Hi and welcome to Module 35 of Two Dimensional Dynamics. Today, we're going to go ahead and calculate the kinetic energy for bodies moving in 2-D Planar Rigid Body Motion. And so this is where we left off last time, for bodies we said we could have translational kinetic energy, and rotational kinetic energy. And we're going to work on how to calculate that today. And so let's start off with a wheel. A no slip wheel, rolling on a flat surface. And so the wheel goes along like this, and the wheel, the mass center is translating, and the wheel is also rotating. So it'll have translational kinetic energy and rotational kinetic energy. If it was not rotating, it would only have translational kinetic energy. If it was pinned at the center and just rotating, it would only have rotational kinetic energy. But since it's doing both, it has both translational and rotational kinetic energy. Now by kinematics, we know that for a no slip wheel, the velocity of the mass center is equal to the angular velocity of the wheel times its radius. And, we can look up in a table or a reference that the mass moment of inertia, about the z-axis for a cylinder, is one half mR squared. And so I can go ahead and substitute those in now for my expression for kinetic energy. I substitute omega R for v sub c, and I substitute one half mR squared for the mass moment inertia and I arrive at this. And so, the answer for the kinetic energy of a no slip uniform rolling wheel or cylinder is shown here. All right, now let's go ahead and look at an unbalanced wheel. And so, here's my wheel rolling, but instead of having the, it's not uniform, instead of having the mass center at the geometric center of the wheel, the mass center is, is away from that. And so, I've got that wheel that I used in my previous module that rolled up the hill. And you can see now, you should get a clue as to how I made it roll up the hill, because I have an unbalance here so it shifts the mass center off from the geometric center of the body. And so to describe, since I can't describe it as a uniform cylinder, one half mR squared, to describe what the mass moment of inertia is, I'm going to use something called the Radius of Gyration. So the mass moment of inertia, will be mass times what we'll call the Radius of Gyration squared. And I want to give you some feel for what the Radius of Gyration is. It's the distance from point c, the mass center at which a point mass would have the same mass moment of inertia. And so, as an example, let's go back and look at a uniform cylinder first. And so here's my uniform cylinder. I know that the mass moment of inertia is one half mR squared. I can also express that as one half m times its radius of gyration squared. And, the Radius of Gyration would be a distance, of the radius of the wheel, divided by the square root of 2. It would be the location that a point mass of the same mass of, as the uniform cylinder would be away from the center, or the point c for the uniform cylinder or wheel. And so let me give you one more example. Here's another wheel. It has a mass moment of inertia. And, if I wanted to express it in terms of a Radius of Gyration, I'd have a point mass that would have all the mass of the wheel itself at a point, and it would spin from the center here again. And the distance from the center out to that point mass, would be the Radius of Gyration. It would have the same mass momentum inertia as the entire wheel itself. And so that just gives you a feel for the Radius of Gyration. Okay, so let's continue on with this unbalanced wheel. Since I no longer am operating at the geometric center of the wheel, I'm going to have to look for the velocity at the mass center, which is, is, we're going to have to use our relative velocity equation. So I'm going to say, v of Q, which is the geometric center, plus omega R from the geometric center to the mass center, will equal the velocity of the mass center. And, so I've put in my values here. And then I'm going to go ahead and do the the steps of multiplying it out, and when I get the mathematics, I find that vc squared is equal to this. And, I'll let you go through that on your own. It's just simple math as we we go through, and, and find the magnitude of the velocity, the mass center squared. Now that we have that we also have our our mass moment of inertia expressed in the Radius of Gyration terms. We can substitute into our expression this unbalanced wheel will again have both translational and rotational kinetic energy. And so, here's the expression I arrive at. And, I can simplify it as, as shown. Okay, so that's two examples. Let's do one more example. This is a, a RAFA bar. Rotation about a, a fixed axis. And so I've got a picture of this. [SOUND] Or a model of this. Okay. So here's my RAFA bar. I would have my, if it's a uniform bar, the mass center would be where my han, my finger is pointing. And so, for a uniform bar, the mass center has a velocity. And there is also a rotation about the mass center, and so it's going to have both translational and kinetic energy, if I look at it from the reference of the, the, the mass center of the body. [SOUND] And so, here's my expression for kinetic energy. By kinematics I can say that the velocity of the mass center is l over 2 times omega. If you go into a reference and you look at the mass moment of inertia for a slender bar about its mass center, that's equal to 1 12th ml square. And if I multiply those values out, I would get the kinetic energy as 1 6th mL squared omega squared. Now you might say okay, yeah, but we're this body is actually just doing pure rotation about point A. I, I took it, when I calculated here, I looked at it from the vantage point of point C, which I said okay, C is translating and there's a rotation about C. But, in fact, point A there, there's pure rotation about it. And you could look at the problem that way. And, and I've done that here. If we had looked at the problem from the vantage point of point A, then the velocity of point A, since it's pinned would have 0 velocity. And I sub A, if you go into a, a reference again, I sub A ends up being 1 3rd mL squared, instead of 1 12th mL squared. So it's 1 12th the spin about the, the mass center. It's 1 3rd mL squared, if you spin about an, an endpoint. And so I could use point A instead of the mass center, if I, if I calculate it out, I get the same result for either from either vantage point. So it's whatever you choose to be most comfortable with in these problems. And so that gives you a, a good sense of how to calculate kinetic energy for different types of bodies.
