Question

A disc of radius R has a light pole fixed perpendicular to the disc at its periphery which in turn has a pendulum of length R attached to its other end as shown in figure. The disc is rotated with a constant angular velocity $\omega$. The string is making an angle ${45}^o$ with the rod. Then the angular velocity $\omega$ of 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. ${\left[\frac{\sqrt{2}g}{(\sqrt{2}+1)R}\right]}^{1/2}$

Answer 1

Answer: (D)

${\left[\frac{\sqrt{2}g}{(\sqrt{2}+1)R}\right]}^{1/2}$

$\begin{array}{c}\text{Distance of mass from the axis of Rotation is}\\ d=R+\frac{R}{\sqrt{2}}=\frac{R(\sqrt{2}+1)}{\sqrt{2}}\\ \text{By for a balancing we get-}\\ \begin{array}{cc}& \frac{m{\omega }^{2}d}{\sqrt{2}}=\frac{mg}{\sqrt{2}}\\ & \Rightarrow {\omega }^{2}d=g\\ & \Rightarrow \omega ={\left(\frac{g}{d}\right)}^{1/2}={\left[\frac{\sqrt{2}g}{(\sqrt{2}+1)R}\right]}^{1/2}\\ \text{.}& \end{array}\end{array}$