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\text{Hint: } \vec{\tau}_{\text{net}} = I\vec{\alpha} \text{Step 1: Use the torque equation for the rod.} \text{The rod is hinged at one end and makes an angle } \theta \text{ with the vertical. The weight } mg \text{ acts at the center of mass (at } \frac{l}{2} \text{ from the hinge). The perpendicular distance from the hinge to the line of action of weight is } \frac{l}{2}\sin\theta. \vec{\tau} = I\vec{\alpha} mg \cdot \frac{l}{2}\sin\theta = \frac{ml^2}{3}\alpha \alpha = \frac{3g\sin\theta}{2l} \text{Step 2: Find the acceleration of the free end of the rod assuming it is performing circular motion.} \text{For circular motion, the tangential acceleration at the free end is:} a_T = R\alpha \text{where } R = l \text{ (distance from hinge to free end)} a_T = l \times \frac{3g\sin\theta}{2l} = \frac{3g\sin\theta}{2} \text{At the instant of release, the tangential velocity } v_T = 0\text{, so the radial acceleration is zero. Therefore, the total acceleration of the free end is purely tangential and equals } \frac{3g\sin\theta}{2}.