A Problem Kinematics "Physics for scientists and engineers"

Oct 2019
Hello everyone,
An object of mass (m) sliding on a circular orbit of radius (r) which has a L shaped cross section. There is no friction on the bottom base but there is kinetic friction (μ) between the object and the side wall. At the time (t=0 sec) the magnitude of velocity is Uo.

I need to find the expression of U in terms of r, μ, t, Uo.

The answer is (rUo)/(r+μtUo)

How is that proven?
Jan 2019
magnitude of the net force acting on the object in the tangential direction is $f_k =ma_T = m \dfrac{du}{dt}$

$f_k = \mu N$

the normal force acting on the object's side is equal to the centripetal force

$f_k = \mu \dfrac{mu^2}{r}$

$ma_T = \mu \dfrac{mu^2}{r}$

since $a_T$ acts opposite in direction to $u$

$\dfrac{du}{dt} = -\dfrac{\mu}{r} u^2$

separate variables

$\dfrac{du}{u^2} = -\dfrac{\mu}{r} \, dt$


$-\dfrac{1}{u} = -\dfrac{\mu}{r} t + C$

initial velocity is $u_0$ at $t=0$

$-\dfrac{1}{u_0} = 0 + C \implies C = -\dfrac{1}{u_0}$

$-\dfrac{1}{u} = -\dfrac{\mu}{r} t - \dfrac{1}{u_0}$

I leave you to work out the algebra and solve for $u$
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