- #1

Ethidium

- 3

- 1

## Homework Statement

A uniform rod is released on a friction-less horizontal surface from rest. Initially, the rod makes an angle θ = 60° with the horizontal. What is the acceleration of the center of mass of the rod, just after release?

## Homework Equations

Torque = Moment of Inertia × Angular acceleration

Angular acceleration α = dω/dt = d

^{2}θ/dt

^{2}

Mg - N = Ma, where M is the mass of the rod, a is the acceleration about the rod's center of mass and N is the normal reaction.

## The Attempt at a Solution

The only two forces acting on the rod: the normal reaction by the surface and the weight of the rod.

Finding torque about the point of contact with the surface, we get: (M is mass of the rod, L is it's length)

Mg(L/2)cosθ = ⅓(ML

^{2})α ..... (1)

Also if we balance forces, we get: (a is the acceleration of centre of mass of the rod, N is the normal reaction)

Mg - N = Ma ...(2)

Now I realize that the axis of rotation is not fixed and thus wonder how to go about using equation (1). I also cannot use a condition of pure rotation such as a = rα. I'm guessing I will get a second order differential equation by virtue of the fact that α = d

^{2}(θ)/dt

^{2}.

So, Mg(L/2)cosθ = ⅓ML

^{2}(d

^{2}θ/dt

^{2})

⇒ secθ d

^{2}θ/dt

^{2}= (3g)/(2L)

Am I required to solve this differential equation?

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