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← JEE Advanced 2014 · Paper 2
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JEE Advanced 2014 · Paper 2, Physics question 2

A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is

A quarter-circle wire fixed vertically on the ground: A at the top of the arc, B at ground level, the centre of the arc at the ground below A with the 90° angle marked between the vertical and horizontal radii; a bead threaded on the wire near A.
Choose one option
Answer and worked solution

Official answer: D

Bead on the outside of the arc. The centre of the arc is on the ground below A, so the bead moves over the convex side, like a particle sliding on top of a sphere. At angle θ\theta from the vertical, after falling R(1−cos⁡θ)R(1-\cos\theta):
v2=2gR(1−cos⁡θ).v^2=2gR(1-\cos\theta).

Radial equation (towards the centre positive, with NN the wire's force on the bead, outward positive):
mgcos⁡θ−N=mv2R  ⟹  N=mg(3cos⁡θ−2).mg\cos\theta-N=\frac{mv^2}R\implies N=mg(3\cos\theta-2).

  • For cos⁡θ>23\cos\theta>\frac23, N>0N>0: the wire pushes the bead outward, so the bead pushes the wire inward.
  • For cos⁡θ<23\cos\theta<\frac23, N<0N<0: the wire must pull the bead inward (a free bead would leave the arc), so the bead pushes the wire outward.

(D)

Topics:Circular motion dynamicsConservation of mechanical energy
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