A body starts falling from the top of a smooth sphere of radius r. What angle does the bodysubtend at the centre of the sphere when it just loses contact with the sphere and what willbe its velocity then ?

A body starts falling from the top of a smooth sphere of radius r. Wha

1.	A body starts falling from the top of a smooth sphere of radius r. What angle does the bodysubtend at the centre of the sphere when it just loses contact with the sphere and what willbe its velocity then ?
Answer» Solution :Let the ANGLE subtended by the body atthe centre of the sphere when it loses contact with the sphere be `theta`. The body just loses contact with the sphere when the component of gravitational force cannot provide the NECESSARY centripetal force to the body so that it can continuein the circular path. So, according to Fig .1.36. `mg cos theta =(mv^2)/r...(1)` (V= velocity of the body WHENIT just loses contact with the sphere) According tothe principle of conservation of energy, `1/2 mv^2 =mgr (1- cos theta)` [`OC=OA=r, OB=r cos theta BC=OC-OB=r-r cos theta=r(1-cos theta)]` or, `v^2 =2gr(1-cos theta)` From equations(1) and (2) we get, `mg cos theta =m/r*2gr(1-cos theta)` or, `cos theta =2(1-cos theta)` `or, cos theta =2/3 or, theta=cos^(-1)"" 2/3=48.2 ^@` again ,`v^2=2gr(1-cos theta)=2gr(1-2/3)=2/3gr` `therefore v=sqrt(2/3)gr`.