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Rotational Kinematics

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data
    This isn't so much of a problem as a general question. I am trying to find the starting velocity of a spinning ball going upwards (in air, close to earth's surface, only force acting on it is the gravitational force) until its linear velocity reaches zero. I found the initial velocity two different ways (with mechanical energy and with kinematics), and I am getting answers that differ by a constant.


    2. Relevant equations
    [itex]mgh=\frac{7}{10}mv_0^2[/itex]

    [itex]v_f^2=v_0^2-2gh[/itex]
    (where the final velocity is 0)
    3. The attempt at a solution
    When I try to find the initial velocity, I can see that the two differ by a constant. I know that if the ball had no rotational kinetic energy, the equations would line up. However, I thought that the rotational motion would not have an effect on the linear motion of its center of mass. I think that the discrepancy is because the kinematics equation is derived from the conservation of energy of an object that has no rotational motion, but I'm not sure. I appreciate your help!
     
  2. jcsd
  3. Dec 14, 2013 #2
    Yes. If you set [itex]mgh[/itex] = [itex]\frac{1}{2}[/itex]mv2 then you get that the initial velocity is [itex]\sqrt{2gh}[/itex]. This is the same as the result from the kinematics equation when the final velocity is zero.

    Any additional energy that you add on top of the KE will cause the answers to be different. There's no need to include the rotational energy in a treatment of the vertical motion. For the same reasons that you wouldn't include, for example, the internal energy of the particles that make up the ball.
     
  4. Dec 14, 2013 #3
    Oh, wait. I think I just realized my error. When the ball reaches the point where its center of mass has a velocity of zero, the ball still retains its initial angular velocity. Therefore, the terms cancel out in the end, even if you include the rotational kinetic energy in the equation. Thank you for your help!
     
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