How to deal with direction on impluse?

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The discussion focuses on the application of the impulse-momentum theorem in a collision scenario involving a bullet and a bar. The user describes a situation where a bullet with mass m and initial velocity v_0 strikes a bar of mass M, leading to confusion regarding the direction of momentum and the resulting signs in calculations. The user correctly derives the initial angular velocity of the bar as ω = - (3I_b y) / (ML^2), but questions why the result is negative despite being correct. The conclusion emphasizes that the sign of angular velocity is a matter of convention, dependent on 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 [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

[tex] mv_0 = (m+M)V_f[/tex]

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

[tex] MV_f = -m(v_f-v_0) = -I_b[/tex]

then the initial angular momentum of bar can be given by

[tex] L = MV_f y = -I_b y[/tex]

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.
 
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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.
 
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 [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.

Thanks
 

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