# Kinetic friction along a curve (Basic Calculus Question)

1. Aug 7, 2009

### RoyalCat

A mass $$m$$ is traveling inside a horizontal circular tube of radius $$r$$, starting with an initial velocity $$v_0$$
There is friction between the outer wall of the tube and the mass (Friction between the floor and the mass is negligible)

EDIT:
What is the mass' velocity as a function of the distance it's traveled $$v(s)$$
(Sorry for the confusion if anyone happened to read what was written here earlier)

This is a problem that is listed as a solved example in my book, so I only really needs help with one little calculus step.

The first solution is an energy-based one.
The initial energy of the mass is $$\tfrac{1}{2}mv_0^2$$
The friction force acting on the mass : $$f_{friction}=\mu N= \mu \frac{mv^2}{r}$$

At the limit of a very small fraction of the distance along the curve, $$S$$, the force is constant, and so the work of friction along that distance is:
$$dW_f=\mu\frac{mv^2}{r}\,dS$$
This quantity is also equal to the kinetic energy lost by the mass due to friction.
$$\mu\frac{mv^2}{r}\,dS=-dE_k=-d(\tfrac{1}{2}mv^2})$$

Now according to the book: $$-d(\tfrac{1}{2}mv^2})=-mv\,dv$$
Now I can see why that's true, an integration of that expression over $$dv$$ would provide us with the infinitesimal difference in kinetic energy, but I'm really have trouble understanding what the author did here to reach his conclusion, and why it is valid.
If anyone's got a webpage explaining switching between integrands (At least that's what I think it's called) I'll gladly read up on the link.

Anatoli

Last edited: Aug 7, 2009
2. Aug 7, 2009

### kuruman

Without reference to kinetic energy, this is an application of the "product rule"

d(v2)=d(v*v) = vdv+vdv = 2vdv.

3. Aug 7, 2009

### RoyalCat

Ooh, thanks for giving me the exact name, I've struck up on some webpages I think will help me.
Thanks again. :)

Hurr, would I be correct to think of it as doing the following? :
$$d(\tfrac{1}{2}v^2)=\frac{d(\tfrac{1}{2}v^2)}{dv}\, dv=v\, dv$$

Is that a use for the product rule as well? I'm having a hard time finding the $$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}$$ in there. what exactly is my $$dx$$? Can it be just 1? And if so, how should I write it?

I've never had a class in integrals, so I'm just piece-mealing it from different physics questions I've solved, so forgive my ignorance. :uhh:

Last edited: Aug 7, 2009