Work Power and Energy 4 Question 1

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) always radially outwards

(b) always radially inwards

(c) radially outwards initially and radially inwards later

(d) radially inwards initially and radially outwards later

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Answer:

Correct Answer: 2. (c)

Solution:

  1. h=RRcosθ

Using conservation of energy

mgR(1cosθ)=12mv2

Radial force equation is

mgcosθN=mv2R

Here, N= normal force on bead by wire

N=mgcosθmv2R=mg(3cosθ2)N=0 at cosθ=23

So, normal force act radially outward on bead, if cosθ>23 and normal force act radially inward on bead, if cosθ<2/3

Force on ring is opposite to normal force on bead.



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