0
$\begingroup$

Look at this picture: http://i.imgur.com/mY8ShkV.png

Here we have a ball resting on top of a platform (the contact point is marked red). I know the normal vector (marked green). I also know the mass, radius, and current angular and linear velocity of the ball.

Let's say I set the new angular velocity after friction which is always big enough ("infinite") has been applied to: $V_a = V_l \times rN$ where $V_l$ is the linear velocity, $r$ is the radius of the ball and $N$ is the normal vector. Does this math check out, is this correct?

Because this model doesn't work if I leave the ball leaning on the side of the platform where it should obviously roll of the platform and fall. But it doesn't. What is the correct mathematical model for simulating "infinite" friction?

I suppose you will also have this problem when you want to simulate finite friction and the friction doesn't give it's max force.

$\endgroup$
  • $\begingroup$ No it is not correct. What are you trying to get out of this pure rolling (no slip) scenario? $\endgroup$ – ja72 Dec 11 '13 at 22:48
1
$\begingroup$

If C is the center of mass of the ball, and A is the contact point then the velocity of the ball at the contact point is

$$\vec{v}_A = \vec{v}_C + \vec\omega \times (\vec{r}_A-\vec{r}_C) $$

If you have infinite friction then you have no slipping which means $\vec{v}_A =0$. If there was slipping then only the speed in vertical direction should be zero $\hat{n}\cdot\vec{v}_A =0 $, where $\hat{n}$ is the normal direction vector.

I think this is what you are trying to get to with your equation, yet is unclear what you are asking exactly. In your case, the no-slip condition is

$$ \vec{v}_C + \vec{\omega} \times (-\hat{n} r) =0 $$ $$ \vec{v}_C = r \hat{n} \times \vec\omega $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.