MHTCET 25 Apr 2025 Shift 2 Paper

Concept:The swinging rod behaves like a physical pendulum pivoted at one end.At the lowest position, its angular speed is maximum and the rod possesses only rotational kinetic energy.At the highest position, the rod momentarily comes to rest and all this kinetic energy is converted into gravitational potential energy of the centre of mass.Explanation:At the lowest point, the rod is vertical and its angular speed is maximum, equal to $\omega$.For a uniform rod of mass $M$ and length $L$, pivoted at one end, the moment of inertia is:$$I = \frac{1}{3}ML^{2}$$The maximum rotational kinetic energy is:$$K = \frac{1}{2}I\omega^{2} = \frac{1}{2}\left(\frac{1}{3}ML^{2}\right)\omega^{2} = \frac{1}{6}ML^{2}\omega^{2}$$At the highest point of swing, the angular speed becomes zero, so the entire kinetic energy converts into potential energy.If the centre of mass rises by a vertical height $h$, the gain in potential energy is:$$\Delta U = Mgh$$Applying conservation of mechanical energy:$$\frac{1}{6}ML^{2}\omega^{2} = Mgh$$Cancelling the mass $M$ from both sides gives:$$h = \frac{L^{2}\omega^{2}}{6g}$$Answer:The maximum height risen by the centre of mass is $\frac{L^{2}\omega^{2}}{6g}$.This matches option D.