question_answer

A small ball of mass m is allowed to slip down from the top of a hemispherical dome of radius R. At what height h from the lower end the ball will leave contact with the dome?

A) 2R/3                

B) R/2

C) R/4                              

D) R/6

Correct Answer: A

Solution :

[a] If the contact is lost at P, the velocity v would be\[v=\sqrt{2g(R-H)}\](PE lost = KE gained) \[\frac{{{v}^{2}}}{R}=g\cos \theta \] (Centripetal force \[\frac{m{{v}^{2}}}{R}\]) \[\Rightarrow \]\[{{v}^{2}}=2g(R-h)=Rg\cos \theta \] Thus, \[\frac{2g(R-h)}{R}=g\times \frac{h}{R}\left( \because \cos \theta =\frac{h}{R} \right)\] \[\Rightarrow \]\[2g(R-h)=gh\]or\[2gR=3gh\Rightarrow h=\frac{2}{3}R\]