Q. 403.6( 19 Votes )

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

Answer :

Let the radius vector 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)

m/ω^{2} = Nsin Θ …(2)

In triangle OPQ, we have,

sinΘ = I/R

→ I = RSin Θ …(3)

Substituting equation (3) in equation (1) , we have,

m(R SinΘ )ω^{2} = NsinΘ

→ mRω^{2} = N …(4)

Substituting equation 4 in equation 1, we have,

mg = mRω^{2}Cos Θ

CosΘ = g/ Rω^{2} …(5)

Since, CosΘ ≤ 1, the bead will remain at the lowermost point,

for g/ Rω^{2}≤ 1,

i.e. ω≤(g/R)^{1/2}

For ω = (2g/R)^{1/2} …(6)

On equation equations 5 and 6, we have

2g/R = g/RcosΘ,

CosΘ = 1/2

Θ = Cos^{-1}(0.5)

→ Θ = 60°

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The displacement vector of a particle of mass m is given by r (t) = î A cos ωt + ĵ B sin ωt.

(a) Show that the trajectory is an ellipse.

(b) Show that F = −mω^{2}r

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