A uniform rod of mass $m$ and length $2a$ is held inclined to the vertical at an angle $\alpha$ with its lower end in contact with a smooth horizontal table and is released from rest. Prove that when the inclination of the rod to the vertical is $\theta$, the reaction of the table is $$\frac{mg(4-6\cos{\alpha}\cos{\theta}+3\sin^2{\theta})}{(1+3\sin^2{\theta})^2} $$
I got reaction of the table as $\dfrac{mg(4-6\cos{\alpha}\cos{\theta}+3\cos^2{\theta})}{(1+3\sin^2{\theta})^2} $ which is not what we need to prove but I think my answer is correct, If $\alpha=0$ and $\theta=0$ then rod is vertical so reaction must be $mg$ which is satisfy my answer but $\dfrac{mg(4-6\cos{\alpha}\cos{\theta}+3\sin^2{\theta})}{(1+3\sin^2{\theta})^2} $ is not satisfied
Can anyone help me for this? Am I correct?
I got the same result as you.
The motion of the rod can be understood as translation of the center od mass (midpoint of rod) and rotation around the center of mass. As there are no horizontal forces acting on the center of mass, the center of mass only moves vertically. If we denote its height by $y$, the conservation of energy gives $$\frac12 m\dot{y} + \frac12 I\omega^2 + mgy = mgy(0).$$ where the moment of inertia is $I = \frac{ma^2}3$ and $\omega = \dot{\theta}$ is angular rotation around the center of mass. We can note that $y = a\cos\theta$ so $\dot{y} = -a\omega\sin\theta$. Plugging this in we get $$\frac12ma^2\omega^2\sin^2\theta + \frac16ma^2\omega^2 + mga\cos\theta=mga\cos\alpha $$ which implies $$\omega = \sqrt{\frac{6g(\cos\alpha - \cos\theta)}{a(1+3\sin^2\theta)}}.$$
Now, since the table is smooth, the reaction force $\vec{N}$ acts only upwards. Its projection orthogonally to the rod is $N\sin\theta$. The resulting torque $Na\sin\theta$ drives the rotation of the rod. The rotational equation of motion is $$N\sin\theta = I\dot{\omega}.$$ After some calculation, we get $$\dot\omega = \sqrt{\frac{6g}{a}} \frac1{2\sqrt{\frac{\cos\alpha - \cos\theta}{1+3\sin^2\theta}}}\frac{(4-6\cos\alpha\cos\theta +3\cos^2\theta)\sin\theta}{2(1+3\sin^2\theta)^2} \omega = \frac{3g(4-6\cos\alpha\cos\theta +3\cos^2\theta)\sin\theta}{2a(1+3\sin^2\theta)^2}$$ so $$N = I\frac{\dot\omega}{\sin\theta} = \frac{ma^2}3\frac{3g(4-6\cos\alpha\cos\theta +3\cos^2\theta)}{2a(1+3\sin^2\theta)^2} = \frac{mg(4-6\cos\alpha\cos\theta +3\cos^2\theta)}{(1+3\sin^2\theta)^2}.$$