- #1
KFC
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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.
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.