \ast \text{velocity of centre}=\sqrt{2}m/s\left(v_0\right)\end{array}$

$\begin{array}{c}\text{For point}A\to \\ \text{we howe}-\\ V_A=V_0\hat{i}+r\omega \hat{i}\end{array}$

$\begin{array}{cc}& \text{Since}{v}_0=\text{how}\omega =\text{angular velocity}\\ \Rightarrow v_A& =2v_0\hat{i}\end{array}$

$\begin{array}{c}\text{For point}B\to \\ \text{we have -}\\ \begin{array}{cc}{V}_B& =V_0\hat{i}+\left(\mu \omega \right)(-\hat{j})\\ & =v_0\hat{i}-v_0\hat{j}\end{array}\end{array}$

$\begin{array}{c}\text{Now relative velocity of}A\text{wret.}B\text{is}-\\ \begin{array}{cc}{\overline{v}}_{AB}& ={\overline{V}}_A{\overline{V}}_B=2V_0\hat{i}(v_0\hat{i}-v_0\hat{j})\\ & =v_0\hat{i}+v_0\hat{j}\\ \left|V_{AB}\right|=\sqrt{2}{v}_0=\sqrt{2}\times \sqrt{2}m/s& =2m/s\end{array}\\ \text{Henu, option (c) is correct.}\end{array}$