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13. Rotational Inertia & Energy

Moment of Inertia of Systems


Inertia of disc with point masses

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Hey, guys. So let's check out another example of a moment of inertia Problem. So here we have a solid disk, remember, a solid disk has the same moment of inertia as a solid cylinder. And when I tell you the shape I'm telling you which equation to use for I So I of the disk is half M R Square. Um, I'm also given that it has a radius of four. So r equals four in a mass of 10. Okay. And then we're gonna add to small objects on top of it. The fact that it's saying to small objects, and it's not giving me a shape for the object. It's an indication that these were going to be treated as point masses. Okay, And these are the two objects here. So I'm gonna call this, um one. So this is m one, and this is M two. Okay, says the object of the left has a massive 2 kg, and it's placed halfway between the discs center, so the center disc is here in the edge. Now, the distance between the center and the edge is the radius right, which is four. If you are halfway between the center and the edge. Your distances, Half the radius. Okay, so this distance here is half the race. So I'm gonna call this our one, because the distance for mass one and it is half of the radius, which to? And the other guy Thea Other object is 3 kg in mass 3 kg in its place at the edge of the disk. So if you're all the way at the edge, your distance let's call this. Our two is the same as the radius, which is four. Okay, so I'm giving you all the information you need to calculate the systems. Um, moments of inertia. Now, system is a combination of the disk with the masses. So I system is I won, plus I to right object one object to plus I disc And remember, for every one of these, you have to determine Is this a point mass or is this a shape? Well, one and two are point masses. We talked about this here, so I'm gonna write em one r one squared plus m two r two square and the disk is a shape and it has moment of inertia given by half m. R. square, half big M big r squared. Now all we gotta do is plug in the numbers. Very straightforward. So what I'm gonna do is plug the masses the masses air 12 and 10. So I'm sorry to three and 10 23 and 10. So I'm gonna do to three plus half of 10 or squared, okay? And then all we have to do is plug in the ours. This are Here is the radius. It's very straightforward. We know that the radius is a four. So I'm gonna put a four here now, These ours we have to slow down for a little bit. Thes air the distances between the center, um, the axis of rotation, which is in the center and where the object is little arms of distance between the object in the center. And we already had these figured out here. It's two and four. So the 2 kg has a 2 m distance, and the 3 kg has a 4 m distance. Okay, so let's just do this real quick. Um, this is going to be eight. This is going to be I'm three times 16. So that's 48. Andi, this is going to be 80. Okay, so we have eight plus 48 plus 80. And this is going to be 1 36. Um, 1 36 kg. Meter squared. Cool. That's it for this one. Hopefully got it. Let me know if you have any questions.

You build a wheel out of a thin circular hoop of mass 5 kg and radius 3 m, and two thin rods of mass 2 kg and 6 m in length, as shown below. Calculate the system’s moment of inertia about a central axis, perpendicular to the hoop.


A composite disc is built from a solid disc and a concentric, thick-walled hoop, as shown below. The inner disc (solid) has mass 4 kg and radius 2 m. The outer disc (thick-walled) has mass 5 kg, inner radius 2 m, and outer radius 3 m. Calculate the moment of inertia of this composite disc about a central axis perpendicular to the discs.


Three small objects, all of mass 1 kg, are arranged as an equilateral triangle of sides 3 m in length, as shown. The left-most object is on (0m, 0m). Calculate the moment of inertia of the system if it spins about the (a) X axis; (b) Y axis.