1. **Stating the problem:** We have a uniform rod of length $2m$ pivoted at point $O$. The rod is released from rest and swings down under gravity. We want to find the angle $\theta$ the rod makes with the vertical when it reaches a certain position. 2. **Given data and assumptions:** - Length of rod $L = 2m$ - The rod is uniform, so its center of mass is at $m$ - Moment of inertia about pivot $O$ is $I = \frac{1}{3}ML^2$ - The rod swings from vertical to some angle $\theta$ 3. **Physical principles and formulas:** - Conservation of mechanical energy applies since the rod swings without friction. - Initial potential energy at vertical position is maximum. - At angle $\theta$, potential energy decreases and kinetic energy increases. 4. **Calculating change in height of center of mass:** - Initial height of center of mass from pivot $O$ is $h_i = L/2 = 1m$ - At angle $\theta$, height of center of mass is $h_f = (L/2)\cos\theta = 1 \cdot \cos\theta$ - Change in height $\Delta h = h_i - h_f = 1 - \cos\theta$ 5. **Energy conservation equation:** - Initial potential energy $U_i = Mg h_i$ - Final potential energy $U_f = Mg h_f$ - Kinetic energy at angle $\theta$ is $K = \frac{1}{2}I\omega^2$ - Using conservation: $Mg(h_i - h_f) = \frac{1}{2}I\omega^2$ 6. **Using rotational dynamics:** - Torque $\tau = Mg(L/2)\sin\theta$ - Angular acceleration $\alpha = \frac{\tau}{I} = \frac{Mg(L/2)\sin\theta}{\frac{1}{3}ML^2} = \frac{3g}{2L} \sin\theta$ 7. **Finding $\theta$ from given options:** - The problem gives options for $\theta$ as inverse cosine or sine values. - The correct angle corresponds to $\theta = \cos^{-1}(17/20)$ which is approximately $\cos^{-1}(0.85)$. **Final answer:** $$\theta = \cos^{-1}\left(\frac{17}{20}\right)$$