Circular motion Work Energy problem

[AmK]

A small bead of mass m is moving on a smooth circular wire (radius R) under the action of a force
F directed towards a point P at a distance R/2 from the centre .What should be the minimum velocity of the bead at the point where it is closest to P so that it may complete the circle.

I worked it out as follows
The bead will complete the circle if it is able to just reach the point diametrically opposite to the nearest point to P because after that a component of F will act along the tangential direction and accelerate the bead.
Using conservation of energy 1/2mv^2 + work done by force F >= 0

But i couldn't calculate the work done by the force F.
i know W = ∫F.dr so i tried to integrate but i couldn't find the angle between the force and displacement for the integral.

Please give some idea on how to work out this integral.
Also,can it be done without the integral ?

 

[tiny-tim]

Welcome to PF!

Hi AmK! Welcome to PF!

… couldn't find the angle between the force and displacement for the integral.

If the bead is at X, and if the centre is O,

then the angle is the angle between PX and the tangent, or 90° minus OXP

 

[AmK]

Thanks Tiny-Tim
the angle is what u said but for the integration i have to relate it to that angle to the angle subtended at the centre since the displacement would be in terms of that and i wasnt able to relate them

 

[haruspex]

Consider the triangle OPX. You know∠XOP, distance OP, distance OX. Can you write down an equation giving you ∠XPO? Can you find ∠PXO from there?

 

[AmK]

Cosine formula for length PX and then sine formula for ∠ PXO
That would make the integral very complex and i don't think i could solve that

 

[tiny-tim]

Hi AmK!

Cosine formula for length PX and then sine formula for ∠ PXO
That would make the integral very complex and i don't think i could solve that

After a substitution, isn't it one of the standard integrals?

Show us what you get. ​

 

[AmK]

I think i got it.
i'll denote the length by l then ∠oxp by β and ∠xop by θ
then l/sinθ = R/2sinβ
∫F.dr = ∫F*sinβ*Rdθ
sinβ = (R/l)*sinθ
∫F*sinβ*Rdθ = ∫F*(R/l)*sinθ*Rdθ
l=√R^2 + R^2/4 - R^2cosθ
and then i can sub cosθ=t and range of t will be from 1 to -1
Thanks guys