Question

Two points $A$ and $B$ on a disc have velocities $v_1$ and $v_2$ at some moment. Their directions make angles ${60}^o$ and ${30}^o$ respectively with the line of separation as shown in the figure. The angular velocity of the disc is:

• A. $\frac{\sqrt{3}{v}_1}{d}$
• B. $\frac{v_2}{\sqrt{3}d}$
• C. $\frac{v_2-v_1}{d}$
• D. $\frac{v_2}{d}$

Answer 1

The correct option is D $\frac{v_2}{d}$


For rigid body, seperation between two point remains the same.
Therefore, $v_1\cos {60}^{\circ }=v_2\cos {30}^{\circ }$
$\frac{v_1}{2}=\frac{\sqrt{3}{v}_2}{2}\Rightarrow v_1=\sqrt{3}{v}_2$

${\omega }_{disc}=\left|\begin{array}{c}\frac{v_2\sin {30}^{\circ }-v_1\sin {60}^{\circ }}{d}\end{array}\right|=\left|\begin{array}{c}\frac{\frac{v_2}{2}-\frac{\sqrt{3}{v}_1}{2}}{d}\end{array}\right|$
$=\left|\begin{array}{c}\frac{v_2-\sqrt{3}\times \sqrt{3}{v}_2}{2d}\end{array}\right|=\frac{2v_2}{2d}=\frac{v_2}{d}$
$\Rightarrow {\omega }_{disc}=\frac{v_2}{d}$