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Softball Energy Question

  1. Jun 27, 2007 #1
    Hi I just need confirmation on a problem.

    Q: A softball pitcher rotates a 0.250-kg ball around a vertical circular path of radius 0.600 m before releasing it. The pitcher exerts a 30.0-N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 15.0 m/s. If the ball is released at the bottom of the circle, what is its speed upon release?

    A: PEg= mgh
    F=ma; F= 30.0 N, m= .250 kg, a= 30.0 N/.250 kg= 120 m/s^2

    v^2= vo^2 + 2ax
    v^2= (15m/s)^2 + 2*(120m/s^2)*.6pi.
    v= 26.0
    Thanks so much.
     
  2. jcsd
  3. Jun 27, 2007 #2

    G01

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    The force acts parallel to the ball's motion at all points in the circle. So, does the force do any work?
     
    Last edited: Jun 27, 2007
  4. Jun 27, 2007 #3
    no but how does that change anything?
     
  5. Jun 27, 2007 #4

    G01

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    A force acting PARALLEL to motion WILL do work. Now you should see how this changes the situation. Gravity is not the only force accelerating the ball. I would recommend going about this problem from an energy perspective. If you knew the balls kinetic energy you could find its speed, correct? Can you find the kinetic energy of the ball when it is released now that you know all of the forces accelerating it?
     
  6. Jun 27, 2007 #5
    How can I find the kinetic energy of the ball when it is released? Sorry if I am asking too much.
     
  7. Jun 27, 2007 #6

    G01

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    From you first post, I can see that you know that the gravitational potential energy is turned into kinetic energy. Now, if you can find how much energy is added to the ball from the force, you'll have it's total kinetic energy upon release. Start here:

    [tex]K_{final}=K_{initial}+U_{gravitational} + W[/tex]

    See how far you can get now.
     
  8. Jun 28, 2007 #7
    so according to your equation, i should get the following equation:

    KEf = (.5)(.250)(15.0^2) + (.250)(9.8)(2pi.6) + (30.0)(1.2)

    Im not sure what change of x is.
    Thank you very much for helping.
     
  9. Jun 28, 2007 #8

    G01

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    The change in x will be the distance the force acts on it, which would be half the circumference of a circle.

    Also, I noticed a mistake you have. You have the change in height in the potential term equal to the circumference. This is not the case. What is the change in height from the top of the circle to the bottom? This will be the change in height of the ball.

    Other than this, you seem to be on the right track. Do you understand where I got that equation from?
     
  10. Jun 28, 2007 #9
    Ok i got the answer. but im interested in where you got the equation from.
     
  11. Jun 28, 2007 #10

    G01

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    It's just comes from the definition of work.

    The work done on an object will be equal to the total change in mechanical energy:

    [tex]W=\Delta K +\Delta U[/tex]

    [tex]W = K_f -K_i + U_f - U_i[/tex]

    If you say the final potential energy is 0, you should be able to solve for Kf and get the equation we used above. I'm glad you asked! It shows you care about understanding the actual physics of the problem, not just plugging and chugging.
     
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