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Q. 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 $ \omega$. If the angular momentum of the system, calculated about O and P are denoted by $ L_o$ and $ L_p $ respectively, thenPhysics Question Image

IIT JEEIIT JEE 2012System of Particles and Rotational Motion

Solution:

Angular momentum of a particle about a point is given by $L = r \times P = m(r \times v )$
For $L_0$
$ | L |= (mvrsin \theta ) = m (R \omega ) (R )sin 90^\circ $ = constant
Direction of $ L_0 $ is always upwards. Therefore, complete $ L_0 $ is constant, both in magnitude as well as direction.
For $L_p$
$|L _P |= (mvrsin \theta ) = (m) (R \omega) (l)sin 90^ \circ = (mRl \omega)) $
Magnitude o f $L_P$ will remain constant but direction o f $L_P$ keeps on changing.

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