A spherical bowl of radius R rotates about the verical diameter with angular velocity omega.The bowl contains a small object inside and in absence of friction, this object takes up a position inside the bowl such that its radius vector makes an angle theta with the vertical (see figure). Then

A spherical bowl of radius R rotates about the verical diameter with a

1.	A spherical bowl of radius R rotates about the verical diameter with angular velocity omega.The bowl contains a small object inside and in absence of friction, this object takes up a position inside the bowl such that its radius vector makes an angle theta with the vertical (see figure). Then
Answer» `omega=sqrt(g//r COS theta)` `omega=2 pi g//r` `omega = sqrt((g cos theta)/r)` `omega=sqrt(r cos theta xx g )` Solution :Reaction N has component in VERTICAL direction equal to the weight of the object . The horizontal component of N provides the centripetal force. Now { `omega` can be CALCULATED }

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A spherical bowl of radius R rotates about the verical diameter with angular velocity omega.The bowl contains a small object inside and in absence of friction, this object takes up a position inside the bowl such that its radius vector makes an angle theta with the vertical (see figure). Then

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