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Homework Help: What angular speed is the stick spinning after the collision?

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A uniform stick 1.0 m long with a total mass of 220 g is pivoted at its center. A 3.2 g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s

    With what angular speed is the stick spinning after the collision?

    2. Relevant equations

    I*wi=I*wf

    I of stick = (1/12)ML^2


    K= (1/2)(Mb*Vf^2+(1/2)Iw^2

    3. The attempt at a solution

    (1/2)*Mb*Vi^2 = (1/2)(Mb*Vf^2+(1/2)Iw^2

    i use the above equaltion but it wrong
    anyone got any idea how to solve this XD
    please help me ^_^
     
  2. jcsd
  3. May 4, 2008 #2

    alphysicist

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    Hi saturn67,

    Since this is not an elastic collision, kinetic energy is not conserved and so

    (1/2)*Mb*Vi^2 = (1/2)(Mb*Vf^2+(1/2)Iw^2

    is not true.

    What is conserved in this collision?
     
  4. May 4, 2008 #3
    angular momentum conserved?

    I*wi = I*wf

    but bullet don't have angular momentum?

    so 0=I*wf

    so it dont make sense >.<
     
    Last edited: May 4, 2008
  5. May 4, 2008 #4

    alphysicist

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    Yes, that's right. So your first equation (written out for two separate objects)

    [tex]I_{1} \omega_{1,i} +I_{2}\omega_{2,i} =I_{1} \omega_{1,f} +I_{2}\omega_{2,f}[/tex]

    should lead to the answer. What do you get?
     
  6. May 4, 2008 #5
    how to relate I1*w1i to bullet??

    i know I2w2i= 0 since stick is at rest
     
  7. May 4, 2008 #6

    alphysicist

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    We can treat the bullet as a point particle. What is the moment of inertia of a point particle at the point where it hits the stick? We know its linear speed; how is that related to its angular speed?
     
  8. May 4, 2008 #7
    humm

    soo L= R cross P

    p= mv
    so L=m1*v1*r

    is right?
     
  9. May 4, 2008 #8

    alphysicist

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    Yes, that's a good way to do it. So you know:

    [tex]
    I_1 \omega_{1,i} = m_1 v_1 r_1
    [/tex]

    and you can do the same thing for the angular momentum of the bullet after the collision. Once you find [itex]I_2\omega_{2,f}[/itex], you can then solve for [itex]\omega_{2,i}[/itex].

    (If you had done them separately you would find [itex]I=m r^2[/itex] and [itex] \omega=v/r[/itex], which would have given the same answer of mvr.)
     
  10. May 4, 2008 #9
    ok thank alot alphysicist that is a big help XD

    yep i got it

    w= 4.8 rad/sec
     
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