How can a mass slide up the inside of a rotating cone? 


So I solved this problem in the picture for the maximum value of the period and got
$$T = 2\pi\sqrt{\frac{h\tan{\beta}}{g}\frac{\sin{
\beta+\mu_s\cos{\beta}}}{\cos{\beta}-\mu_s\sin{\beta}}}$$
with friction acting up the slope. The minimum value of $T$ occurs when the friction acts down the slope. What I don’t understand is how this is possible since friction opposes motion and if you resolve the forces in the direction of the slope the net force does not act upwards, so how could the mass move up the slope? My tutor said the inertia of the mass carries it up the slope. Could someone please give me a more quantitative explanation of this?
 A: Here is a diagram of the forces for motion slow enough so that the friction acts up the slope

The normal force provides the centripetal acceleration towards the middle of the circle, but also an upward force that can match the weight.
If the rotation is very fast $N$ has to be large to provide the necessary centripetal force, so the upward force from $N$ can be bigger than the weight - and make the mass move upwards (this also means outwards if the mass is still in contact with the slope).
If the rotation is fast, $F$ in the diagram above is pointing down the slope and there seems to be no force that could move the mass to the right.
However looking from above, the mass would move naturally move in a straight line, in the direction of the green arrow (below), if the normal contact force and friction forces were absent.
So as your tutor says, it's the inertia that helps it move right (and up the slope).  This would usually cause it to lose kinetic energy and slow down, but the friction acting on the mass from the spinning cone helps it maintain its speed.

A: To determine the direction in which friction will act, you need to consider the situation without (radial) friction. In other words, assume that the block is constrained to move along a frictionless radial groove on the cone.
Without friction, there are only two forces acting on the block, which are gravity and the centrifugal force. There is exactly one value of $T$ where they cancel and the block does not move radially. If $T$ is greater than this value, the centrifugal force wins over gravity and the block is pulled outward (and upward). Therefore, if friction were present, it would act downward to oppose this motion. The opposite would be true if $T$ were lesser than that value.
