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Substituting the angular velocity and moment of inertia

  1. May 2, 2012 #1
    Hello, I need some help verifying whether what I did is correct.

    1. The problem statement, all variables and given/known data

    Attached is a screenshot of the problem.

    2. Relevant equations

    Angular momentum [itex]L = \vec{r} × m\vec{v} = mvr \sin{\theta}[/itex]
    Conservation of angular momentum.

    3. The attempt at a solution

    This is my solution:

    a) Before the collision, the block's angular momentum is zero (since it is at rest), and the angular momentum of the bullet is [itex]l mv \sin(\frac{\pi}{2}) = mvl[/itex], since the bullet's velocity vector and the block's direction vector to the pivot are perpendicular, so the total angular momentum of the system before the collision is [itex]mvl[/itex].

    b) The bullet and block stick together, so from conservation of angular momentum the bullet + block system must have angular velocity [itex]mvl[/itex]. This means that:

    [itex]m v l = (m + M) v_f l[/itex]

    Where [itex]v[/itex] is the bullet's initial velocity and [itex]v_f[/itex] is the velocity after collision.

    So [itex]v_f = \frac{m}{m + M} v[/itex]

    And [itex]v_f = lw[/itex], so [itex]w = \frac{v_f}{l}[/itex] the angular momentum after the collision must then be:

    [itex]w = \dfrac{\dfrac{m}{m + M} v}{l}[/itex]

    c) Before the collision, the kinetic energy is purely linear, the kinetic energy is the block at rest is zero, and the kinetic energy of the bullet is [itex]\frac{1}{2} m v^2[/itex]. After the collision, the kinetic is purely rotational, with angular velocity [itex]w[/itex] (found above) and mass [itex]m + M[/itex].

    Because the rod is massless, the system's center of mass is the block's center of mass. So after the collision, the system is rotating about the pivot, so the system's moment of inertia is:

    [itex]I = (m + M) l^2[/itex]

    So its rotational kinetic energy must be [itex]K = \frac{1}{2} I w^2[/itex]

    Substituting the angular velocity and moment of inertia in, we get [itex]K = \frac{1}{2} (m + M) l^2 \left ( \dfrac{\dfrac{m}{m + M} v}{l} \right )^2[/itex]

    Which simplifies to to [itex]K = \frac{1}{2} \frac{m^2}{m + M} v^2[/itex]

    So before the collision, [itex]K = \frac{1}{2} m v^2[/itex], and after the collision, [itex]K = \frac{1}{2} \frac{m^2}{m + M} v^2[/itex]. Clearly the fraction [itex]\frac{m^2}{m + M}[/itex] will always be less than [itex]m[/itex] so kinetic energy is not conserved: the collision is inelastic.

    ----

    So did I get this right? I generally suck at physics (although good at math) so I'm pretty sure I messed up at some point, but the results I get kind of seem to work, so I don't know :mad:
     

    Attached Files:

  2. jcsd
  3. May 2, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Your methods and results look fine.
     
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