A stone of mass m tied with a thread is rotating along a horizontal circular path (force of gravity is neglected). The length of the thread is decreasing gradually in such a manner that the angular momentum of the stone remains constant with respect to the centre of the circle. If the tension in the thread is T= Ar^(n) , where A = constant , r = instantaneous radius of the circle, then find the value of n.

A stone of mass m tied with a thread is rotating along a horizontal ci

1.	A stone of mass m tied with a thread is rotating along a horizontal circular path (force of gravity is neglected). The length of the thread is decreasing gradually in such a manner that the angular momentum of the stone remains constant with respect to the centre of the circle. If the tension in the thread is T= Ar^(n) , where A = constant , r = instantaneous radius of the circle, then find the value of n.
Answer» Solution :If the instantaneous angular velocity of the stone is `omega`, then angular momentum , L`=Iomega= mr^(2)omega`= constant (according to the problem ) or, `omega = (L)/(mr^(2))` Here the tension in the thread provides the necessary centripetal FORCE for rotation. So, `T = Ar^(n) = momega^(2)R= m. (L^(2))/(m^(2)r^(4))r= (L^(2))/(m) r^(-3)` =`Ar^(-3)(A=(L^(2))/(m)="constant")` `:."" n=-3`.