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Derivative of momentum with respect to time

  1. Oct 31, 2013 #1
    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.
     
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  3. Oct 31, 2013 #2

    Pythagorean

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    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?
     
  4. Oct 31, 2013 #3
    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?
     
  5. Oct 31, 2013 #4

    Pythagorean

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    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).
     
  6. Oct 31, 2013 #5
    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
     
  7. Oct 31, 2013 #6

    Pythagorean

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    Well, you can derive it from p=mv using the definition of the unit vector.
     
  8. Oct 31, 2013 #7

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    you would also have to consider that:

    [tex]\vec{p} = m\vec{v}[/tex]

    and that

    [tex]\vec{v} = x\hat{x} + y\hat{y}[/tex]
     
  9. Oct 31, 2013 #8
    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
  10. Oct 31, 2013 #9

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    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?
     
  11. Oct 31, 2013 #10
    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
     
  12. Nov 5, 2013 #11

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    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 [itex]\frac{d \hat{p}}{dt}[/itex] 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?
     
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