00:01
In this question we're going to find the moment of inertia of this rod of negligible mass with three masses attached to it, and then we are going to find the angular velocity when the rod is allowed to swing around the pivot point a.
00:18
So to get our moment of inertia, we need to take the sum of each mass times its distance from the axis of rotation.
00:31
So we're going to have an m at a distance d squared, or i forgot to say, it's mr squared.
00:37
And then we're going to have another m on the right side at a distance d squared.
00:41
And then we're going to have a mass at a distance of 2d squared.
00:47
So our moment of inertia, this is part a not part d, there we go, is going to be 2md squared plus 4md squared for a total of 6md squared.
01:06
And now when we allow this rod to swing, we are going to apply conservation of energy to this.
01:17
And real quick, i'm going to sketch in, let me put part b here, i'm going to sketch this in.
01:27
And we're going to have this rod that drops down like this, right? and we're going to notice that the mass on the left is actually going to lift to a distance d above where it is now.
01:42
The innermost mass on the right side is going to drop a distance d.
01:51
And then our outermost mass is going to drop a distance 2d.
01:58
So when we apply conservation of energy, we have gravitational potential energy at the start equals the rotational kinetic energy at the end.
02:08
And obviously the maximum angular velocity is going to be when it swings through the bottom through this vertical position, just in case that wasn't clear, that omega max will be here.
02:22
So we can say that the two masses on the right are elevated above where they're going to be.
02:29
So we can have mgd plus mg times 2d.
02:39
And then, oh, i forgot...