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. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-5 Laws Of Motion



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.

OP = R = Radius of the circle

N = Normal reaction
The respective vertical and horizontal equations of forces can be written as:
mg = N Cosθ(i)
mlω2 = Sinθ(ii)



In ΔOPQ, we have:
Sin θ = l / R
l = R Sinθ … (iii)
Substituting equation (iii) in equation (ii), we get:
m(R Sinθ) ω2 = N Sinθ
mR ω2 = N(iv)
Substituting equation (iv) in equation (i), we get:
mg = mR ω2 Cosθ
Cosθ = g / Rω2    …(v)
Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω2 ≤ 1, i.e., for ω ≤ (g / R)1/2
For ω = (2g / R)1/2  or ω2 = 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)  =  600