A small ball of mass m suspended from the ceiling at a point O by a thread of length l moves along a horizontal circle with a constant angular velocity `omega`. Relative to which points does the angular momentum M of the ball remain constant? Find the magnitude of the increment of the vector of the ball's angular momentum relative to the point O picked up during half a revolution.

Explanation

(a) The ball is under the influence of forces `vecT` and `mvecg` at all the moments of time, while moving along a horizontal circle. Obviously the vertical component of `vecT` balance `mvecg` and so the net moment of these two about any point becomes zero. The horizontal component of `vecT`, which provides the centripetal acceleration to ball is already directed toward the centre (C) of the horizontal circle, thus its moment about the point C equals zero at all the moments of time. Hence the net moment of the force acting on the ball about point C equals zero and that's why the angular mommentum of the ball is conserved about the horizontal circle.
(b) Let `alpha` be the angle which the thread forms with the vertical.
Now from equation of particle dynamics:
`Tcosalpha=mg` and `T sin alpha=momega^2l sin alpha`
Hence on solving `cos alpha=(g)/(omega^2l)` (1)
As `|vecM|` is constant in magnitude so from figure.
`|DeltavecM|=2Mcos alpha` where
`M=|vecM_i|=|vecM_f|`
`=|vecr_(bo)xxmvecv|=mvl` (as `vecr_(bo)_|_vecv`)
Thus `|DeltavecM|=2mvlcosalpha=2momegal^2sin alphacosalpha`
`=(2mgl)/(omega)sqrt(1-((g)/(omega^2t))^2)` (using 1).