A disc of mass $1\ \text{kg}$ and radius $R$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of the disc. A body of the same mass as that of the 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 $4\sqrt{\dfrac{x}{3R}}\ \text{rad s}^{-1}$ where $x$ = ______.
Numerical value type. Enter your answer.
Answer: 5
The body falls through $2R$. Moment of inertia: $I = \dfrac12(1)R^2 + (1)R^2 = \dfrac32R^2$.
Taking $g = 10\ \text{m s}^{-2}$:
$$1\times g\times2R = \frac12\cdot\frac32R^2\omega^2\ \Rightarrow\ \omega^2 = \frac{8g}{3R} = \frac{80}{3R}$$
$$\omega = \sqrt{\frac{16\times5}{3R}} = 4\sqrt{\frac{5}{3R}}\ \Rightarrow\ x = 5$$
Solution by Sreeraj P, M.Sc Physics