Question

A small block of mass m slides on a horizontal surface where friction between the block and the surface is negligible as it travels around the inside wall of a hoop of radius R. The coefficient of friction between the block and the wall is μk. The block is given an initial speed vo tangent to the hoop and allowed to travel along the inside edge. The velocity of the block at time is:_____

Answer 1

v = v_o/(1 + (tv_o(μ_k)/R))

Step-by-step explanation:

We are told that the coefficient of friction between the block and the wall is μ_k.

Now, formula for the frictional force will be;

F_fric = -μ_k × mv²/R

mv²/r is used because it is a centripetal motion.

Now, we know that F = ma = m(dv/dt)

Thus;

m(dv/dt) = -μ_k × mv²/R

m will cancel out to give;

dv/dt = -μ_k × v²/R

Rearranging to get;

dv/v² = -μ_k × (1/r)dt

We are told the initial velocity is v_o.

Thus, let the final velocity be v.

The initial time is 0 and thus the final time will be represented by t.

Thus;

(v,v_o)∫dv/v² = (t, 0)∫-μ_k × (1/R)dt

Integrating this we have;

-1/v - (-1/v_o) = -((μ_k)/R) × (t - 0)

-(1/v) + (1/v_o) = -t(μ_k)/R

Rearranging to get;

1/v = (1/v_o) + t(μ_k)/R

Simplifying the right hand side gives;

[R + tv_o(μ_k)]/Rv_o

Thus;

1/v = [R + tv_o(μ_k)]/Rv_o

v = Rv_o/[R + tv_o(μ_k)]

Simplifying further gives;

v = v_o/(1 + (tv_o(μ_k)/R))