1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Newton's second law dp/dt version?

  1. Jun 30, 2015 #1
    Hello,

    I am confused by the momentum version of newtons second law...

    So we know
    [tex]\bar{F}=m\bar{a}=m\left(\frac{d\hat{v}}{dt}\right) [/tex]
    and that
    [tex]\bar{\rho}=m\bar{v}=m\left(\frac{d\bar{x}}{dt}\right)[/tex]

    so is

    [tex]\frac{d\bar{p}}{dt}=m\frac{d\left(\frac{d\bar{x}}{dt}\right)}{dt}[/tex]

    What I mean is this bit [tex]\frac{d\left(\frac{d\bar{x}}{dt}\right)}{dt}[/tex] somehow equal to [tex]\bar{a}[/tex]

    Thanks
     
  2. jcsd
  3. Jun 30, 2015 #2

    Filip Larsen

    User Avatar
    Gold Member

    Yes, since acceleration is the (first order) time derivative of velocity and velocity is the (first order) derivative of position, the acceleration is said to be the second order derivative of position, which can be written as ##a = \frac{d^2x}{dt^2}##

    You can read more about other notations for higher derivatives on [1]

    [1] https://en.wikipedia.org/wiki/Derivative
     
  4. Jun 30, 2015 #3
    Force F = dp/dt, this result summarizes newton's first/second law, to prove this, we know that F = ma = m*dv/dt, mass is invariant so we can treat it as a constant, which yields to m*dv/dt = d(m*v)/dt = dp/dt, If no force is acting on an object F = 0 = dp/dt, p is constant from which it follows newton's first law and momentum conservation, Good luck :p
     
  5. Jul 1, 2015 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Using the product rule:

    [tex]\vec{F} = \frac{d\vec{p}}{dt}=\frac{d}{dt}(m\vec{v}) = m\frac{d\vec{v}}{dt} + \frac{dm}{dt}v[/tex]

    If m is constant, dm/dt = 0 so:

    [tex]\vec{F} = m\frac{d\vec{v}}{dt} = ma = m\frac{d}{dt}v = m\frac{d}{dt}(\frac{d\vec{x}}{dt})[/tex]

    AM
     
    Last edited: Jul 5, 2015
  6. Jul 5, 2015 #5
    Thanks guys, forgot about the chain rule for differentiation.

    So in general, whenever there is a [tex]\frac{d^{2}}{dx}[/tex] then it can be thought of two separate derivatives, each giving their own result. But the [tex]^2[/tex] means it skips the first result and we go right to the second?
     
  7. Jul 5, 2015 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Actually, I misspoke. It is the product rule not the chain rule. I have corrected the error.

    The ##\frac{d^{2}x}{dt^2}## signifies the second derivative with respect to time - the derivative with respect to time of (the derivative with respect to time of x).

    AM
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Newton's second law dp/dt version?
Loading...