Horizontal rotating rod with another mass attached

[zinc79]

Homework Statement

A uniform rod AB, of length (2a) and mass (m) has a particle of mass (0.5m) attached to B. The rod is smoothly hinged at (A) to a fixed point and can rotate without resistance in a vertical plane. It is released from rest with AB horizontal. Find, in terms of (a) and (g), the angular acceleration of the rod when it has rotated through an angle of (pi/3).

Homework Equations

C = I*(alpha) = (Sum of torques)

alpha = angular acceleration
I = Moment of Inertia

The Attempt at a Solution

I of rod at (A) = (1/3)ma^2 + 1ma^2 = (4/3)ma^2
I of particle at (A) = 2(0.5m)(2a)^2 = 2ma^2
therefore total I at (A) = (10/3)ma^2

thus, (total I)*(alpha) = (mg*a) + (0.5mg * 2a)
which gives (alpha) = 3g/5a

The answer is (apparently) correct.

But the question I asked myself later is, how does this equation apply at all when the total torque is not constant? The wight acts downwards all the time, which means that the force, due to weight, on the rod + particle is perpendicular only at first when they were horizontal. As the object rotates downwards, the torque, i.e. (perpendicular force) * (distance of force from pivot) decreases as the perpendicular force acting becomes a decreasing fraction of the weight. This means that the angular acceleration is not constant.

If I am correct that the method applied above is faulty, then should I use energy equations of the initial and final state to find the final angular velocity? Is that correct? And even if I do find the final angular velocity, what form of angular acceleration am I supposed to find anyway when it (alpha) is not constant to being with?

 

[Doc Al]

As the rod falls, the torque and resulting angular acceleration are not constant. That's why you were asked to calculate the angular acceleration at one particular point. There's nothing wrong with the method--you can use it to find the angular acceleration at any point in the motion.

(Don't confuse angular acceleration with angular velocity.)