Question

A horizontal disk is rotating with angular velocity 'w' about a vertical axis passing through its centre. A ball is placed at the centre of groove and pushed slightly. The velocity of ball when it comes out of the groove –

• A. $\frac{\sqrt{3}}{2}$wR
• B. $\frac{\mathbf{\omega R}}{\mathbf{2}}$
• C. wR
• D. $\frac{\mathbf{\omega R}}{\sqrt{\mathbf{2}}}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}}{2}$wR

Answer 2

Let us consider motion of ball with respect to disk

Net force along groove = mw²r sin q

= mw²r $\frac{x}{r}$

= mw²x

$\therefore$ma = mw²x

̃ v $\frac{dv}{dx}$ = w²x

̃ v = $\frac{\omega R}{2}$

${\overrightarrow{v}}_{Ball,Ground}$ = ${\overrightarrow{v}}_{Ball,Disk}$+

${\overrightarrow{v}}_{Ball,Ground}$

\ ${\overrightarrow{v}}_{Ball,Ground}$

=$\left\{ (\omega R)^{2} + \left( \frac{\omega R}{2} \right)^{2} + 2(\omega R)\left( \frac{\omega R}{2} \right)\cos{}120{^\circ} \right\}^{\frac{1}{2}}$=$\frac{\sqrt{3}}{2}\omega R$