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Conservation of angular momentum

  1. Mar 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A uniform stick 1.00 m long with a total mass of 270 g is pivoted at its center and is initially stationary. A 30 g piece of clay is thrown at the stick midway between the midpoint of the stick (pivot) and one end. The clay piece is going at 50 m/s and sticks to the stick. What is the angular velocity of the stick after the collision?

    2. Relevant equations
    Conservation of angular momentum.
    Li = Lf

    3. The attempt at a solution
    I = moment of inertia for stick
    ms = mass of stick
    mb = mass of bullet
    v = velocity of the bullet
    r = distance between the pivot and where the bullet hit = 0.5 m
    Iω(rod) + (mb)vr = Iωf(rod) + (mb)ωf
    Iω(rod) = 0
    ωf = ((mb)vr)/(I + (mb))
    I = (1/12)(ms)r^2
    ωf = 14.286 rad/s

    I assumed the angular velocity is the same for the mass and the stick, since they are stuck together.

    Thanks.
     
  2. jcsd
  3. Mar 22, 2016 #2
    You are correct in saying that the r value is the distance between the pivot and the point where the clay hits. However, the r value is not 0.5 meters here. The entire stick is 1.00 m, and the clay sticks at the point halfway from the pivot to the end. The pivot is in the middle.
     
  4. Mar 22, 2016 #3
    Your conservation equation has an error. Corrected, it stands as: [itex] m_b v r = (I + m_b r^2)\omega_{f} [/itex]

    The stick in the initial configuration has no angular momentum. All the angular momentum in the final configuration is contributed by the bullet.

    [EDIT]: Also, how is r = 0.5 m. If the stick is pivoted about the centre and the origin of the coordinate system has been chosen to be this point, then r is midway between the pivot/origin and the end (r=0.25 m)
     
  5. Mar 22, 2016 #4
    So,
    ωf = ((mb)vr1)/(I + (mb)) , where r1 = 0.25 m (half of the half...)
    I = (1/12)(ms)r^2 (r = 0.5 m (half of the stick)
    ωf = 10.526 rad/s
     
  6. Mar 22, 2016 #5
    EDIT: NVM read it wrong

    How did you get [itex] m_b r^2 [/itex]?
     
  7. Mar 22, 2016 #6
    The angular momentum of an object is Iω. I is the moment of inertia. The moment of inertia of a point mass is mr2.
     
  8. Mar 22, 2016 #7
    In your initial post, you wrote :

    The second term on the right side of this equation is incorrect. It should be [itex] m_b \omega_f r^2 [/itex]. Remember, angular momentum has dimensions M L^2 T^(-1).
     
  9. Mar 22, 2016 #8
    Oh ok, I didn't think you could assume that clay is a point mass. If you do that, the angular velocity if 50 rad/s... Is that correct?
     
  10. Mar 22, 2016 #9
    This is not the value I am finding for the final angular momentum.
    Make sure to use mbvr=(I+mbr2f and to use I as the moment of inertia of the rod, which you previously said, was (1/12)mrL2.
    Also make sure to use all lengths in meters and masses in kilograms so you get the correct final units.

    edit: the assumption that the clay is a point mass is a common occurrence in these types of problems. If a bullet or "piece" of clay is hitting something and the problem doesn't give you a value for its moment of inertia, it is usually safe to assume that they want you to treat it as a point mass.
     
  11. Mar 22, 2016 #10
    Is it 15.38 rad/s?
     
  12. Mar 22, 2016 #11
    That's what I get:thumbup:
     
  13. Mar 22, 2016 #12
    Thanks.
     
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