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Bullet Strikes Wheel - find rotations/second

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    The wheel shown in the sketch below, is comprised of two 50.0 cm long thin rods of negligible mass and a thin metal ring with a mass of 10.00 kg, mounted on a very low friction bearing. A 100 gram lead weight is shot horizontally at the stationary wheel with an initial speed of 50.0 m/s and sticks to the rim of the wheel. What is the rotational speed of the wheel plus lead mass immediately after the collision? Give your answer in revolutions/second.


    2. Relevant equations
    I = mr^2
    L = Iω
    KE = 1/2 mv^2
    KErot = 1/2 Iω^2

    3. The attempt at a solution
    1/2 mv^2 = 1/2 Iω^2
    1/2(.1 kg) (50 m/s)^2 = 1/2( 10.1 kg )(.25 m)^2 ω^2
    ω^2 = (125 kg * m^2 / s^2)/ .315625 kg * ms^2)
    ω = 19.9 / s
    Rotations = ω/2pi = 19.9/2pi = 3.16 rev/s

    The answer is .315 RPS so I'm not sure what I'm doing wrong
     
  2. jcsd
  3. Apr 25, 2012 #2

    Doc Al

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    Staff: Mentor

    Energy is not conserved--there's an inelastic collision taking place.

    But what is conserved?
     
  4. Apr 25, 2012 #3
    Is linear momentum/angular momentum conserved? I thought you couldn't go between the two
     
  5. Apr 25, 2012 #4

    Doc Al

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    Staff: Mentor

    One is, one isn't. Which?
    What do you mean?
     
  6. Apr 25, 2012 #5
    I'd assume linear momentum isn't conserved because the momentum absorbed would be redirected to the earth, but doesn't the bullet have no angular momentum? I tried to think of treating it as a tangential force but there's no acceleration
     
  7. Apr 25, 2012 #6

    Doc Al

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    Staff: Mentor

    Good.
    Sure it does. L = r X p.
     
  8. Apr 25, 2012 #7
    Since r is the vector position relative to the origin, it would be .25m (where the origin in the center of the wheel, and the radius is .25m) and the p would be linear momentum, m*v, .1kg * 50 m/s = 5 kg * m/s

    It's a tangent so the ABsin(theta) is simply AB, and that would be (5 kg * m/s)(.25m) = 1.25 m^2/s

    L= Iw, I is MR^2 for a thin cylindrical shell

    1.25 m^2/s = (10.1 kg)(.25 m)^2 w
    w = 1.98 rad/s
    1.98 rad/s / 2pi = .315 rotations per second

    Thanks for help.
     
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