Question 159 of 280

A disc of mass $1kg$ and radius $R$ is free of rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $4\sqrt{\dfrac{x}{3R}}rad{s}^{-1}$ where $x=$ _____ .

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Answer: 5

Let the angular speed of disc be $\omega$.

Using conservation of mechanical energy
$mg2R=\dfrac{1}{2}{I}_{disc}{\omega }^{2}+\dfrac{1}{2}{I}_{particle}{\omega }^{2}$
Where, $I$ is moment of inertia.
$\Rightarrow mg2R=\dfrac{{\omega }^{2}}{2}\left[\dfrac{{m}^{2}}{2}+m{R}^{2}\right]$
$\Rightarrow mg2R=\dfrac{{\omega }^{2}}{2}\left(\dfrac{3}{2}m{R}^{2}\right)$
$\Rightarrow \dfrac{3}{4}{\omega }^{2}=\dfrac{2g}{R}$
$\Rightarrow {\omega }^{2}=\dfrac{8g}{3R}$
Thus, angular speed is $\omega =\sqrt{\dfrac{80}{3R}}rad{s}^{-1}$
Given $\omega =4\sqrt{\dfrac{x}{3R}}$
Comparing both, $16\dfrac{x}{3R}=\dfrac{80}{3R}$
Therefore, the value of $x=5$.