Question

A disc of radius $R$ has a light pole fixed perpendicular to the disc at the circumference which in turn has a pendulum of length $R$ attached to its other end as shown in the figure. The disc is rotated with a constant angular velocity $\omega$. The string is making an angle ${30}^0$ with the rod. The angular velocity $\omega$ of the disc is :

• A. ${\left(\frac{\sqrt{3}g}{R}\right)}^{1/2}$
• B. ${\left(\frac{\sqrt{3}g}{2R}\right)}^{1/2}$
• C. ${\left(\frac{g}{\sqrt{3}R}\right)}^{1/2}$
• D. ${\left(\frac{2g}{3\sqrt{3}R}\right)}^{1/2}$

Answer 1

The correct option is D ${\left(\frac{2g}{3\sqrt{3}R}\right)}^{1/2}$

The bob of the pendulum moves in a circle of radius
$(R+Rsin{30}^0)=\frac{3R}{2}$
Force equations :
$Tsin{30}^0=m\left(\frac{3R}{2}\right){\omega }^2$
$Tcos{30}^0=mg$
$\Rightarrow tan{30}^0=\frac{3}{2}\frac{{\omega }^{2}R}{g}$
$=\frac{1}{\sqrt{3}}\Rightarrow \sqrt{\frac{2g}{3\sqrt{3R}}}$