Uniform rotating rod about a point at one end of the rod

[Kennedy]

Homework Statement

A uniform rod (mass = 5.0 kg, length = 0.3 m; the rotational inertia of the rod relative to the axis through its centre of mass is I = (1/12) ML^2 ) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60° below the horizontal? (a) 15.5 rad/s^2 (b) 24.5 rad/s^2 (c) 28.5 rad/s^2 (d) 29.5 rad/s^2 (e) 33.5 rad/s^2

2. Homework Equations
T = Iα

The Attempt at a Solution

I believe that I have almost solved this problem all the way through, but I'm missing something, and my brain is telling me that it has something to do with the parallel axis theorem. So, I know that the only force that is responsible for the rotation of the rod is due the gravity, but since we want to know the torque at exactly the moment that the rod makes an angle of 60 degrees with the horizontal, the force of gravity acting on the rod at exactly this instant is sin60(5)(9.8) = 42.44 N. Of course, because this is a rod of uniform mass, this force acts on the centre of mass of the rod. So, the torque would be 42.44(0.15)(sin30) = 3.18 N*m, because 30 degrees is the angle between the force vector and the position vector of the rod. Now, using T = Iα, it should be fairly easy to figure out the angular acceleration, but I'm having some problems calculating the rotational inertia. The rotational inertia formula is given for when the point of rotation is through the centre of mass of the rod (in this case right in the middle), but I don't have anything telling me the rotational inertia for when the point of rotation is at one end of the rod. I assume that my professors would want me to use the parallel axis theorem somehow, but I don't know how to go about doing this. As of right now, I have the rotational inertia through the centre of mass of the rod to be (1/12)(5)(0.3^2) = 0.0375 kg*m^2.

 

[Orodruin]

sin60(5)(9.8) = 42.44 N

Where did the sine come from? The gravitational force itself does not depend on the angle. Its component orthogonal to the moment arm does (as taken into account by the sine in the expression for the torque), but not the force itself.

I assume that my professors would want me to use the parallel axis theorem somehow, but I don't know how to go about doing this.

What about the parallel axis theorem confuses you? You have the moment of inertia about the centre of mass, which makes the parallel axis theorem directly give you the moment of inertia about any other point.

 

[Kennedy]

Where did the sine come from? The gravitational force itself does not depend on the angle. Its component orthogonal to the moment arm does (as taken into account by the sine in the expression for the torque), but not the force itself.What about the parallel axis theorem confuses you? You have the moment of inertia about the centre of mass, which makes the parallel axis theorem directly give you the moment of inertia about any other point.

Okay, so maybe that's where I screwed up. The force of gravity is 5(9.8) = 49 N regardless of the angle of the rod. So, the torque would be 5(9.8)(sin30)(0.15) = 3.675 N*m. Maybe I do understand the parallel axis theorem, then. So, I already know that the rotational inertia about the centre of mass, which is 0.0375 kg*m^2. Using the parallel axis theorem, I need to take the rotational inertia about the centre of the rod, and add the distance between the "new" point of rotation and the "old" point of rotation. Which yields, 0.0375 + (5)(0.15^2) = 0.15 kg*m^2. Which means that the angular acceleration at this exact moment is 3.675/0.15 = 24.5 rad/s^2. Is my logic and reasoning correct?

 

[Orodruin]

Maybe I do understand the parallel axis theorem, then. So, I already know that the rotational inertia about the centre of mass, which is 0.0375 kg*m^2.

I strongly suggest not plugging in any numbers until you have the final expression. Plugging in numbers unnecessarily will only lead to obstructing the actual physics from you and make it more difficult to find errors. Furthermore, it will lead to unnecessary computations and factors that may end up cancelling will enter twice and possibly lead to computational errors each time. Keep everything symbolically as long as you can, always.

and add the distance between the "new" point of rotation and the "old" point of rotation

The parallel axis theorem says nothing about "new" and "old" points. It says something about the relation of the moment of inertia relative to the centre of mass and the moment of inertia relative to some other point. Do not use it blindly without understanding this. As it just happens, your "new" and "old" were appropriately assigned in such a way that they became the appropriate ones. Whether or not you actually understood this or arrived at it by pure luck is difficult to judge. I therefore suggest that you consider the parallel axis theorem again with this in mind to make sure that you understood what is going on.

The end result is correct, but had you done it without inserting numbers at each step you would have also learned that the actual mass of the rod is irrelevant. This is also something that could have been inferred from dimensional analysis. The only combination of $m$, $\ell$, and $g$ that has the appropriate dimension is $g/\ell$ and the only question is what the prefactor is (i.e., you could immediately say that $\alpha = Cg/\ell$ for some dimensionless constant $C$ and then reduce your problem to finding $C$).