MIT OpenCourseWare http://ocw.mit.edu 8.01 Physics I: Classical Mechanics, Fall 1999 Please use the following citation format: Walter Lewin, 8.01 Physics I: Classical Mechanics, Fall 1999. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/termsMIT OpenCourseWare http://ocw.mit.edu 8.01 Physics I: Classical Mechanics, Fall 1999 Transcript – Lecture 24 With our knowledge of torque... calm down. With our knowledge of torque and angular momentum, we can now attack rolling objects which roll down a slope. For instance, the following... I have here a cylinder or it could be a sphere, for that matter, and this angle is beta. I prefer not to use alpha because that's angular acceleration, and there is this friction coefficient with the surface, mu, and this object is going to roll down and you're going to get an acceleration in this direction, a. And I will evaluate the situation when we have pure roll. That means the object is not skidding and is not slipping. What is pure roll? If here is an object, the cylinder is here with radius R, and I'm going to rotate it like this and roll it in this direction, the center is called point Q. Once it has made a complete rotation, if then the point Q has moved over a distance 2pi R, then we call that pure roll. When we have pure roll, the velocity of this point Q, and the velocity of the circumference, if you can read that-- I'll just put a c there-- are the same. In other words, vQ is then exactly the same as v circumference, and v circumference is always omega R. This part always holds, but for pure roll, this holds. You can easily imagine that if there is no friction here, that the object could be standing still, rotating like crazy, but Q would not go anywhere. So then we have skidding and we have slipping and then we don't have the pure roll situation. If the object is skidding or slipping, then the friction must always be a maximum here. If the object is in pure roll, the friction could be substantially less than the maximum friction possible. Now I would like to calculate with you the acceleration that a cylinder would obtain.When it pure rolls down that slope, it has mass M, it has length l and it has radius R. And I would like you to use your intuition and don't be afraid that it's wrong. I'm going to roll down this incline two cylinders. They're both solid, they have the same mass, they have the same length but they're very different in radii and I'm going to have a race between these two. Which one will reach the bottom first? So I repeat the problem. Two cylinders, both solid, same length, same mass, but one has a larger radius than the other. There's going to be a race. We're going to roll them down, pure roll. Which one will will? Win win? Will win? Who thinks that the one with the largest radius will win? Who thinks the large with the smallest radius will win? Who thinks there will be no winner, no loser? Wow, your intuition is better than mine was. We'll see how it goes. Keep in mind what your vote was and you will see it come out very shortly. Okay, let's put all the forces on this object that we know. This one is Mg. And we're going to decompose that into one a longer slope, which is Mg sine beta, and one perpendicular to the slope. We have done that a zill... zillion times now. And this one equals Mg cosine beta. Then there is right here a normal force, and the magnitude of that normal force is Mg cosine beta, so there is no acceleration in this direction and then we have a frictional force here for which I will write F of f. There is an angular velocity at any moment in time, omega, which will change with time, no doubt. And then this center point Q, which is the center of mass, is going to get a velocity v and that v will also change with time. And the v of the point Q which changes with time is v of the circumference, because that is the condition of pure roll. That equals omega R. This is always true, but this is only true when it is pure roll. I take the time derivative.The derivative of the velocity of that point Q is, per definition, its acceleration, so I get a equals omega dot times R, and that equals alpha R, alpha being the angular acceleration. So this is the condition for pure roll. Now I'm going to take the torque about point Q. When I take the torque about point Q, N has no effect because it goes through Q, and g has no effect because it goes through Q, so there's only one force that adds to the torque. If this radius is R, the magnitude is RF and the direction is in the blackboard. But I'm only interested in the magnitude for now, so I get R times the frictional force. This must be I alpha, I being the moment of inertia for rotation about this axis through point Q, times alpha, but I can replace alpha by a/R. So I get the moment of inertia about Q times a/R. And this is my first equation, and I have as an unknown the frictional force, and I have as an unknown a, and so I cannot solve for both. I need another equation. The next equation that I have is an obvious one, that is Newton's second law: f = MA. For the center of mass, I can consider all the mass right here at Q. We must have f = MA. And so M times the acceleration of that point Q, which is our goal, by the way, equals this component, equals Mg sine beta. That is the component downhill. And minus Ff, the frictional force, which is the component uphill, and this is my equation number two. Now I have two equations with two unknowns. So I can solve now. I can eliminate Ff and I will substitute for Ff in here this quantity divided by R. And so I get Ma equals Mg sine beta minus moment of inertia about point Q, times a divided by R squared. And now notice that I've eliminated F of f, and so now I can solve for a. So I'm going to get... I bring the a's to one side. So I get a times M plus moment of inertia divided by R squared equals Mg sine beta, and a now we have.I multiply both sides with R squared. I get MR squared g sine beta upstairs, and downstairs I get MR squared plus the moment of inertia about that point Q. This is my result, and all I have to put in …
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