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(C) We treat the contact point as a hinge.Applying the work-energy theorem:$W_g = \Delta K.E.$$Mg \left( \frac{L}{2} (1 - \cos 60^{\circ}) \right) = \

Vedclass · Accepted Answer

$1, 3, 4$

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(C) We treat the contact point as a hinge.Applying the work-energy theorem:$W_g = \Delta K.E.$$Mg \left( \frac{L}{2} (1 - \cos 60^{\circ}) ight) = \frac{1}{2} I \omega^2$$Mg \left( \frac{L}{2} \cdot \frac{1}{2} ight) = \frac{1}{2} \left( \frac{ML^2}{3} ight) \omega^2$$\frac{MgL}{4} = \frac{ML^2}{6} \omega^2 \Rightarrow \omega = \sqrt{\frac{3g}{2L}}$ (Statement $3$ is correct).Radial acceleration of $C.M.$: $a_r = \left( \frac{L}{2} ight) \omega^2 = \frac{L}{2} \cdot \frac{3g}{2L} = \frac{3g}{4}$ (Statement $1$ is correct).Using $	au = I \alpha$ about the contact point:$Mg \left( \frac{L}{2} \sin 60^{\circ} ight) = \left( \frac{ML^2}{3} ight) \alpha$$\alpha = \frac{3g}{2L} \sin 60^{\circ} = \frac{3g}{2L} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}g}{4L}$ (Statement $2$ is correct).Net vertical acceleration of $C.M.$: $a_v = a_r \cos 60^{\circ} + a_t \cos 30^{\circ}$$a_v = \left( \frac{3g}{4} ight) \left( \frac{1}{2} ight) + \left( \alpha \frac{L}{2} ight) \cos 30^{\circ} = \frac{3g}{8} + \left( \frac{3\sqrt{3}g}{4L} \cdot \frac{L}{2} ight) \frac{\sqrt{3}}{2} = \frac{3g}{8} + \frac{9g}{16} = \frac{15g}{16}$.Applying $F_{net} = Ma_v$ in the vertical direction:$Mg - N = M \left( \frac{15g}{16} ight) \Rightarrow N = \frac{Mg}{16}$ (Statement $4$ is correct).Thus,statements $1, 3, 4$ are correct.

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