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Ethidium
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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 = d2θ/dt2
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(ML2)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 = d2(θ)/dt2.
So, Mg(L/2)cosθ = ⅓ML2(d2θ/dt2)
⇒ secθ d2θ/dt2 = (3g)/(2L)
Am I required to solve this differential equation?
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