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Rotation angular velocity of a turntable

  1. Nov 21, 2014 #1
    I've spent at least an hour trying to figure this out, but can't seem to figure out how to solve this.

    1. The problem statement, all variables and given/known data

    A 170g , 30.0-cm-diameter turntable rotates on frictionless bearings at 66.0rpm . A 25.0g block sits at the center of the turntable. A compressed spring shoots the block radically outward along a frictionless groove in the surface of the turntable.

    What is the turntable's rotation angular velocity when the block reaches the outer edge?

    2. Relevant equations
    Conservation of momentum: Ii*wi = If*wf

    3. The attempt at a solution
    I(turntable) = Mr^2 = (.170 kg)(.15 m)^2 = 3.825*10^-3 kg m^2
    I(block)o = mr^2 = 0
    I(block)f = mr^2 = (.025 kg)(.15 m)^2 = 5.265*10^-4 kg m^2
    wi = ((66 rpm)(2pi))/60 = 6.91 rad/s
    I(turntable+block)f = 3.825*10^-3 + 5.265*10^-4 = 4.3875*10^-3 kg m^2

    (3.825*10^-3)(6.91) = (4.3875*10^-3)wf
    0.1826 = (4.3875*10^-3)wf
    wf = 6.02 rad/s -- wrong

    Initially, I tried using conservation of energy, and ended up with 5.52 rad/s. The program told me "not quite" and suggested a rounding error, so I think 5.52 might be close to the right answer, but I can't seem to figure out why I'm not getting it. Any help would be much appreciated!
     
    Last edited: Nov 21, 2014
  2. jcsd
  3. Nov 21, 2014 #2
    I think you should recheck the equation of ##I_{turntable}##.
     
  4. Nov 21, 2014 #3
    Would it just be 1/2*(.170)(.15)? That's what I had at first, but I wasn't sure if that was correct or not.
     
  5. Nov 21, 2014 #4
    If I use Iturntable = (0.5)(0.17)(0.15)2, I get Iturntable = 1.9125 * 10-3 kg m2. Plugging this in, I get a final answer of 5.42 rad/s. Does this seem right?
     
  6. Nov 21, 2014 #5
    Believe yourself.
     
  7. Nov 21, 2014 #6
    By the way, due to the error generated by calculating, recommend you to calculate at the last pace.
    You can get the best result.
     
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