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Assume the following scenario of rolling without slipping:

enter image description here

A small disk starts rolling w/o slipping at $\vec{r}_{CM}=-9R\hat{x}$ (scenario A) and reaches the bottom of the large hollow disk (scenario B).

I was told that there is no friction in scenario B, but how is this possible?
In scenario A - I was told that there is friction.

As far as I know, when a wheel rolls w/o slipping, it always has friction in the opposite direction of its movement.

Further explanation:

At the bottom of the disk, the following equation apply: $$(-R\hat{y})\times (-f_s\hat{x}) = I \alpha$$

So $f_s=\mu_s\cdot N$ and $N=m\omega^2\cdot 9R$, so $N > 0$ (because that $\omega > 0$).

How would you resolve the conflict?

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  • $\begingroup$ Say you're holding a tire connected to an axle and you spin it up so that it's spinning at, say, one revolution per second. Then without yet letting it touch the ground, you get on a truck and accelerate at a velocity where you travel one circumference length of the tire every second. When you drop the tire, its angular momentum and momentum both do not need to change at all. So the ground doesn't actually have to exert any tangential friction force at all. Does this image help? $\endgroup$ – user12029 Feb 24 '15 at 17:02
  • $\begingroup$ kind of.. I added explanation to my original post, plz review it.. $\endgroup$ – Dor Feb 24 '15 at 17:26
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As far as I know, when a wheel rolls w/o slipping, it always has friction in the opposite direction of its movement.

Here you go wrong . In pure rolling (that is no slipping ) the bottom point is at rest wrt ground . So there is no kinetic friction acting on it as there is no relative motion .But static friction acts .
In your question gravity accelerates the motion and makes the velocity of the contact point change . Hence kinetic friction starts to act whose direction need not be always opposite to the movement but opposite to the relative motion of the contact point .

Check this out :
* Rolling

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