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/ω2 = 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 )ω2 = NsinΘ
→ mRω2 = N …(4)
Substituting equation 4 in equation 1, 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≤(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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