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#Actualdeekr

Posted 11 February 2013 - 03:07 AM

The torque is given by R x F, not F x R.

 

In the z-direction: N=mg. In the forward direction: ma=-uNv^=-umgv^=f, where f is the force due to friction.

The torque on the contact point is then R x f = -umgr (r^ x v^) = I dw/dt, so dw = -(5ug)/(2r) dt (r^ x v^).

 

If Ww^ (magnitude and direction) is the current angular momentum, the update after euler integration is [ W w^ - ((5ug)/(2r)) (r^ x v^) dt ].

 

There is more analysis that could be done, but check if this sign change in the cross product does anything for you.


#1deekr

Posted 11 February 2013 - 03:05 AM

The torque is given by R x F, not F x R. Also, I think the coefficient of friction should be less in the slipping case (i.e. less traction). Given the same normal force, a higher coefficient means more friction, but the ball doesn't roll more when it's slipping.

 

In the z-direction: N=mg. In the forward direction: ma=-uNv^=-umgv^=f, where f is the force due to friction.

The torque on the contact point is then R x f = -umgr (r^ x v^) = I dw/dt, so dw = -(5ug)/(2r) dt (r^ x v^).

 

If Ww^ (magnitude and direction) is the current angular momentum, the update after euler integration is [ W w^ - ((5ug)/(2r)) (r^ x v^) dt ].

 

There is more analysis that could be done, but check if this sign change in the cross product does anything for you.


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