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Question

A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoing circular motion in the x-y plane with centre at O and constant angular speed ω. If the angular momentum of the system, calculated about O and P are denoted by LO and LP respectively, then


A
LO and LP do not vary with time.
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B
LO varies with time while LP remains constant.
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C
LO remains constant while LP varies with time.
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D
LO and LP both vary with time.
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Solution

The correct option is C LO remains constant while LP varies with time.
Angular momentim about O :
As we know, angular momentum L=m(r×V)
Here r=OA. Since mass is undergoing circular motion,V will be in the tangential direction of a circle in the plane of circular motion.


LO=m(OA×V)
Angular momentum will be in z-direction (perpendicular to the plane of circular motion).
After some time, mass is at B. Velocity will be in tangential direction. Hence
LO=m(OB×V) has the same magnitude and direction.
i.e LO remains constant.

Angular momentum about point P :
Here, r=PA
and angular momentum will be L=m(PA×V)
V will be in tangential direction in the circular plane. Hence, the direction of angular momentum will be as shown in figure. (by right hand thumb rule).


After time t, when particle is at B, r=PB and velocity vector will be in tangential direction.
So, angular momentum L=m(OB×V) and the direction of angular momentum (from right hand rule ) as shown in the figure.

Hence, magnitude of angular momentum will be same about point P, but direction is changing.
Angular momentum about point O is constant, but about point P it is varying.

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