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Homework Help: Find the final angular speed of the target

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data

    A fairgorund target consists of a uniform circular disk of mass M and radius a that can turn freely about its diamerter which is fixed in a vertical position. Iniitally the target is at rest . A bullet of mass m is moving with speed u along a horizontal line at right angles to the target . The bullet embeds itself in the target at a point distance from the rotation axis . Find the final angular speed of the target.(the moment of inertica of the disk about its roation axis is Ma^2/4




    2. Relevant equations

    Equations for rotational energy

    3. The attempt at a solution

    T_trans., initial + T_rot., initial, + V_initial=T_rot, final . There is no translational energy now that the bullet is embedded into the block.

    1/2*m*u^2++0+0=1/2(Ma^2/4
    m*b^2/4)*omega^2? Is my equation correct? There is no potential energy.
     
  2. jcsd
  3. Jan 29, 2009 #2

    tiny-tim

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    inelastic collision

    Hi pentazoid! :smile:

    Nooo … energy is not usually conserved in collisions …

    and certainly not when one body embeds itself in the other!

    Hint: momentum is always conserved in collisions …

    and that includes angular momentum. :wink:
     
  4. Jan 29, 2009 #3
    Re: inelastic collision

    yes I know energy isn't conserved because the collision is not elastic since the bullet becomes embedded into the block.

    L_initial=L_final

    m*v_bullet+0=(M+m)v_new?

    and then I can plug my v_new into my energy equation?
     
  5. Jan 29, 2009 #4

    tiny-tim

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    No, that's linear momentum …

    you need angular momentum. :wink:

    (and what energy equation? :confused:)
     
  6. Jan 29, 2009 #5
    right , how silly of me.

    m*v_bullet*a=(m+M)*v_final*(a+b)? Would I then plug v_final into my energy equation.

    a being the horizontal distance and a+b being the horizontal distance plus the radius of the disk?
     
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