A hemispherical bowl of radius 0.1 m is rotating about its own axis [which is vertical] with an angular speed w . A particle of mass 10^-2 kg on the

Question

the frictionless inner surface of the bowl is also rotating with same w . The particle is at a height h from the bottom of the bowl. Then the relation between h and w is

[A] h=w^2/g

[B] h=R/2

[C] h=R-g/w^2

[D NOTA

Answer 1

a) Along x-direction, NSinq = m (AC) w2 

=>NSinq = m (RSinq) w2 N = mRw2 --------------- (1)

 Along y-direction, NCosq = mg -------------- (2) 

Using (1) and (2) mRw2Cosq = mg