A thin circular loop of radius R rotates about its vertical diameter with an angular frequency . Show that a small bead on the wire loop remains at its lowermost point for g R . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for = 2g R ? Neglect friction.

Let the radius vector joining the bead with the centre make an angle , 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: mg=N ....(i) ml =N ...(ii) In , we have: = l R l=R ...(iii) Substituting equation (iii) in equation (ii), we get: m(R ) ^2=N mR ^2=N ...(iv) Substituting equation (iv) in equation (i), we get: mg=mR ^2 = g R ^2 ...(v) Since 1, the bead will remain at its lowermost point for g R ^2 1 , i.e., for ( g R )^ 1/2 For = ( 2g R )^ 1/2 or ^2= 2g R ...(vi) On equating equations (v) and (vi), we get: 2g R = g R =1 2 = ^ -1 (0.5)=60^

A thin circular loop of radius R rotates about its vertical diame…

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Question

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency $\omega.$ Show that a small bead on the wire loop remains at its lowermost point for $\omega\le\sqrt{\frac{\text{g}}{\text{R}}}.$ What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega=\sqrt{\frac{2\text{g}}{\text{R}}}?$ Neglect friction.

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Answer

Let the radius vector joining the bead with the centre make an angle $\theta,$ 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:
$\text{mg}=\text{N}\cos\theta\ ....(\text{i})$
$\text{ml}\omega=\text{N}\sin\theta\ ...(\text{ii})$
In $\triangle\text{OPQ},$ we have:
$\sin\theta=\frac{\text{l}}{\text{R}}$
$\text{l}=\text{R}\sin\theta\ ...(\text{iii})$
Substituting equation (iii) in equation (ii), we get:
$\text{m}(\text{R}\sin\theta)\omega^2=\text{N}\sin\theta$
$\text{mR}\omega^2=\text{N}\ ...(\text{iv})$
Substituting equation (iv) in equation (i), we get:
$\text{mg}=\text{mR}\omega^2\cos\theta$
$\cos\theta=\frac{\text{g}}{\text{R}\omega^2}\ ...(\text{v})$
Since $\cos\theta\le1,$ the bead will remain at its lowermost point for $\frac{\text{g}}{\text{R}\omega^2\le1},$ i.e., for $\omega\le\Big(\frac{\text{g}}{\text{R}}\Big)^{1/2}$
For $\omega=\Big(\frac{2\text{g}}{\text{R}}\Big)^{1/2}\ \text{or}\ \omega^2=\frac{2\text{g}}{\text{R}}\ ...(\text{vi})$
On equating equations (v) and (vi), we get:
$\frac{2\text{g}}{\text{R}}=\frac{\text{g}}{\text{R}\cos\theta}$
$\cos\theta=\frac{1}{2}$
$\therefore\theta=\cos^{-1}(0.5)=60^\circ$