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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:
mg = Ncosθ ... (i)
mlω2 = Nsinθ … (ii)
In ΔOPQ, we have:
`sin theta = l/R`
`l = Rsin theta`...(iii)
Substiting equation (iii) in equation (ii) we get
m(Rsinθ) ω2 = Nsinθ
mR ω2 = N ... (iv)
Substituting equation (iv) in equation (i), we get:
mg = mR ω2 cosθ
`cos theta = g/(Romega^2)` ...(V)
Since cosθ ≤ 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^@`
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