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.)

 

[Moetasim]

I would point out that the method used in the solution is correct. The equation used, (total I)*(alpha) = (mg*a) + (0.5mg * 2a), is the equation for the sum of torques, which is valid for any rotating object regardless of the torque being constant or not. The key is that the equation takes into account the changing torque by using the total moment of inertia, which includes the mass and distribution of mass of both the rod and the attached particle.

Using energy equations to find the final angular velocity is also a valid approach, but it may be more complicated and time-consuming compared to using the sum of torques equation. Additionally, finding the final angular velocity does not necessarily provide information about the angular acceleration, which is what the question is asking for.

The angular acceleration in this case is not constant, but it is still a useful and valid quantity to calculate. It represents the rate of change of the angular velocity, and can be used to predict the motion of the object in the future. In fact, most objects do not have constant angular acceleration, but this does not make the concept any less useful or applicable.

In summary, the method used in the solution is correct and valid, and the equation used (sum of torques) is applicable to any rotating object, regardless of the torque being constant or not. Using energy equations or finding the final angular velocity may also be valid approaches, but they may be more complicated and may not provide the same information as the angular acceleration.