A smooth hemispherical bowl ` 30 cm` diameter, rotates with a constant angular velocity ` omega`, about its vertical axis of symmetry Fig. 2 (APC) . 2 (a) . A particle at (P) of weighing ` 5 kg` is observed to remain at rest relative to the bowl at a height `10 cm` above the base. Find the magnitude of the force exerted by the bowl on the particle and speed of rotation of the bowl.
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Explanation

Let (R ) be the force ezxerted by the bowl on the particle. It will be as a rectional force ( R) Fig. 2 (APC). 2 (b0`
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Various forces acting on the particle at (P) will be ltbRgt (i) Weight mg acting vetically downward
(ii) Normal reaction (R ) acting along (PO) Resolving (R ) into two rectangular components, we have (R ) sin theta` acts vertically upwards and ` R cos theta` carts horizontally along (PQ). When the particle is just at rest, then
` R sin theta = mg`
or ` R= (ntg0/(sin theta) = (mg)/(OQ//OP)= (5 xx 9.8) /((0.05 //0 .15)`
As the horzontal camponent ` R cos theta ` will provide the required centripetal force, so
` R coa theta = mr omega^2`
or ` omega = sqrt ( R cos theta )/( mr)) = ( sqrt 15 xx 9.8 xx r// 0. 15)/ (mr))`
= 14 rad//s`.