A disc of mass 1 kg and radius R is a free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A particle of same mass that of disc is fixed at the highest point of the disc .now the system is released, when the particle comes to lowest position, its angular speed will be 4x3R rads-1 where x is equal to (g =10 m/s2)

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A disc of mass 1 kg and radius R is a free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A particle of same mass that of disc is fixed at the highest point of the disc .now the system is released, when the particle comes to lowest position, its angular speed will be $4 \sqrt{\frac{x}{3 R}}$ rads^-1where $x$ is equal to (g =10 m/s²)

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Detailed Solution

$\begin{array}{l} mg (2 R) = \frac{1}{2} (I + {mR}^{2}) ω^{2} \\ 4 (1) gR = (\frac{{mR}^{2}}{2} + {mR}^{2}) ω^{2} \\ 4 gR = \frac{3}{2} R^{2} ω^{2} \\ ω = \sqrt{\frac{8 g}{3 R}} = \sqrt{\frac{80}{3 R}} \Rightarrow ω = 4 \sqrt{\frac{5}{3 R}} \Rightarrow x = 5 \end{array}$