Hi, this is module 21 of Two-Dimensional Dynamics. Today, we're gonna continue on and instead of having a wheel rolling on a thick straight line, we're gonna have a wheel rolling on a thick plane curve or curved surface. So here's a picture of the situation. We've got a wheel rolling to the right with angular velocity of theta dot, the radius of the wheel is again little r. And so we're going to actually use normal and tangential components or coordinates for this particular system, and the reason I do that is because the wheel is rolling along a curved path and so it'll be more convenient to use those coordinates. You'll recall our relative velocity equation, and again, I'm going to go from the point of known velocity, which is the point of contact here, the instantaneous point of zero velocity, or a center of zero velocity. And we're gonna move to our unknown, which is the velocity of the center of the wheel. And so let's go ahead and write that equation, and I've got v of C = v of the known, which is v of I plus theta dot K, and that's crossed with now in, we're going from our known to our unknown, and so we've got v of C = 0 for the velocity of the instantaneous center + theta dot K. And again we're using normal and tangential components. So going from I to C we're going a distance of little r in the e normal direction. So this is little r in the e normal direction. And so I've got v of C equal to, magnitude is gonna be r theta dot. Let's find the direction now. Using the right hand rule, from my IJ coordinate system, again I've got K pointed into the board. I'm now gonna cross K with e normal. So K is down into the board. I'm crossing it with I, e sub n by the right hand rule, that means that the direction now is e tangential. And so this is going to be e tangential. And that makes sense. The velocity is tangent to the path and its magnitude, which we call the speed, is equal to r theta dot. Okay, let's continue on. So here is the velocity of the center of the wheel now. We want to find the acceleration of the center of the wheel. How do I do that? As before, we're gonna go ahead and take the time derivative, and so we get the acceleration of C is equal to. r is a constant, but theta dot will change, or can change. So that's theta double dot, e tangential. Now, this time, e tangential will also change direction as the wheel rolls along. And so, it changes with respect to time, so I've gotta use the product rule, and I've got plus r theta dot times the derivative of e tangential. And you'll recall, you'll need to recall from an earlier lesson at the beginning of the course, that we developed a relationship that the derivative of e tangential is equal to s dot over rho e normal. And so let's go ahead and substitute that in so we have a of C = r theta double dot e tangential + r theta dot s dot over rho, e normal. s dot, however, as we said earlier, is the magnitude of the velocity or equal to r theta dot, and so I can substitute that in and I get a sub C, the acceleration of the center of the wheel, is equal to r theta double dot e tangential + now, since s dot is r theta dot ,we have an r theta dot, that's gonna be r theta dot squared over the radius of curvature to the center of the wheel, e normal. So now we have acceleration of the center of the wheel, which includes both a tangential component and a radial or normal component, directed radially inward. Okay, so far we know that the velocity of the instantaneous center is zero. We've found what the velocity of the center of the wheel is. We also found the acceleration of the center of the wheel, and so we now wanna move on and find the last unknown point, which is the acceleration of the point of contact, or point I. And to do that we're gonna use our relative velocity equations going from our known, which is now the acceleration of the center of the wheel, to our unknown, which is the acceleration of the point of contact. So let's go ahead and write that equation, and I've got the acceleration of I is equal to the acceleration of C plus theta double dot K crossed with r now from the known to the unknown, we know the acceleration at C, we don't know it at I, so this is gonna be r from C to I, minus theta dot squared r from C to I. r from C to I Is going to be little r for the radius in the negative e normal direction. This is -r in the e normal direction. And so I've got a sub I, oh let's put it in over here, too. -r e sub normal. So I've got a at the point I is equal to, let's go ahead and substitute in a at the center of the wheel. So that's gonna be r theta double dot e sub tangential plus the quantity r theta dot squared over rho e normal, and then I've got K, which is into the board again, crossed with e normal, so by my right hand rule it's in the e tangential direction. Its magnitude is gonna be r theta double dot. And so we've got minus r times theta double dot. So this is gonna be minus r theta double dot, and we said, K cross e normal is e tangential. And then I've got plus, cuz I got a negative times a negative, r theta dot squared e normal. Okay, now I see that actually this term and this term cancel. And so what I'm left with is the acceleration of the point of contact for a wheel rolling on a fixed plane curve is equal to, if I simplify that, I get (1 + r over rho) r theta dot squared e normal. And that's where we'll leave off, and we'll come back next time and work a problem with a wheel.