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Instantaneous Veloctiy, kinda hard Help

  1. Nov 12, 2007 #1
    Initially, a 2.00kg mass is whirling at the end of a string in a circular path of .750 m on a horizontal frictionless surface with a tangential speed of 5 m/s. The string has been slowly winding around a vertical rod, and a few seconds later the length of the string has shortened to .250 m. What is the instantaneous speed of the mass at the moment the string reaches a length of .250m.

    I = mr^2 but i must be missing a formula because that won't work because it would give me a smaller number and it obviously speeds up the shorter the string gets.
     
  2. jcsd
  3. Nov 12, 2007 #2

    Doc Al

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    Staff: Mentor

    angular momentum

    What they undoubtedly want you to assume is that angular momentum is conserved. How do you compute the angular momentum of a point mass?

    [Note well: Unfortunately, this problem is bogus since the string makes an angle with the radius as it wraps around the pole and angular momentum is not conserved. No work is done on the system so its speed cannot change. But I suggest that you ignore that inconvenient fact and carry on.]
     
  4. Nov 12, 2007 #3
    L=Iw? I'm still kind of lost...
     
  5. Nov 12, 2007 #4

    Doc Al

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    Staff: Mentor

    That's a perfectly OK way to approach it (but not the only way). What's I? What's w? (How does w relate to tangential speed?)

    [tex]I_1 \omega_1 = I_2 \omega_2[/tex]
     
  6. Nov 12, 2007 #5
    v/r =w and I = mr^2 so...

    I = 2kg*.750^2 = 1.125
    5m/s/.750 = 6.667
    L=1.125*6.667 = 7.5

    I=2kg*.250^2 =.125

    7.5/.125 = 80 m/s

    Where did i mess up?
     
  7. Nov 12, 2007 #6

    Doc Al

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    Staff: Mentor

    (1) Recheck your arithmetic.
    (2) L/I = w, not v. (But w = v/r.)
     
  8. Nov 12, 2007 #7
    Awesome! I got it thanks a lot!!
     
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