Question

A loop rolls on a horizontal ground without slipping with a linear speed $2\text{m/s}.$ Speed of a particle $\text{P}$ on the circumference of the loop at angle ${60}^{\circ }$ is

• A. $1\text{m/s}$
• B. $2\text{m/s}$
• C. $3\text{m/s}$
• D. $5\text{m/s}$

Answer 1

The correct option is B $2\text{m/s}$



From the figure, the speed of particle $\text{P}$ is,

$\left|\begin{array}{c}\overrightarrow{V_p}\end{array}\right|=\left|\begin{array}{c}\overrightarrow{V_1}\end{array}\right|+\left|\begin{array}{c}\overrightarrow{V_2}\end{array}\right|$

Here, $\overrightarrow{V_2}=V$ and

$\overrightarrow{V_1}=V(\because \text{Rolling without slipping})$

$\therefore \left|\begin{array}{c}\overrightarrow{V_p}\end{array}\right|=V_p=\sqrt{V^2+V^2+2V^2\cos ({180}^{\circ }-\theta )}$

$=\sqrt{2V^2(1-\cos \theta )}$

$=\sqrt{2V^2\times 2{\sin }^2\frac{\theta }{2}}$

$=\sqrt{4V^2{\sin }^2\frac{\theta }{2}}$

$\therefore V_P=2V\sin \frac{\theta }{2}$

Substituting the values, we have

$V_P=2\times 2\times \sin {30}^{\circ }=2\text{m/s}$

Hence, $\left(B\right)$ is the correct answer.