# What causes a ball to rotate? [now with actual content]

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I've recently started work on a very basic physics engine. Right now, I'm trying to figure out which forces are responsible for causing a ball, going down a slope, to roll. The basic setup is shown in the picture below. The following is what I see happening. It's obviously wrong, so I'm hoping somebody could point out to me what mistakes I'm making and could help me correct them. At time 1, the ball is placed on a slope. Gravity acts on the center of the ball. The normal force has magnitude m*g*cos(theta) and acts on the contact point between ball and slope. A frictional force with magnitude coeff*m*g*cos(theta) acts on the contact point. The direction of this force is tangential with the slope. Since the tangential component of gravity (not pictured here) is bigger than the frictional force, the ball will start sliding down the slope. The frictional force will now generate a torque that causes the ball to start rolling. This torque can be calculated as the cross product of the frictional force and the vector between center and contact point. (I'm ignoring inertia, since it's just a constant.) At time 2, the ball has come down the slope and is now rolling along a flat plane. Once again there is gravity, a normal force and friction. Since theta is 0, the frictional force is now equal to coeff*m*g and is hence bigger than it was at time 1. This in turn causes the torque to be bigger than it was at time 1, which causes the ball to rotate faster. However, the frictional force also causes the ball to slow down. So, according to my - obviously wrong - calculations the ball would start rotating faster, while slowing down. Please tell me what I'm missing here. edit: forum acting weird... apparently this post didn't have any content at first [Edited by - Shai on May 15, 2008 7:58:56 PM]

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Uh... angular momentum? Can you be more specific, perhaps?

Edit: Okay, that's better. What's even weirder is that your last edit time is still an hour before I posted, yet all I saw was a blank post.

[Edited by - Ezbez on May 16, 2008 4:35:49 PM]

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The ball can only accelerate if it's toppling over the contact point. This acceleration is constant, but the normal reaction reduces it accordingly.
Sorry I can't quite explain my thoughts on this without coding it myself!

friction.

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Quote:
 Original post by ibebrettfriction.

To be more specific, I think what happens is friction prevents the ball from sliding.

So, at the first instant, the ball is motionless. However, from your diagram, you can see that the centre of mass of the ball is forward of the contact point. Thus, the ball will start to move as a result (I suppose this comes down to the fact that there is a net torque on the ball). In moving, the friction at the contact point of the ball will prevent slippage; thus the ball rolls.

Note: it's been a while since I've taken physics, so the above may not be 100% accurate.

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At time 2, the ppoint of the ball that is in contact with the ground is not moving with respect to the ground, so the friction force is zero.

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It's friction. As said above, with a perfect sphere and no friction, the centre of mass is aligned with the contact point and collision impulse vector (going along the normal of collision), thus no angular momentum is applied.

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Solved it :). Thanks guys.

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