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!

Homework Help: Validity of Newton's 2nd Law in accelerating ref. frame and constant v.

  1. Jan 27, 2013 #1
    First of all I'm not sure if this is the right forum for this problem.

    1. The problem statement, all variables and given/known data
    Given that [itex]\overline{F}[/itex]=m[itex]\overline{a}[/itex] is valid in the lab frame S, show that:
    (a) it is also valid in a moving frame S' with a constant velocity relative to S,
    but (b) invalid in a moving frame with a changing velocity (ie, a frame S' accelerating relative to S).
    Assume the force and mass are invariant

    2. Relevant equations
    The Galilean transformation.
    Newtons 2nd law

    3. The attempt at a solution
    This is a non-required problem in the first hw set for an applied modern physics (sophomore level) course I'm taking. We have only had one lecture and I don't have the textbook so I'm not sure if the conclusions I've drawn are correct

    So we have two reference frames S and S', with respective coordinates (x,y,z,t) and (x',y',z',t'). But we can use just (x,t) and (x',t') and draw the same conclusions. S' is moving relative to S with a constant velocity [itex]\overline{v}[/itex] in part (a) and with an acceleration [itex]\overline{κ}[/itex] in part (b). I believe t=t' must also be assumed as the concept of absolute time still exists in Newtonian mechanics.

    For part (a)
    In S
    [itex]\overline{F}[/itex]=m(d2x/dt2)=m(d[itex]\overline{u}[/itex]/dt) (Where [itex]\overline{u}[/itex] is the velocity measured in S)

    The Galilean transformation gives this relationship between [itex]\overline{u}[/itex] and [itex]\overline{u'}[/itex] (velocity as measured in S')


    In S'

    if [itex]\overline{u}=\overline{u'}+\overline{v}[/itex] then [itex]\overline{a'}[/itex]= ([itex]d(\overline{u'}+\overline{v})[/itex]/dt)) should equal the same acceleration [itex]\overline{a}[/itex] measured in S. Thus if the force and mass are invariant as well, the 2nd law holds for a moving reference frame with some constant velocity [itex]\overline{v}[/itex].

    For part (b) when S' is accelerating relative to S

    Similarly in S



    Thus ##\overline{F}=m\overline{a'}+m(d\overline{κ}/dt)## where ##(d\overline{κ}/dt)## is the jerk of ##\overline{κ}##

    I believe this shows that the 2nd Law is not valid for an accelerating reference frame.

    I would greatly appreciate any corrections and comments you can provide!
  2. jcsd
  3. Jan 29, 2013 #2


    User Avatar
    Homework Helper

    It is not true: The velocity transforms as


  4. Jan 29, 2013 #3
    If that is the case then how is the result different from part (a)?
  5. Jan 29, 2013 #4


    User Avatar
    Homework Helper

    The velocity of the frame of reference changes with time: its time derivative is equal to κ

  6. Jan 29, 2013 #5
    Ah I Think I understand now.

    ## \overline{u'}=\overline{u}-\overline{v}##

    ## \frac{d\overline{u'}}{dt} = \frac{d\overline{u}}{dt}-\overline{κ} ##

    ##\overline{F}=m\frac{d\overline{u'}}{dt}+ m \overline{κ} ##

    Is this correct?
  7. Jan 29, 2013 #6


    User Avatar
    Homework Helper

    Yes, there is a new force in the accelerating frame, which does not belong to any interaction between bodies. That force pushes you forward in a braking car.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook