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Kinematics and rigid body simulation

  1. Jul 10, 2010 #1
    Hello, I have a question regarding how I'm simulating motion using the common time integrating approach to simulating rigid bodies.

    I have net forces and solve for position from those:
    acceleration = force / mass
    velocity = initial velocity + acceleration * time
    position = initial position + velocity * time

    but if you combine the above and distribute you get something that matches the kinematic equation, but is missing the one half term. Where'd it go?

    position = initial position + (initial velocity + (force / mass) * time) * time)
    position = initial position + initial velocity * time + (force / mass) * time^2

    which does not equal the kinematic equation
    y = y0 + v0*t + (1/2)a*t^2

    Where'd the one half go?
     
  2. jcsd
  3. Jul 10, 2010 #2
    That applies only if the velocity is constant. Since you have an acceleration, velocity is not constant & you cannot use these expressions.

    Do you know calculus? (simple integration). It can very easily be proved with calc.
     
  4. Jul 10, 2010 #3

    Cleonis

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    Gold Member

    As graphene points out, the velocity isn't constant.

    Here is a short route to arriving at y = 1/2*a*t^2

    If the velocity is constant then in diagram the distance covered is proportional to the area of a rectangle. Draw a diagram with time on the horizontal axis and velocity on the vertical axis. So the line representing the velocity at each point in time is horizontal. As time progresses the rectangle stretches in horizontal direction. In other words, the distance covered is given by the area that is enclosed between the horizontal axis and the line that represents the velocity.

    Now the case of uniform acceleration.
    Then the line representing the velocity has a uniform slope. Interestingly, the total distance covered is still the area that is enclosed between the horizontal axis and the line that represents the velocity. The enclosed area is a rectangular triangle, and the surface area of a triangle is (1/2)*width*height.

    At t=0 the velocity is 0,
    At tend the velocity has climbed to a*tend

    The area of the enclosed triangle is (1/2)*a*tend^2
     
  5. Jul 10, 2010 #4
    These equations should be good for constant acceleration, and since I'm doing fixed time steps changes acceleration should always be constant. I've taken calculus and physics and have the books and follow the derivation of the kinematic equation.

    y'' = a
    y' = a*t + c1
    y = (1/2)a*t^2 + c1 * t + c2

    where c1 = initial velocity and c2 = initial position

    Looking in my physics book I see they define the kinematic equation in terms of average velocity

    avgvel = (initial velocity + velocity) / 2

    Which I think is where the one half is coming from, but do I need to find average velocity if I'm doing 16ms time steps? I'll start looking into the difference of average velocity / velocity to see if I can figure things out.
     
  6. Jul 10, 2010 #5
    Alright Cleonis's explaination relating it to triangle area made things mentally click for me, I need to use the average velocity term. Thanks!
     
  7. Jul 10, 2010 #6

    Cleonis

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    Gold Member

    By coincidence a calculation that works with average velocity can end up looking the same. Don't be distracted by that: the relation:

    y = (1/2)*a*t^2

    is exact. Whenever the acceleration is uniform it is exact for every point in time.

    Your purpose is not to average the velocity; your purpose is to use the exact velocity at each point in time. In this particular case it ends up looking the same.
     
  8. Jul 11, 2010 #7
    Last edited by a moderator: Apr 25, 2017
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