# Deriving Orbital Motion from Collisions in Travel along Chords

1. Aug 8, 2014

### LucidLunatic

1. The problem statement, all variables and given/known data

Inscribe N chords of equal length within a circle of radius r. A body of mass m travels at constant speed v along the chords and is perfectly reflected in each collision with the circle (there is no momentum change tangent to the circle). Show that the radial momentum change per collision is Δp =mvsinθ, where θ is the angle between tangent and chord.

2. Relevant equations
Δpr = pr1-pr0

3. The attempt at a solution
We can write
p0 = mvsinθ \hat{r} + mvcosθ\hat{\theta}
p1 = -mvsinθ \hat{r} + mvcosθ\hat{\theta}

therefore Δp = -2mvsinθ.

This seems like a fairly simple problem, so I'm not sure where I'm going wrong. The later steps of this problem then show that the net centripetal force obeys F = mv^2/r as N→∞, but that depends on getting the correct answer here. What am I missing? Note: this is me going through problems in a book, not an actual homework assignment.

2. Aug 8, 2014

### haruspex

I agree with your Δp (except the sign depends on how you care to define +ve direction), and from it I obtain F = mv2/r.

3. Aug 8, 2014

### rude man

Dumb EE's observation:

The above calculatios (posts 1 and 2) were made under the assumption that the radial component of momentum is also conserved. But that is not stated.

Let M = mass of circle and V be its + radial velocity after collision. Then if the system is isolated form the environment (e.g. on a frictionless horizontal plane):

Assume MV = +3mv sinθ

p1 = +mv sinθ
p2 = -mv sinθ + MV = -mv sinθ + 3mv sinθ = 2mv sinθ
so that p2 - p1 = 2mv sinθ - mv sinθ = mv sinθ.

In other words the problem has no solution unless the physical aspects of the circle are defined:

circle's mass
system isolated form the environment or not?
etc.

4. Aug 9, 2014

### haruspex

I presumed the exercise related to kinetic theory of gases, so the mass of the circle is effectively infinite.

5. Aug 9, 2014

### rude man

I agree, that makes sense.