A disc of mass 1 kg and radius R is free to 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 of fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, it angular speed will be 4x3Rrad⁡s−1, where x = __

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A disc of mass 1 kg and radius R is free to 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 of fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, it angular speed will be $4 \sqrt{\frac{x}{3 R}} rad s^{- 1}$ , where x = ___. (g = 10 ms^–2)

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answer is 5.

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Detailed Solution

Loss in P.E = Gain in K.E.
$\begin{array}{l} 2 mgR = \frac{1}{2} [\frac{1}{2} {mR}^{2} + {mR}^{2}] w^{2} \\ 2 mgR = \frac{1}{2} \times \frac{3}{2} {mR}^{2} w^{2} \\ w^{2} = \frac{8 g}{3 R} \\ w = \sqrt{\frac{8 g}{3 R}} = 4 \sqrt{\frac{g}{2 \times 3 R}} \\ \Rightarrow x = \frac{g}{2} = 5 \end{array}$