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Equations of motion - trouble with signs

  1. Jan 8, 2014 #1
    I seem to be getting into a bit of a mess with my signs when using Newton's second law in classical mechanics. Here's an example (I am fine with completing the question, it's just when I look at alternative ways of setting up axes and solving it, things change):

    1. The problem statement, all variables and given/known data
    A ball of mass m is projected vertically upward at velocity vo. The ball experiences an air resistance force (in addition to gravity) of the form -αv2 where α>0 is constant and v is the velocity, and reaches a maximum height h before it returns back to the point of projection.
    Write down the equations of motion of the ball during its upward and downward journeys.
    Obviously the question doesn't end there but this is the only relevant part.


    2. Relevant equations
    F=md2x/dt2


    3. The attempt at a solution
    Ok, so when we're going up, lets say I take my x axis going upwards. Then I have a particle of mass m, with mg and αv2 as the downward forces. So I write -αv2-mg=md2x/dt2 and solve, which works for the remainder of the question.

    However what if I say my x axis points downwards now. Then I have my forces down too, i.e in the x direction, so these would be positive. So I write αv2+mg=md2x/dt2 and these give different solutions. Clearly I'm missing something fundamental here (although I'm sure I've always worked in this way and not had problems) - making the RHS negative obviously works but I can't see why I'd do that (besides the acceleration is clearly downwards anyway so wouldn't that mean it would be negative in the first instance and positive now?).

    Likewise going down, x axis upwards, I have αv2 going up, mg going down, so I write αv2-mg=md2x/dt2, which gives the correct results. Then x axis downwards, forces still in the same direction so I have mg-αv2=md2x/dt2 which again gives a different solution to above.

    Can anybody explain what is wrong, thanks :)
     
    Last edited: Jan 8, 2014
  2. jcsd
  3. Jan 8, 2014 #2

    rock.freak667

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    You would need to find the resultant of these two forces in the direction of motion. In the first case, you'd be correct.

    In the second case, the positive y-axis is downwards, however your motion is still going up, so you'd have

    av2+mg = -m d2x/dt2
     
  4. Jan 8, 2014 #3

    haruspex

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    You've used x and y interchangeably, but passing over that...
    You should get equivalent solutions for the two equations. If x = f(t) is a solution for the first equation then x = -f(t) is a solution for the second equation. If you are getting different solutions please post your derivations.
     
  5. Jan 8, 2014 #4

    HallsofIvy

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    You said you taking the y axis upward in the first paragraph and downwards in the second. What does that have to do with "x"?

    If that was simply a typo then you have [itex]md^2y/dt^2= -a(dy/dt)^2- mg[/itex] in the first paragraph and [itex]md^2y/dt^2= a(dy/dt)^2+ mg[/itex]. I see no reason for those to give different solutions.

     
  6. Jan 8, 2014 #5
    (edited out y for x, sorry!)...
     
  7. Jan 8, 2014 #6
    Ok, I'm actually solving for x in terms of v, then finding the maximum height h from that by putting v=0. This sort of blinded me from the fact everything was fine because the equations were both different - and then very stupidly I put h in for both of the solutions whereas for the choice of the x axis going downwards I obviously need x=-h. Thanks anyway :)
     
  8. Jan 8, 2014 #7
    Hmm, now I've found my two equations actually do give equivalent results, I'm not sure that this could possibly be correct...
     
  9. Jan 8, 2014 #8

    haruspex

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    No, both equations in the OP were correct, except for the switch between y and x.
     
  10. Jan 8, 2014 #9
    If the resistive force was linear instead, i.e αv, would this make things more complicated? For example, when the ball moves upwards:

    If have my x axis upwards, I'd have -mg-αv=md2x/dt2. If I had it downwards would I have mg+αv=md2x/dt2 or mg-αv=d2x/dt2, because now v would be a negative quantity...
     
  11. Jan 8, 2014 #10

    haruspex

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    On reflection, there were two deficiencies in my first response:
    1. I should have mentioned that your original equations are only valid while the object is moving upwards.
    2. The sign for g depends on what exactly you mean by 'g' in the equation. Based on the OP equations, you were taking g to stand for a positive number (in whatever units). My preference, though, is to have it stand for 'the acceleration due to gravity', and let the value be negative when appropriate. With that convention, your equations would have been
    ##m\ddot x = m g - \alpha v^2## with up positive (g taking a negative value), and
    ##m\ddot x = m g + \alpha v^2## with up negative (g taking a positive value)
    To take into account the possibility that the object is moving downwards:
    ##m\ddot x = m g - \alpha v|v|##
    and that works whichever sign convention you use for up/down.
    See if you can adapt that to the linear drag case.
     
  12. Jan 8, 2014 #11

    rock.freak667

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    I was mighty confused then! Ignore my post :redface:
     
  13. Jan 11, 2014 #12
    Sorry for the late reply. Yes, I now understand that I should use -mαv whatever my choice of the axes. Thanks.

    Sticking with the theme of signs, I'm having some issues with the rocket equation. I'm deriving the velocity as a function of time for a descending rocket. The result I should obtain is

    v=v0+gt+uln(mf/mi)

    where mf is the final mass, mi is the initial mass, u is the relative speed between rocket and the fuel it releases, v0 is the the speed at t=0, and v is the speed after a time t.

    Now I can obtain this if:
    I take positive downwards and consider the scene from an inertial frame on, say, the ground. At time t consider a rocket with mass M and speed v downwards. At time t+Δt consider the rocket with mass M-Δm and speed v+Δv downwards. A small piece of fuel has mass Δm and speed u+v downwards. Then let ΔP be the change in momentum, so

    ΔP=MΔv+uΔm.

    Divide through by Δt and let t→0 so

    dP/dt=Mdv/dt+udm/dt.

    dm/dt=-dM/dt so

    dP/dt=Mdv/dt-udM/dt.

    But dP/dt=Mg so this becomes

    Mg=Mdv/dt-udM/dt
    dv/dt=g+(u/M)dM/dt

    Solving by integrating between t=0 when the speed is v0 and the mass mi to a time t when the velocity is v and the mass mf, and we get the desired result.

    However, if I instead say that when the rocket emits the fuel at t+Δt, its velocity becomes v-Δv, then the whole thing messes up... I can't see why this shouldn't be a valid method of solving it, as in reality the rockets velocity really does decrease if I have positive downwards, rather than increase as in my above method.
     
    Last edited: Jan 11, 2014
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