How to deal with direction on impluse?

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The discussion centers on the application of the impulse-momentum theorem in a scenario involving a bar and a bullet. The user struggles with the direction of momentum, leading to incorrect sign assignments in their calculations. They correctly derive the initial angular velocity of the bar post-collision but note that their result appears negative when it should be positive. The confusion arises from the convention used for defining the direction of rotation, as the impulse on the bullet is negative while the impulse on the bar is positive. Ultimately, the sign of angular velocity is a matter of convention related to the chosen direction of rotation.
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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 v_0 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 I_b, find the intial angular velocity of the bar.

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

<br /> mv_0 = (m+M)V_f<br />

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

<br /> MV_f = -m(v_f-v_0) = -I_b<br />

then the initial angular momentum of bar can be given by

<br /> L = MV_f y = -I_b y<br />

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

L = I\omega

we find that

\omega = \frac{L}{I} = - \frac{3I_b y}{ML^2}

the result (the value) is correct, but it should be positive. I have no idea where is the mistake come from.
 
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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 \omega 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.
 
PhanthomJay said:
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 \omega 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.

Thanks
 
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