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A smooth hemispherical bowl 30 cm diame...

A smooth hemispherical bowl ` 30 cm` diameter potates with a constatn angular velocity `omega`, about its vertical axis of symmetry. A particle at (P) of weighing ` 5 kg`, is onserved to remantn at rest telative to the bowal at a height ` 10 cm` above teh base, Fig. 1 (CF) . 34. The magnitude fo speed of rotation of the bowl is
.

A

` 15 rad//s`

B

` 23 rad//s`

C

` 14 rad//s`

D

` 16 rad//s`

Text Solution

Verified by Experts

The correct Answer is:
C
Let` R` be the force exerted by the bowl in the particle. It will be as a reactional force (R ). Various forces acting on particle at (P) will be (i) weight
Mg (downwards ) (ii) ` R` (Normal reaction ) along (PO).
.
Resolving (R ) into tis two rectangular components we have ` R sin theta ` acts vertically wpards and ` R cos thea ` acts horizontally along ` PQ` As horizontal component ` R cos theta ` will proved required centripetal force.
` :. ` R cos theta = ma_c`
[a_c rarr centripetal accleration]`
` R cos theta mr omega^2 ` or ` omega= sqrt (R cos theta)/(mr)`
` cos theta = theta r/(OP) = r/(0.15)`
[:. R sin theta = mg
R = (mg)/(sin theta)= (mg)/(OQ//OP) = ( 5 xx 9.8)/(0.05 //0 15/
(R = 15 xx 9.8 N)]`
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