; (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.