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 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 1

Let the radius vector, R joining the bead with the centre makes 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 ......…(1)
mlω² = Nsin Θ ….....(2)

In triangle OPQ, we have,
sinΘ = I/R
so, l = RSin Θ ......…(3)
from equations (1) and (3) , we have,
m(R SinΘ )ω² = NsinΘ
so, mRω² = N ......…(4)

from equations (1) and (4), we have, mg = mRω²Cos Θ

CosΘ = g/ Rω² ....…(5)

Since, CosΘ ≤ 1, the bead will remain at the lowermost point,
for g/ Rω² ≤ 1,
i.e. ω ≤ $\sqrt{(g/R)}$ [ hence proved ]

For ω =$\sqrt{\frac{2g}{R}}$  …....(6)
On equation equations 5 and 6, we have
2g/R = g/RcosΘ,
CosΘ = 1/2
Θ = Cos^-1(1/2) =>  Θ = 60°

Answer 2

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic and (c) non periodic motion? Give period for each case of periodic motion (ω is any positive constant) a. sign ωt - cos ωt b. sin² ωt c. 3 Cos (π/4 - ωt) d. cos ωt + cos 3ωt + cos 5ωt e. exp(-ω²t²) f. 1 + ωt + ω²t