A hemispherical bowl of radius R=0.1 m is rotating about its own axis (which is vertical) with an angular velocity `omega` A particle of mass 0.01 kg on the friction less inner surface of the bowl is also rotating with same `omega` The particle is at a height h from the bottom of the bowl.
(a) Obtain the relation between h and `omega` What is the minimum value of `omega` needed in order to have a non zero value of h ?
(b) It is desired to measure g using this set up, by measuring h accurately. Assuming that r and `omega` are known precisely, and that the least count in the measurement of h is `10^(-4)` m. What is the minimum possible error `Deltag` in the measured value of g ? `g = 9.8 m//s^(2)`

Explanation

The forces acting on the particle are reaction and weight , so for vertical equilibrium of the particle

`N "cos" theta = "mg" ............(i) `
And for circular motion in horizontal plane
N sin `theta = "mr" omega^(2) " " ..... (ii)`
(a) Dividing Eqn. (ii) by (i)
tan `theta = (r omega^(2))/(g) = (R ("sin"theta) omega^(2))/(g) `
`" " ["as" r = R "sin" theta]`
or cos `theta = (g)/(R omega^(2)) , i.e., (R-h)/(R) = (g)/(R omega^(2)) [ "as cos" theta = (R - h)/(R)]`
or `h = R - (g// omega^(2))`
So for `h gt 0 , R - (g//omega^(2)) gt 0`
i.e., `omega gt sqrt((g//R))` so `omega_("min") = sqrt((9.8 // 0.1)) = 7sqrt2 ` rad/s
(b) From Eqn. (iii) , g = `omega^(2)(R-h)`. So for a given `omega` and R
`dg = -omega^(2)dh`, i.e., `(Delta g)_("min")= - (omega^(2))_("min") (Deltah)_("min")`
So substituting the value of `omega_("min")` and `(Deltah)_("min")`
`(Deltag)_("min") = -98 xx 10^(-4) = -9.8 xx 10^(-3) m//s^(2)`