#### Question

A thin circular loop of radius *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 `omega <= sqrt(g/R)` .What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for `omega = sqrt("2g"/R)` ?Neglect friction.

#### Solution 1

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*cos*θ* ... (*i*)

*mlω*^{2} = *N*sin*θ* … (*ii*)

In ΔOPQ, we have:

`sin theta = l/R`

`l = Rsin theta`...(iii)

Substiting equation (iii) in equation (ii) we get

*m*(*R*sin*θ*) *ω*^{2} = *N*sin*θ*

*mR** ω*^{2} = *N* ... (*iv*)

Substituting equation (*iv*) in equation (*i*), we get:

*mg* = *mR ω*^{2} cos*θ*

`cos theta = g/(Romega^2)` ...(V)

Since cos*θ* ≤ 1, the bead will remain at its lowermost point for `g/(Romega^2) <= 1` i.e for `omega <= sqrt(g/R)`

For `omega = sqrt((2g)/R)` or `omega^2 = ((2g)/R)` ..(vi)

On equating equations (*v*) and (*vi*), we get:

`(2g)/R = g/(Rcos theta)`

`cos theta = 1/2`

`:. theta = cos^(-1) (0.5 ) = 60^@`

#### Solution 2

Let the radius vector joining the bead to the centre of the wire make an angle `theta` with the verticle downward dirction. if N is normal reaction, then from fig.

`mg = N cos theta` ....(i)

`mromega^2 = N sin theta` ...(ii)

or `m(R sin theta) omega^2 = N sin theta`

or`mRomega^2 = N`

or `cos theta = g/(Romega^2)`

As |cos theta| <= 1, therefore bead will remain at its lowermost point for

`g/(Romega^2) <= 1 or omega <= sqrt(g/R)`

When `omega = sqrt((2g)/R)` from equation iii

`cos theta = g/R(R/"2g") = 1/2`

`theta = 60^@`