Question

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.

Answer 1

Angular momentum does not remain constant from any point in inertial frame.
$Tcos\theta =mgTsin\theta =m{\omega }^{2}r=m{\omega }^{2}lsin\theta \Rightarrow m{\omega }^{2}lcos\theta =mg\Rightarrow cos\theta =\frac{mg}{m{\omega }^{2}l}=\frac{g}{{\omega }^{2}l}Inhalfarevolutionthedirectionoftheangularmomentumisreversedchangeinangularmomentum=2mvl=2m\omega l^{2}sin\theta =2m\omega l^2\sqrt{1-{\left(\frac{g}{{\omega }^{2}l}\right)}^2}$