### Q.40:- 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 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.

**Answer:-**

Let the radius vector joining the bead with the centre make an angle *θ*, with the vertical downward direction.

*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 Sin*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) ^{1/2}* or

*ω*

^{2}

*= 2g / R …..*

**(vi)**

On equating equations

**(v)**and

**(vi)**, we get:

2

*g*/

*R*=

*g*/

*RCos*θ

*Cos*θ = 1 / 2

∴ θ =

*Cos*

^{-1}(0.5) = 60

^{0}