Question

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

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

Answer 1

Answer: (A)

Let us consider the motion of the ball with respect to disk


$Netforcealonggroove=m{\omega }^{2}rsin\theta$
$=m{\omega }^{2}r\frac{x}{r}$
$=m{\omega }^{2}x$
$\therefore ma=m{\omega }^{2}x$
$\Rightarrow v\frac{dv}{dx}={\omega }^{2}x$
$\Rightarrow {\int }_0^{v}vdv={\int }_0^{R/2}{\omega }^{2}xdx$
$\Rightarrow \frac{v^2}{2}=\frac{{\left(\omega \right)}^2}{2}{\left(\frac{R}{2}\right)}^2$
$\Rightarrow v=\frac{\omega R}{2}$
${\overrightarrow{v}}_{Ball,Ground}={\overrightarrow{v}}_{Ball,Disk}+{\overrightarrow{v}}_{Disk,Ground}$
$\therefore \left|{\overrightarrow{v}}_{Ball,Ground}\right|$
$={\left(\left\{{\left(\omega R\right)}^2+{\left(\frac{\omega R}{2}\right)}^2+2\left(\omega R\right)\left(\frac{\omega R}{2}\right)cos120^{\circ }\right\}\right)}^{\frac{1}{2}}$
$=\frac{\sqrt{3}}{2}\omega R$