A uniform hollow disk has two pieces of thin light wire wrapped around its outer rim and is supp

A uniform hollow disk has two pieces of thin light wire...

Key Concepts

Energy Conservation

This concept involves the transformation of gravitational potential energy into kinetic energy as the object moves. In problems like this, the loss in gravitational potential energy is equated to the gain in both translational and rotational kinetic energies, allowing one to solve for the speed or other parameters of the system.

Rolling Without Slipping

Rolling without slipping is a kinematic constraint that establishes a fixed relationship between the translational velocity of the center of mass and the angular velocity about the center of mass (expressed as v = ?R). This constraint is crucial in analyzing systems where the object rolls without skidding, ensuring that the point of contact is momentarily at rest with respect to the surface.

Moment of Inertia

The moment of inertia quantifies how an object's mass is distributed relative to its axis of rotation, affecting its resistance to changes in rotational motion. This concept is essential in problems involving rotational dynamics since it directly influences the rotational kinetic energy and the dynamics of how energy is partitioned during the motion.

Transcript

00:01 Okay, so in this question, we have a hollow cylinder that has a reasonably thick wall.

00:09 It is hung to the ceiling by two wires that are wrapped around it.

00:15 So i'm going to draw a quick picture.

00:17 So we have, you know, a reasonable thickness here that's, you know, not negligible.

00:26 It's hung by two wires to the ceiling like so.

00:32 And then suddenly one of them snaps and it rolls without slipping from the other one and we're supposed to figure out what this what the speed of its center of mass is after the cylinder has dropped 1 .2 meters so there's actually quite a bit going on here and we're not giving a whole lot of information so i'm going to give us the variables that we have to work with.

01:05 So we have the change in height, h, which is 1 .20 meters.

01:10 We have the two radii of the cylinder.

01:15 We have the internal radii radius are 1, which is 30 centimeters or 0 .3 meters.

01:26 And it's hollow inside of that one.

01:29 So if i go to the picture, the actual material is this part.

01:34 So it's like a donut kind of.

01:39 And then the exterior radius, r2, is 50 centimeters.

01:47 Got a little bit ahead of myself.

01:48 It's 50 centimeters, which is 0 .5 meters.

01:54 And so those are our givens.

01:56 And we're supposed to use conservation of energy to figure out what the speed of the center of mass will be once it's dropped, 1 .2 meters.

02:10 So, i mean, we're given, we're told pretty clearly.

02:14 Let's use conservation of energy.

02:17 So what we need to consider here is what energy is being converted into its kinetic energy.

02:24 And that would be gravitational potential energy.

02:26 So we can say that mgh is equal to its change in kinetic energy.

02:33 And the thing is, when it's rolling, it has both translational and rotational kinetic energy.

02:39 So there's going to be two terms for the change in kinetic energy.

02:42 So our equation then becomes mgh is equal to one half i omega squared for the rotational energy and one half mvcm for the velocity of the center of mass squared.

03:01 So now the next thing that we really need is we need the formula for the rotational inertia, the moment of inertia of this kind of donut shape.

03:14 And so the formula for that, i'm going to go ahead and just write it down here.

03:18 I is going to be equal to one -half times m times r2 squared plus r1 squared.

03:28 So we can go ahead and plug that into our equation.

03:31 So we get m -g -h is equal to one -half times one -half, m times r2 squared, plus r1 squared and then we add to that one half m vcm squared so the first thing that we can do here is notice that every term has an m in it so we can go ahead and get rid of that which is great because we weren't given a mass for this object so nice to know that we don't need it and then we can start simplifying things a little bit so oh i forgot the omega squared in this term we we do need an omega squared there, otherwise things will not work.

04:18 So let's go ahead and start simplifying this.

04:20 The main way i'm going to be doing this is this middle term for the rotational energy.

04:25 So 1 half times 1 half is obviously 1 fourth, but i'm going to go ahead and distribute the 1 4th omega squared to the r2 and r1 squared.

04:35 So then our equation is gh is equal to 1 4th r2 squared omega squared plus 1 4.

04:47 R1 squared omega squared plus one half vcm squared.

04:57 So remember we're trying to solve this equation for vcm.

05:01 We know g because it's a you know it's a constant.

05:05 It's an established constant.

05:06 We know h because it's given to us and we know r2 and r1 but we don't know what omega is.

05:17 And so you might instant.

05:19 Try to find omega but the thing is we what we need to remember is that the relationship between we need to remember the relationship between vcm and omega so vcm just a little aside here vcm is going to be equal to omega times the outer radius which in this case is r2 so what we can do is we can actually find ways of substituting r2 and r1 times omega to be vcm, which will get us closer to solving this problem.

06:00 Now, it's very obvious how we can do this with r2 times omega, because that is just by definition, vcm...

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