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Pinball machine problem

  1. May 6, 2006 #1
    I used PEsp = KE, which I got 16 = (0.5)(0.1)(v^2), and got 17.89, for my linear speed, and got that wrong. I know that once I get my linear speed, I plug that into an equation to get angular speed, but I would like to know how to get this linear speed correct. Thanks!
     
  2. jcsd
  3. May 6, 2006 #2

    nrqed

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    2.5 meters?? Wow, that's a huge pinball machine!:biggrin:

    You have forgotten to include the kinetic energy of rotation. You must add to the final kinetic energy [itex] {1 \over 2} I \omega^2[/itex]and use [itex]\omega = {v \over r} [/itex]
     
  4. May 6, 2006 #3
    So PEsp = KE, 16 = (1/2)(0.1)(v^2) + (1/2)(0.1)(0.1^2)(v/0.1)?
     
  5. May 6, 2006 #4

    nrqed

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    I am confused by your second term! First, you need to compute the moment of inertia I of a sphere (you seemed to have use the mass instead of I!) Also, the (v/r) must be squared. And I don't know what you (0.1)^2 means.
     
  6. May 6, 2006 #5
    Okay, my mistake! So is velocity = 12.649 m/s, and angular speed is 126.49 rad/s? I got this wrong...I dunno why.
     
    Last edited: May 7, 2006
  7. May 7, 2006 #6
    Bump! Anyone able to help me finish off this problem?
     
  8. May 8, 2006 #7

    andrevdh

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    All your numerical values, except the mass seems rather suspicious. A ball with a diameter of 20 centimeters going 2.5 meters up in a pinball machine?
    [tex]P_{ela}=P_{grav}+K_{lin}+K_{rot}[/tex]
    the elastic potential energy is converted to gravitational potential energy and linear and rotational kinetic energy giving
    [tex]kx^2=2mgh+mv^2+I{\omega}^2[/tex]
    then use
    [tex]v=\omega r[/tex]
    to subs for [itex]v[/itex] and solve for [itex]\omega[/itex]
     
  9. May 8, 2006 #8
    What is x?
     
  10. May 9, 2006 #9

    andrevdh

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    The first equation in my post states that the stored elastic energy in the spring is converted into the gravitational potential energy of the ball, the linear kinetic energy of the ball and the rotational kinetic energy of the ball.
     
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