Question

FIGURE P8.54 shows a small block of mass m sliding around the inside of an L-shaped track of radius r. The bottom of the track is frictionless; the coefficient of kinetic friction between the block and the wall of the track is µ_k,The block's speed is v_o at t_o =0 Find an expression for the block's speed at a later time t.

Short answer

The expression for the velocity at time t is $v_t=\frac{\left(v_{o}r\right)}{(v_o{\mu }_{k}t+r)}$

Step-by-step solution

Given Information

mass= m

radius= r

The bottom of the track is frictionless

coefficient of kinetic friction between the block and the wall of the track is µ_k,

speed is v_o at t_o =0

Explanation

Centripetal force is given by

$F_c=\frac{m{v}^2}{r}$
where m = mass, v = velocity and r = radius.
Frictional force is µ_kN

Where µ_k,= coefficient of friction and N is Normal force

From the Newton's second law, we know

$$\begin{gathered} m\frac{dv}{dt}=-{\mu }_k\left(\frac{m{v}^2}{r}\right) \\ \frac{dv}{dt}=-\frac{{\mu }_k{v}^2}{r} \end{gathered}$$

Now integrate and find v_t

_{$$\begin{gathered} {\int }_{v_0}^{v_t}\frac{-dv}{v^2}={\int }_0^t\frac{{\mu }_{k}dt}{r} \\ \frac{1}{v_t}-\frac{1}{v_0}=\frac{{\mu }_{k}t}{r} \\ \frac{1}{v_t}=\frac{{\mu }_{k}t}{r}+\frac{1}{v_0} \\ {v}_t=\frac{v_{o}r}{v_o{\mu }_{k}t+r} \end{gathered}$$}