The uniform rod AB of weight W is released from rest when be

The uniform rod AB of weight W is released from rest when beta = 70degree. Assuming that the friction force is zero between end A and the surface, determine immediately after release the angular acceleration of the rod. the acceleration of the mass center of the rod, the reaction at A.

Solution

Here, it needs to be understood that the rod will suffer no horizontal force, hence once it is released, the centre of mass of the rod will come down vertically and the left end will start to rotate so as to ensure that when it hits the ground, the rod is flat.

Now, while falling down the rod will have a torque about its left end due the gravitational pull acting of the COM.

We will use this torque to find the angular acceleration of the rod.

Part A.) Now, W(L/2)(Cos70) = (WL²/3g)a

or, a = (3g/2L)Cos70 is the required angular acceleration of the rod.

Part B.) Now the angular acceleration of the rod to be a means that the linear acceleration of the centre of the rod

would be aL/2 or (3g/4)Cos70

Part C.) We can write for the vertical direction that: W - N = (W/g)(Linear acceleration)

or, N = W - (W/g)(3g/4)Cos70

or, N = W(1 - 0.75Cos70) is the required normal reaction