- #1

StephenP

- 5

- 0

## Homework Statement

You push an object of mass m slowly, partway up a loop-the-loop track of radius R, starting from the bottom, where the normal force to the track is vertically upward, and ending at a point a height h< R above the bottom. The coefficient of friction between the object and the track is a constant [tex]\mu[/tex].

Show that the work you do against friction is [tex]\mu mg(2rh-h^2)^{1/2}[/tex]

## Homework Equations

[tex]W=\vec{F}\cdot\vec{x}[/tex]

[tex]F_f=\mu mg [/tex]

## The Attempt at a Solution

I assume that since the mass is being moved 'slowly', I can ignore the usual equations of circular motion, centripetal force, etc.

Further, I noted that at point "A", the normal force is exactly equal/opposite to mg, and that at point "B", the normal force is exactly 0. (at h=r).

I further assumed (

**not sure if this is valid**) that the normal force would decrease linearly from mg to 0 as h goes from 0 to r.

Using the basic equation of a line, I get:

[tex]F_n(h) = \frac{-\mu mg}{r}(h) + \mu mg[/tex]

This is where I more or less run out of steam. I know that I ought to integrate it, but for the life of me I can't figure out how to integrate it "along" the displacement (which would be given by the circle). Simply taking the indefinite integral wrt h yields:

[tex]\mu mg(\frac{2rh-h^2}{2r})+C[/tex]

Which is rather similar, but obviously significantly different. I guess what I need is a last nudge to figure out what I'm doing wrong, I feel very close but have just gotten frustrated working on this for so long :(

-Stephen