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Dynamics Module - Kinematics Problem

  1. Dec 18, 2013 #1
    Hi,

    I have an basic kinematics problem I can't seem to grasp -- hopefully someone can shed some light on it for me?

    The acceleration, a, of an object is equal to a=s, where s is the
    distance from the start position. Sketch a graph representing velocity
    against distance, s, for the object if, at t=0, the velocity v=u and s=0.

    [7 Marks]


    This is particularly frustrating, as it appeared in a very early tutorial and the rest of the course has been fine...

    Another example of essentially the same problem is:

    The angular acceleration, α, of an object is given by α = 2ω, where ω is the angular velocity.

    Sketch a graph representing angular velocity against angle, θ, rotated from the start position for the object if, at t = 0, the angular velocity ω = ωo and θ = 0.

    [5 Marks]



    Thanks for any help!
     
  2. jcsd
  3. Dec 18, 2013 #2

    Simon Bridge

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    Welcome to PF;

    a=s doesn't actually mean anything - the dimensions don't match for a start.

    It could be trying to say that the acceleration, in this case, is proportional to the displacement.
    In that case, it is not a constant and therefore this is not a kinematics problem. It's an exercise in drawing a v-s graph.

    You need to figure the function v(s) - which you can approximate by starting at t=0 and using small time-steps δt during which the acceleration is treated as a constant. Find s(0), s(δt), s(2δt), etc. and the same for v. Plot them against each other.
     
  4. Dec 18, 2013 #3

    collinsmark

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    Hello MechEngJordan

    Welcome to Physics Forums! :smile:

    Let's just start with the first one.

    So you have a differential equation,

    [tex] \ddot s = s [/tex]
    What are the solution to this type of differential equation? (Hint: it's a second order differential equation, so you can expect two solutions.) :wink:
     
  5. Dec 18, 2013 #4
    Thanks for the reply.

    I had assumed that the question was saying that a is proportional to s, and after looking back at the lecture notes for this tutorial, I have been playing around with the idea that if

    [tex]a = s [/tex]

    then

    [tex]v\frac{dv}{ds}=s[/tex]

    Rearranging and integrating:

    $$\int_{u}^{v} vdv = \int_{0}^{s} sds$$

    and solving for v:

    [tex]v= \sqrt{u^2+s^2}[/tex]

    But I'm not sure how much further that has gotten me...
     
  6. Dec 18, 2013 #5
    Hi,

    Thanks for the welcome and reply!

    This module hasn't used a second order ODE treatment for the problems -- so I hope that isn't what I'm expected to do... haha!
     
  7. Dec 18, 2013 #6

    collinsmark

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    Ummm, :uhh: I'm not sure what you did there, but that's not quite right. [Edit: actually, that might still be right. But there's a different approach.]

    Well, you're going to have to solve second order ODEs for these types of problems. So you might want to review your differential equations when you get the chance.

    But I'll give you a push for this problem. The independent solutions to the differential equation

    [tex] \ddot s = s [/tex]
    are
    [tex] s_1(t) = Ae^{t} [/tex]
    [tex] s_2(t) = Be^{-t} [/tex]
    Since the differential equation is linear, the general solution is the linear sum of the independent solutions,

    [tex] s(t) = s_1(t) + s_2(t) [/tex]

    So now apply your initial conditions and solve for A and B.

    (And while you're at it, you'll also benefit from reviewing the definitions of your hyperbolic functions such as sinh() and cosh().)
     
    Last edited: Dec 18, 2013
  8. Dec 18, 2013 #7

    AlephZero

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    You were doing OK with post #4. A common trick to integrate this type of equation is multiply both sides by 2s'. Then you have
    ##s'' = s##
    ##2s's'' = 2s's##
    ##d/dt(s'^2) = d/dt(s^2)##
    ##s'^2 = s^2 + C##
    which is what you got in #4.
    You can find C from the initial conditions: ##s' = u## when ##s = 0##.

    You have an equation connecting velocity ##s'## and distance ##s##. The question says "sketch a graph of velocity against distance." You don't need to do any more calculus - just draw the graph.

    Finding ##s## and ##s'## as functions of time, as in post #6, doesn't really help here.

    Note: If you go on to learn about the Lagrangian formulation of mechanics, you will find out that this "trick" to integrate the equation isn't really a trick. It comes from the physics of the situation - it's not just a bit of magic math.

    Another note: If you have studied simple harmonic motion, that is a very similar differential equation, but with a sign changed: ##s'' + s = 0##.
     
    Last edited: Dec 19, 2013
  9. Dec 19, 2013 #8
    Thanks. I'll be sure to brush up on differential equations and hyperbolic functions so as to better tackle problems of this sort.

    That makes a lot of sense, thank you.

    That approach from the Lagrangian formulation is really helpful.
     
  10. Dec 19, 2013 #9

    collinsmark

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    Oh, right. Velocity vs. distance. For some reason I kept reading that as velocity or distance vs. time. Sorry about that. Nevermind. :redface:
     
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