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Question

Two particles of mass 'm' each are attached to a light rod of length 'd', one at its centre and the other at a free end. The rod is fixed at the other end and is rotated in a plane at an angular speed ω. Calculate the angular momentum of the particle at the end with respect to the particle at the centre.


A

12mωd2^j

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B

14mωd2^j

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C

13mωd2^j

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D

23mωd2^j

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Solution

The correct option is B

14mωd2^j


The situation is shown in figure. The velocity of the particle A with respect to the fixed end O is vA=ω(d2) and that of B with respect to O is vB=ωd. Hence the velocity of B with respect to A is vBvA=ω(d2).The angular momentum of B with respect to A is, therefore,

L=mvr=mω(d2)d2=14mωd2

Along the direction perpendicular to the plane of rotation i.e. j


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