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Conservation of Kenetic Energy: Rotational Style

  1. Apr 21, 2008 #1
    For my physics engine, I have a pretty simple formula to deal with collisions.

    Essentially, in a collision, both Momentum and Kinetic Energy have to be conserved. Thus, m1v1 + m2v2 = m1u1 + m2u2 and m1v1^2 + m2v2^2 = m1u1^2 + m2u2^2, and, given m1, m2, v1, and v2, we can solve for u1 and u2 (final velocities of each object) like this:

    u1 = (m1-m2)/(m1+m2)v1 + (m2+m2)/(m1+m2)v2
    u2 = (m1+m1)/(m1+m2)v1 + (m2-m1)/(m1+m2)v2

    And this all works well and good (I think)

    However, I am having trouble doing the same thing for rotation: what's the equivalent formula for rotation? Are there rotational equivalents for mass, velocity, momentum, and kinetic energy, and how can I use a formula similar to the one above to, given rotational momentum, calculate rotational velocities after a collision?
     
  2. jcsd
  3. Apr 21, 2008 #2
    p=I*w and KE=.5*I*w^2
    where I is the moment of intertia and w is the angular frequency. p should be replaced with L to show that it is angular momentum.

    L=m*(r cross v). the moment of inertia is the intergral of the distance (from the axis of rotation) squared over the entire mass
     
    Last edited: Apr 21, 2008
  4. Apr 21, 2008 #3
    in a rotational system you have to replace conservation of momentum with conservation of angular momentum, but conservation of energy still works (you might have to add a potential term, depending on the situation).
    Angular momentum L = mV x R (mass times velocity cross radius), if its circular motion then the cross product turns into simple multiplication L = mVR; keep in mind that the cross product gives you a ang mom vector perpendicular to the velocity and radius (use the right hand rule to get the sign correct + or -)
     
  5. Apr 21, 2008 #4
    i'm pretty sure its r cross v. point your hand away from the center and curl your fingers couterclockwise for an up moment and clockwise for a down moment.

    nate, also notice that angular momentum can be written L=r cross p
     
  6. Apr 21, 2008 #5
    totally, sorry for the mistake. (also note that RxP = - PxR , which often leads to terms canceling in many-body systems)
     
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