00:01
Hello, everyone.
00:02
In this problem, we're asked to find the period for a situation where we have a hoop that is suspended around its radius at some point, and then it's set off to oscillate about that point.
00:15
So first what we want to do is find the moment of inertia of this system.
00:22
And the reason we want to do that is because essentially what we're going to do is we're going to analyze the torque on the system to see how the angle that is suspended by the hoop with the center of mass and the pivot changes over time, which is going to allow us to find what the period is.
00:39
So first of all, what is the moment of inertia around the center of mass? so the pivot is at the center of mass for the hoop, which is if it's a uniform hoop, then it's exactly in the center, then this is given by mr squared.
00:52
Now, by the parallel axis theorem, if we then translate this pivot to the edge, or over to the sides or like actually onto the hoop, then we need to add an extra md squared accounting for that change where d is going to be the radius of the hoop, because that's how much we moved it.
01:12
So then the total moment of inertia is going to become, it's going to become i equals to m r squared from the center of mass one plus mr squared for the change.
01:23
And so we have that i is equal to 2mr squared.
01:28
So given that, we can now kind of look at a system and say, okay, if i pivoted my hoop over there and i kind of set it to oscillate, like i draw it out to some angle, some initial angle theta, then this is just happening, right? so there's a pivot there.
01:43
And then because it's an extended object, it's a rigid body now, it's not just a point, then it has a weight that is acting in the middle and it's pointing downwards.
01:52
And this is enclosing an angle of theta with this same line from the pivot and the center.
01:57
Of mass as the theta we had defined over here.
02:01
So then this has a component that is going to be acting towards like along this line, right? which is along the line of the pivot to the center of mass or to the center or to the point where any force is acting, which is all in this case is only the weight.
02:18
So if that's true, then this weight is going to have a component that is parallel to this axis, which means that it's not going to produce any torque.
02:26
But the other component, which is point perpendicular to that axis is going to produce a norm.
02:31
So what is that toward going to be? well, the component of the force that's acting that way is going to be minus mg sine theta.
02:39
It's minus because essentially theta increases in this direction.
02:46
So if i'm looking at the theta component, which is, you know, it's either going around the circle.
02:51
So if this is the angle or this is the direction of increasing theta half, so theta heads growing.
03:00
Then this weight is pointing opposite to that, right? this component of the weight is pointing opposite to that direction.
03:06
So that's why i get a minus sign.
03:09
Okay, so then given that, we can now write down our torque equation.
03:12
So we have that some of the torques acting on the object is equal to its moment of inertia times its angular acceleration...
