# Rod slipping against block due to gravity? [closed]

A uniform rod of mass $m$ and length $l$ is pivoted at point O. The rod is initially in vertical position and touching a block of mass M which is at rest on a horizontal surface. The rod is given a slight jerk and it starts rotating about point O. This causes the block to move forward. The rod loses contact with the block at $\theta = 30^\circ$. All surfaces are smooth.

1. The value of $M/m$?
2. The velocity of block when the rod loses contact with the block?
3. The acceleration of center of mass of rod, when it loses contact with the block?
4. The hinge reaction at O on the rod when it loses contact with the block?

Here is my attempt at the problem

The component of angular speed of the tip of the rod in horizontal direction is same as that of the block.

Let V be the speed of the block when rod loses contact with block. So,Horizontal component of angular speed of the tip of the rod = $\omega l\cos(60)=V$

So, $\omega l/2=V$ Thus, $\omega = 2V/l$

Applying Conservation of Energy,

Loss in Potential Energy of Rod = Gain in Rotational Kinetic Energy of Rod + Gain in KE of Block

$$\frac{Mgl}{2}(1-\sin60^\circ) = \frac{1}{2}\frac{ml^2}{3}(ω ^2) +\frac{1}{2}MV^2$$

$$\frac{Mgl}{2}(1-\sin60^\circ) = \frac{1}{2}\frac{ml^2}{3}\frac{4V ^2}{l^2} +\frac{1}{2}MV^2$$

$$\frac{Mgl}{4} = \frac{2mV^2}{3} +\frac{1}{2}MV^2$$

The forces on the rod are

Hinge force in x direction = $H_{x}$ Hinge force in y direction = $H_{y}$ Force due to gravity = mg Normal contact force from the block = $N$

The force on the block are

Normal contact force from the rod =$N$

Now, I am not able to proceed further.

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## closed as off-topic by Brandon Enright, tpg2114, jinawee, Qmechanic♦Nov 28 '13 at 0:55

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Is there friction here? – ja72 May 14 '13 at 12:11
All surfaces are smooth – Tanya Sharma May 16 '13 at 12:55

You already solved the first two parts. The acceleration of the centre of mass is probably calculated as follows: $a_c=\frac{\alpha l}{2}$ and substitute $\frac{3mg\cos(30)}{ml^2}$ for alpha.
For the last part, split the angular acceleration of the centre of mass in two directions. For the vertical component:- $\frac{(mg−H_y)}{m}=3mg\cos(30)∗l∗\frac{√3}{4ml^2}$ and in horizontal direction the weight component dissapears $\frac{H_x}{m}=\frac{3mg\cos(30)}{4ml^2}$