A solid cylinder of mass \(M\) and radius \(R\) rolls down an inclined plane of length \(L\) and height…

1. \(\sqrt{2 g h}\)
2. \(\sqrt{\frac{3 g h}{4}}\)
3. \(\sqrt{\frac{4 g h}{3}}\)
4. \(\sqrt{4 g h}\)

Solution

The given situation is shown in the following figure.
When cylinder reaches at bottom, then its whole potential energy is converted into its rotational kinetic energy and linear kinetic energy of its centre of mass. $\begin{aligned} \text{Hence, } & M g h=\frac{1}{2} I \omega^{2}+\frac{1}{2} M v_{\text{COM}}^{2} \\ &=\frac{1}{2} \times \frac{M R^{2}}{2} \times\left(\frac{v_{\text{COM}}}{R}\right)^{2}+\frac{1}{2} M v_{\text{COM}}^{2} \\ & \quad\left[\because I=\frac{M R^{2}}{2} \text{ and } \omega=\frac{v_{\text{COM}}}{R}\right] \\ &=\frac{M v_{\text{COM}}^{2}}{4}+\frac{M v_{\text{COM}}^{2}}{2} \\ & \Rightarrow \quad M g h=\frac{3}{4} M v_{\text{COM}}^{2} \end{aligned}$ $\begin{aligned} \Rightarrow & v_{\text{COM}}^{2}=\frac{4 g h}{3} \\ \Rightarrow & v_{\text{COM}}=\sqrt{\frac{4 g h}{3}} \end{aligned}$

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

Practice more Rotational Motion questions on Aicharya