- #1

Kennedy

- 70

- 2

## 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

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.