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# two particles of mass m each are attached to a light rod of length d, one at its center and the other at free end B.the rod is fixed at the other end O and is rotated in a plane at an angular speed w (omega). calculate the angular momentum of the particle at the end with respect to the particle at the center?In this question particle B and particle A are rotating on the same rod so they will have same angular speed omega. So angular speed of B wrt A should be 0 therefore angular momentum is 0 but there is something else written in the solution image.

Eshan
3 years ago
Dear student,

When you are using the logic$\dpi{80} \omega_1=\omega_2$, then the angular speed you talk about is about the center O. Thus the relative angular velocity of 2 with respect to 1 about center is always zero. Hence the angular displacements of the two about the center is same all the time.
However the angular displacement of the second with respect to first would not be zero all the time since we can note that the second particle would actually rotate about the first as it moves from right hand side of it to left hand side. Hence in this question, we can ensure, that the relative angular vecloity about the first point itself as been asked.
Lunatic
13 Points
3 years ago
Let me clear you about angular momentum, it is the rate of change of angle turned. So A and B are on the same rod so they will turn same angle in the same time therefore angular velocity of A wrt B is 0 therefore I m correct

Lunatic
13 Points
3 years ago
Thank you, actually I was wrongly interpreting the question