# Homework Help: Derivative of momentum with respect to time

1. Oct 31, 2013

### MAC5494

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

Hi, I don't really have a particular problem that needs to be solved, I'm just interested in why dPhat/dt = v/R. It's explained very poorly in my physics book, or I'm just not smart enough to understand it the way they explained it.

2. Oct 31, 2013

### Pythagorean

dp/dt = F, where F is force.

So there's no way it could equal v/R if v is speed and R is a distance, the units don't work out. What are v and R?

3. Oct 31, 2013

### MAC5494

Sorry, I explained that poorly. I don't mean the derivative dPhat/dt = v/R, i want to know why dPhat/dt is equal to v/R

pvector = pmag*phat

so by using the product rule you get:

dpvector/dt = (dpmag/dt)*phat + pmag * (dphat/dt)

(dpmag/dt)*phat is the rate of change of the magnitude of momentum, or the force parallel.

pmag * (dphat/dt) is the rate of change of the direction of the momentum, or the force perpendicular.

It says in my book that, lim Δt approaches 0, magnitude(Δphat/Δt) = magnitude(dphat/dt) = vmag/R

So I want to know why that is?

4. Oct 31, 2013

### Pythagorean

Ah, yes, dphat is dimensionless, I see. I think you can imagine v is tangential velocity of an object rotating around some circle with radius R with your coordinate system at the center of the circle. So if you have a huge circle, R is going to be very big and your tangential force will be smaller since the object turns at a slower rate with a bigger circle (given a constant tangential velocity, v).

5. Oct 31, 2013

### MAC5494

Okay, I get some of that, but I'm still having a hard time conceptualizing how those 2 quantities are equal. I get that the velocity will be pointing in the direction of the Phat, but I don't understand how one derives velocity/radius from dPhat/dt.

I'm basically trying to understand it, because it makes more sense to truly understand rather than memorize that dPhat/dt = v/R

6. Oct 31, 2013

### Pythagorean

Well, you can derive it from p=mv using the definition of the unit vector.

7. Oct 31, 2013

### Pythagorean

you would also have to consider that:

$$\vec{p} = m\vec{v}$$

and that

$$\vec{v} = x\hat{x} + y\hat{y}$$

8. Oct 31, 2013

### MAC5494

How do you derive it? I've been messing around with it for a little while and I can't seem to figure it out.

Phat = p/pmag = m*vvector/m*vmag = vvector/vmag

dt = dp/Fnet

(vvector/vmag)*Fnet/dp

Am I doing this right? I'm kind of lost at that point.

I'm not sure I've seen that bottom equation before.

Last edited: Oct 31, 2013
9. Oct 31, 2013

### Pythagorean

hrm. I'm not getting there with my approach either. I'll have to rethink it. can you define v and R for me and tell me the general physical set up?

10. Oct 31, 2013

### MAC5494

Well v is just velocity, and R is the radius of a kissing circle.

For instance, if an asteroid had a circular orbit around a planet, then the speed would be constant. Only the direction would be changing, so you'd want to find the rate of change of the direction of momentum, which is |p|*(dpvector/dt) which = |p|*(|v|/R) = (m*|v^2|)/R

11. Nov 5, 2013

### Pythagorean

So the derivation is lots of algebra so I didn't carry it through.

However, if you have a vector and you break it into magnitude and direction, then $\frac{d \hat{p}}{dt}$ is just "change in direction of vector per time".

So if you imagine a particle orbiting around in a circle, it's going to turn faster as it's tangential velocity is higher (it's going faster around the circle so it's turning faster) and, given a fixed v, it's going to turn faster on a smaller circle. So v/r seems to fit qualitatively, no?