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How to deal with direction on impluse?

  1. Dec 31, 2008 #1


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    I am using impulse-momentum theorem in problem solving. But I am quite confusing about deal with the direction of momentum so always get the wrong sign.

    For example, assume a long thin bar with mass M and length L hanging from a fixed frictionless point A at the ceiling, the bar is stay at rest. Now a bullet with mass m and initial velocity [tex]v_0[/tex] moving horizontally towards the bar and hit it at point B (the distance b/w A and B is y). Finally, the bullet embed into bar and then moving together with it. The instantaneous horizontal impulse when it hit the bar is [tex]I_b[/tex], find the intial angular velocity of the bar.

    Since the system's total momentum is conserved, we can write

    mv_0 = (m+M)V_f

    and the change of the momentum of the bullet is the impulse

    MV_f = -m(v_f-v_0) = -I_b

    then the initial angular momentum of bar can be given by

    L = MV_f y = -I_b y

    After collision, the bar (and the bullet) move around pivot A, the moment of inertia about A is [tex]I=ML^2/3[/tex] (ignore the mass of bullet). With the help of following equation

    [tex]L = I\omega[/tex]

    we find that

    [tex]\omega = \frac{L}{I} = - \frac{3I_b y}{ML^2}[/tex]

    the result (the value) is correct, but it should be positive. I have no idea where is the mistake come from.
  2. jcsd
  3. Dec 31, 2008 #2


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    Since the impulse on the bullet is negative, it follows from Newton 3 that the impulse on the rod must be equal but opposite, that is, positive. Bottom line, however, is that whether one calls [tex] \omega [/tex] positive or negative, it's rather a question as to whether the rotation is clockwise or counterclockwise. The sign choice is largely a matter of convention.
  4. Dec 31, 2008 #3


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