; (n is the pressure between the ball and the inner wall of the cylinder)

N= centripetal force =mv? ÷R；

So a=μv? ÷R；

Let's use calculus: dv/dt=μv? ÷R；

Separation variable: dv/v? =μdt/R

Bilateral integral ∫(dv/v? )=∫(μdt/R)+c

So:-1/v=μt/R+c, and then when t=0, the ball is V0, and c =-1/v0;

So the velocity of the ball at any moment is v = v = rv0/(r+μ v0t); -The first problem is solved.

This is the second one:

From v=RV0/(R+μV0t):

ds/dt = RV0/(R+μV0t)；

Transposition ds = rv0dt/(r+μ v0t);

Two-sided integrated:

As above, we know that s=Rln(R+μV0t)/μ (the minus sign is omitted because it is a distance).

Typing is not easy. If you are satisfied, please adopt it.