# How to deal with direction on impluse?

1. Dec 31, 2008

### KFC

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

$$mv_0 = (m+M)V_f$$

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 $$I=ML^2/3$$ (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.

2. Dec 31, 2008

### PhanthomJay

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.

3. Dec 31, 2008

Thanks