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## Homework Statement

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 θ = 30°. 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?

Ans 1) 4:3

2)√(3gl)/4

3) (3g)/4

4)(mg/4)j

## Homework Equations

## The Attempt at a Solution

The 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 speed of the tip of the rod = V

Component of speed perpendicular to length of the rod =[itex]V_{\bot}[/itex] = VSin30°=V/2

Angular speed of the rod at the instant block leaves contact with the rod =[itex]\frac{V_{\bot}}{l}[/itex] = [itex]\frac{V}{2l}[/itex]

Applying Conservation of Energy ,

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

[itex]\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}(ω ^2) +\frac{1}{2}MV^2 [/itex]

[itex]\frac{Mgl}{2}(1-Sin60°) = \frac{1}{2}\frac{ml^2}{3}\frac{V ^2}{4l^2} +\frac{1}{2}MV^2 [/itex]

[itex]\frac{Mgl}{4} = \frac{mV^2}{24} +\frac{1}{2}MV^2 [/itex]

Is my approach correct ? Now ,how am I supposed to proceed further ? one more equation is required to solve the problem ...