__Question 5.40__:*R*rotates about its vertical diameter with an angular frequency ω

*.*Show that a small bead on the wire loop remains at its lowermost point for ω ≤ √

*g/R*. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = √

*2g/R*? Neglect friction.

__Solution__:Let the radius vector joining the bead with the centre make an angle

*θ*, with the vertical downward direction.

OP =

*R*= Radius of the circle

*N*= Normal reaction

The respective vertical and horizontal equations of forces can be written as:

*m*g =

*N C*os

*θ*... (

*i*)

*mlω*

^{2}=

*N*

*S*in

*θ*… (

*ii*)

In ΔOPQ, we have:

Sin θ = l / R

*l*=

*R S*in

*θ*… (

*iii*)

Substituting equation (

*iii*) in equation (

*ii*), we get:

*m*(

*R S*in

*θ*)

*ω*

^{2}=

*N S*in

*θ*

*mR*

*ω*

^{2}=

*N*... (

*iv*)

Substituting equation (

*iv*) in equation (

*i*), we get:

*mg*=

*mR ω*

^{2}Cos

*θ*

Cos

*θ = g / R*

*ω*

^{2}...(v)

Since cos

*θ*≤ 1, the bead will remain at its lowermost point for

*g / R*

*ω*

^{2}≤ 1, i.e., for

*ω*≤ (g / R)

^{1/2}

For

*ω = (2g / R)*or

^{1/2}*ω*

^{2}= 2g / R .....(vi)On equating equations (

*v*) and (

*vi*), we get:

2g / R = g / RCos θ

Cos θ = 1 / 2

∴ θ = Cos

^{-1}(0.5) = 60

^{0}

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