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Darragh

Perpetual motion of a cylinder - with friction..

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Darragh    308
Hi everybody, Lately I've been trying to understand some of the forces acting on rotating cylinder shaped objects in order to better understand how wheels work. I've considered the following scenario and it's left me puzzled, because it does not seem to make sense. I know this sort of situation could never happen in the real world (because things are not perfect) but let me explain anyhow... Consider the following idealized situation: We have perfect cylinder travelling along a perfectly flat surface, lets say an infinite plane. The cylinder has a radius 'R', a linear velocity 'U' and an angular velocity 'W'. It makes contact with the ground at a point called 'C', which is directly underneath it's center of mass. Now consider for a moment that the only external force (or impulse in this case) acting on the Cylinder is friction (no aerodynamic drag forces). Lets say our simulation uses the following simple equation (per frame) to calculate a velocity loss for friction: F=KV Where F = Loss of velocity due to friction Where K = Coefficient of dynamic friction (i.e amount of friction) Where V = The velocity of the contact point 'C' on the sphere This is a velocity loss, but we could get to an impulse or force easily if we wanted to. Now the velocity of the contact point 'C' will be a combination of both linear and rotational velocity, and we will need to calculate this in order to calculate friction at that particular point. If a positive angular velocity means a clockwise rotation, and positive linear velocity means movement to the right, then to get the velocity of the contact point we would do the following: (note we are completly ignoring the Y/Z axis for this problem, 1 degree of freedom) V = U - R*W Where V = Velocity of contact point (for the x-axis) Where U = Linear velocity of object along x-axis Where R = Radius of the cylinder Where W = Angular velocity of the cylinder Simple enough? The velocity of the contact point is just a combination of rotational and linear velocity. Clockwise rotation will also produce a velocity in the opposite direction of the X axis at the contact point (because it is tangental to the rotating object). So consider if we plug in the following numbers: R = 10 (radius) W = 10 (angular velocity) U = 100 (linear velocity) If we calculate the velocity of the contact point using these numbers, we get a velocity of 0! V = 100 - (10*10) So in this case we get absolutely no friction, because the velocity of the contact point is said to be 0. Even if we calculated friction due to rotational and linear velocity separately, we would still end up with a net friction of 0- and torques would also balance. So in essence we have a situation where the two friction forces, due to rotation and linear movement cancel each other out- resulting in no friction. In a system with no other forces this would cause perpetual motion! Is this correct? Or am I missing something? It does not seem to make sense that an object could experience no friction even though it is moving and rotating at the same time. Here's a (crappy) diagram if that wasn't quite clear:

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Geometrian    1810
Quote:
Original post by Darragh
Now the velocity of the contact point 'C' will be a combination of both linear
and rotational velocity
Wrong. Think about it--"C" is directly under the center of mass, which moves only according to the linear velocity.
-G

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Rattenhirn    3114
The answer is rolling resistance.

You can observe the change from slide friction to roll friction very nicely when playing billiards. When the ball is struck, first if accelerates quickly, then it decelerates quickly due to friction. Then there's a point where it suddenly stops decelerating quickly and starts to roll gently.

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Darragh    308
Quote:
Original post by Geometrian
Quote:
Original post by Darragh
Now the velocity of the contact point 'C' will be a combination of both linear
and rotational velocity
Wrong. Think about it--"C" is directly under the center of mass, which moves only according to the linear velocity.
-G


I probably should be clearer on that. 'C' is not at the center of mass in this case, it is beneath it (as in the diagram). If it were precisely at the center of mass then that would be true, however since it is not 'C' is also moving at a certain velocity due to rotation.

Quote:
Original post by Rattenhirn
The answer is rolling resistance.

You can observe the change from slide friction to roll friction very nicely when playing billiards. When the ball is struck, first if accelerates quickly, then it decelerates quickly due to friction. Then there's a point where it suddenly stops decelerating quickly and starts to roll gently.


Ah... I knew there had to be something missing here. Thanks for the pointer!

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oliii    2196
That is basically the essence of grip. an ideal rolling tire will have no friction at the point of contact, and will roll perpetually. It means the angular velocity cancels the linear velocity exactly.

When you add torque to the tire (brake / accelerate), the velocity at the contact point becomes <> 0, and it generates a friction force that will change the momentum of the tire (and the car), accelerating or decelerating the body. As you stop applying a torque to the tire (no throttle / braking), the tire will then attempt to reach equilibrium, by matching the angular velocity to the linear velocity and then (ideally) produce zero friction at the point of contact.

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taliesinnz    231
Are you not calculating the friction twice, once on the angular velocity, and one on the linear.

since:
linear velocity = anglear velocity * radius

I will be just dealing with the angular velocity, and when friction reduces that velocity the cylinder slows down as a result.

Also by looking at the point C on you diagram. This point has no linear velocity in relation to the ground. When it is in contact with the ground, it does not move, the tube just rolls around it; making v=0.

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Darragh    308
Quote:
Original post by taliesinnz
Are you not calculating the friction twice, once on the angular velocity, and one on the linear.

since:
linear velocity = anglear velocity * radius

I will be just dealing with the angular velocity, and when friction reduces that velocity the cylinder slows down as a result.

Also by looking at the point C on you diagram. This point has no linear velocity in relation to the ground. When it is in contact with the ground, it does not move, the tube just rolls around it; making v=0.


No, in this case 'ordinary' linear friction is not being calculated twice. A point on a rotating body not only has the linear velocity of the body but velocity due to rotation also. When dealing with a body that is both rotating and moving you need to take both these things into account when calculating how fast a point on a body is moving. The linear velocity never changes no matter which point on the moving body you sample (at snapshot in time) but velocity due to rotation changes in both magnitude and direction depending on where you are sampling the rotational velocity at; in three dimensions it will be the cross product of a vector to the center of mass with the angular velocity. This is the reason why many physics libraries include a 'velocity at point' style function to compute this for you, because it is a fairly common calculation.

But there is also 'rolling' friction to take into account with rotating bodies (a force caused by deformation) as Rattenhirn mentioned- this was what I was missing. This is not the same as regular linear friction (which is a function of velocity) however... This is a different type of friction.

As for the point not moving, that is not true. Although (in this diagram) we always sample at the same point on the body (in local coordinates) the point is still moving- the diagram just shows a snapshot in time. The next time we sample the point 'C' it will not be the same point, as the previous 'C' will have rolled on- provided of course that the body is rotating.





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