Free fall vs falling on incline of rotating bodies

When you have a sphere and a hoop on an incline, the sphere will always roll down faster because of the smaller moment of inertia. And this is the case no matter the angle of the incline. But what if the angle is $$90^\circ$$ (basically freefall). Shouldn't the sphere again get to the ground faster ((because the angle of incline doesn't matter)? But that is obviously not the case because if two rotating objects are in freefall they hit the ground at the same time. Ignoring air resistance. So what explains this phenomenon?

Oh, and there is no slipping when the sphere and hoop are falling down the incline. I would like to know if a slip or no-slip affects the answer to my previous question. So I would like to hear about that also.

Also the two objects are rotated before placing in the incline. So the friction might help with the rotation but you have already gave the objects an initial rotation.

• But as the angle of incline -> 90 degrees and the normal force -> 0, is there still friction and is the object still rolling? Commented Aug 3, 2020 at 1:12
• @Not_Einstein, I forgot to write it but I just edited my post. The objects are given roatation by you before placed on the incline. Commented Aug 3, 2020 at 1:34

If you are requiring no slipping at all, even in the case of a $$90^\circ$$ incline, then the sphere would still "fall" faster than the hoop. This is because rolling without slipping requires a friction force. Of course, the larger the incline angle the easier it is for slipping to occur, and without any other forces at play at $$90^\circ$$ you will have to have "slipping" (really just falling, with both objects then hitting the ground at the same time). But if you had some other force pushing the objects against the $$90^\circ$$ incline then they would not "fall" at the same rate.
Indeed, one can show that the acceleration of each object is given by $$a=\frac{g\sin\theta}{1+\gamma}$$ where $$\gamma$$ is obtained from the moment of inertia of the object $$I=\gamma mR^2$$. Of course, when $$\theta=90^\circ$$ this acceleration still depends on $$\gamma$$, but this is because we have assumed rolling without slipping in the first place. In reality in order to achieve rolling without slipping on a $$90^\circ$$ incline one would need to apply a force completely perpendicular to the incline to prevent slipping (and a force in this direction will not change the acceleration given above). Without this force you will essentially have no friction force, and hence you will have free fall.
• @Jacob That is the Greek letter "gamma" (hopefully no Greek people here will be offended by you calling their alphabet weird. Greek letters are often used in physics and math). It is just used to simplify general expressions when the objects you are considering are things like uniform hoops, disks, spheres, etc. For example, for the hoop since $I=mR^2$ we would have $\gamma=1$, or for a disk since $I=\frac12mR^2$ we would have $\gamma=\frac12$. Commented Aug 4, 2020 at 11:07
When the angle of incine is $$90^o$$ there can be no friction acting on the objects. That is because kinetic friction between two surfaces is directly proportional to normal force between them. When the incline is vertical, there will be no normal reaction since there is no horizontal component of the falling bodies.