Let the speed of the slider reaching point B be vB, and the kinetic energy theorem applied to the slider from point A to point B is as follows: μmg5R= 12mv2B? 12mv20

Solution: v2B=8gR

When the slider moves from point B, the mechanical energy is constant. If the slider reaches point P at vP, then:12mv2b =12mv2p+mg2r.

Solution: VP = 2gr

Time for the slider to pass through the P hole and return to the platform: t = 2vpg = 4rg.

In order to realize the problem-solving process, we need to satisfy: ωt=(2n+ 1)π.

ω=π(2n+ 1)4gR(n=0， 1，2…)

Therefore, the angular velocity ω of the platform rotation should satisfy the condition ω = π (2n+ 1) 4gr (n = 0, 1, 2 …).