Find the angular acceleration of the rod.

Question

A light rod of length a can turn freely about the end A which is fixed, and the other end B is attached to a particle of mass m. The end B is joined by a string of length a to a small ring C of mass m, which can slide on a smooth horizontal wire through A. The system is released from rest when AC=2a. Show that when CAB=60°, the angular acceleration of the rod is -g/16a

Answer 1

The easiest approach seems to be via conservation of energy, Quite of bit of differential calculus is involved.

Let Θ = the measure of angle CAB. Then the coordinate of the ring with respect to end A is

x = 2 a cos(Θ) , so its velocity is 2 a sin(Θ) ω (where ω is the time derivative of Θ)

The KE of the ring is thus (1/2) m4 a² sin²(Θ) ω²

The moment of inertial of the light rod with the mass m at the end is I = m a²

So the KE of the light rod plus mass is (1/2) m a² ω²

The only PE is that associated with the weight of the ring PE = - mg a sin(Θ)

The total system energy is (1/2) a² ω² + (1/2) m 4 a² sin²(Θ) ω² - mg a sin(Θ)

This is a conserved quantity with initial value of zero. Setting it to zero and solving for ω² gives

ω² = (g/a) sin(Θ) /[1/2 + 2 sin²(Θ) ]

The desired quantity is α , the time derivative of ω. The easiest way to get an expression for α is to differentiate both side of this equation with respect to time and then solve for α. In doing so it is noted that a factor of ω cancels out. This leads to a fairly simple expression involving α. It is:

2 α = (g/a) [ cos(Θ) / ( 1/2 + 2 sin²(Θ) ) - 4 sin²(Θ) cos(Θ) /(1/2 + 2 sin²(Θ) )² ]

All that remains is to substitute Θ = 60 deg. The result for α = - (g/a) (1/16)