- #1
RingNebula57
- 56
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Hello everyone!
I have a problem , to which I do not understand the law of conservation of angular momentum... I searched this problem on the web and it is obvious that I am making the mistake.
So we have a rod of length ##L## and mass ##m## that is lying on a horizontal frictionless table. We hit the rod at one end, perpendicular to the rod, with a ball of mass ##m## that is rolling on the table with initial speed ##v_0##. After they collide elastically , the rod begins rotational and translational motion with angular velocity ##\omega##(about its CM) and velocity ##v_1##(of the CM) , while the ball continues its translation with velocity ##v_2##, in the same direction as the initial velocity ##v_0##.
Now , if we consider the moment of inertia of the rod about its center of mass to be ## I ## , I say that the conservation of angular momentum about the center of mass of the rod is:
## m \cdot v_0 \cdot \frac{L}{2} = m \cdot (v_2 - v_1) \cdot \frac{L}{2} + I \cdot \omega ##
But the solution says:
## m \cdot v_0 \cdot \frac{L}{2} = m \cdot v_2 \cdot \frac{L}{2} + I \cdot \omega ##
##( I = m \cdot \frac {L^2}{12} )##
why ?
Isn't the conservation of angular momentum always relative to the center of mass of the rod?
I have a problem , to which I do not understand the law of conservation of angular momentum... I searched this problem on the web and it is obvious that I am making the mistake.
So we have a rod of length ##L## and mass ##m## that is lying on a horizontal frictionless table. We hit the rod at one end, perpendicular to the rod, with a ball of mass ##m## that is rolling on the table with initial speed ##v_0##. After they collide elastically , the rod begins rotational and translational motion with angular velocity ##\omega##(about its CM) and velocity ##v_1##(of the CM) , while the ball continues its translation with velocity ##v_2##, in the same direction as the initial velocity ##v_0##.
Now , if we consider the moment of inertia of the rod about its center of mass to be ## I ## , I say that the conservation of angular momentum about the center of mass of the rod is:
## m \cdot v_0 \cdot \frac{L}{2} = m \cdot (v_2 - v_1) \cdot \frac{L}{2} + I \cdot \omega ##
But the solution says:
## m \cdot v_0 \cdot \frac{L}{2} = m \cdot v_2 \cdot \frac{L}{2} + I \cdot \omega ##
##( I = m \cdot \frac {L^2}{12} )##
why ?
Isn't the conservation of angular momentum always relative to the center of mass of the rod?
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