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B Are momentum and centripetal acceleration related somehow?

  1. Mar 22, 2017 #1
    I have been wondering, simple question, really: What is the relationship between momentum and centripetal acceleration, if there is one? Is there a relationship in terms of velocity, maybe, or is there none whatsoever?
  2. jcsd
  3. Mar 22, 2017 #2
    An acceleration is a rate of change of velocity, remembering that velocity involves both magnitude and direction. In the usual case, centripetal acceleration is related to the change in velocity associated with the change in direction. When you change the direction of motion, you change the momentum of the body as well. So, yes, they are related.
  4. Mar 22, 2017 #3
    I see. Well, I have one more question. Centripetal Acceleration is proportional to the square of the velocity divided by the radius. How is the square of the velocity related to the velocity in momentum, if it is?
  5. Mar 22, 2017 #4
    Dr.D's answer is neat, but if you're looking for something more I think you may want to clarify what you mean by being related; both of them have a dependency on velocity, yes, but that would make every cinematic quantity somehow related with each other, thus making your question a bit vague.

    Not sure if i get what you are asking here, but consider a point mass in motion along some curve: every single point P on the curve can be locally approximated by a circle of radius [itex]R_P[/itex] (see osculating circle), so at every point you can compute a value for your centripetal acceleration [itex]a_c = \frac{v^2}{R_P}[/itex], so if you remember that linear momentum is [itex]q = mv[/itex] you'll note that if velocity increase by a factor of 2, linear momentum will increase by the same factor, while centripetal acceleration will increase by a factor of 2^2=4.
    Likewise, if you change the centripetal acceleration acting on the body by a factor k (without changing its istantaneous radius of curvature), its speed, and therefore its linear momentum, will change by a factor [itex]\sqrt{k}[/itex]
  6. Mar 22, 2017 #5


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    By a power of 2.
  7. Mar 22, 2017 #6
    I see. Well, beyond the square root of velocity in centripetal acceleration and such, there wasn't much else I was asking for, if anything at all. So, the velocity squared in centripetal acceleration is independent from the velocity in momentum, I take it?
  8. Mar 22, 2017 #7
    No, how can the velocity be independent from itself? I mean, it is the same velocity we are talking about, it just plays a different role in the 2 quantites, momentum being linear in v while centripetal acceleration is quadratic. In fact, as I showed you, with fixed mass and trajectory a change in centripetal acc implies a change in linear momentum and viceversa.
    If what you are asking is "is the [itex]v^2[/itex] in the centripetal acceleration directly derivable from the linear momentum?" then for sure you can find some fancy way to derive it like that(consider a body under action of centripetal force, note that F = dq/dt, etc..) but it would be hardly meaningful with respect to the standard way of defining those quantities.
  9. Mar 22, 2017 #8
    Yeah, that is what I meant to say, the playing a different role part. My bad. Sorry about that.
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