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College Physics Volume 1 Second Edition Answers
Acceleration a=μN÷m

; (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).

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