Circular motion Work Energy problem

In summary, the minimum velocity for the bead at the point where it is closest to P is when its closest to P and has a velocity of just over half of the radius.
  • #1
AmK
4
0
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 ?
 
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  • #2
Welcome to PF!

Hi AmK! Welcome to PF! :smile:
AmK said:
… 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 :wink:
 
  • #3
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
 
  • #4
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?
 
  • #5
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
 
  • #6
Hi AmK! :smile:
AmK said:
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. :smile:
 
  • #7
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
 

1. What is circular motion?

Circular motion is the movement of an object along a circular path, where the object's distance from a fixed point remains constant but its direction changes continuously.

2. How is circular motion related to work and energy?

In circular motion, work and energy are interrelated concepts. The work done on an object in circular motion is equal to the change in its kinetic energy, and the energy of an object in circular motion is constantly changing due to the work done by external forces.

3. Can you provide an example of a circular motion work energy problem?

An example of a circular motion work energy problem is a roller coaster. As the roller coaster car moves along the track, it experiences a constant change in direction, resulting in circular motion. The work done by the force of gravity and the normal force on the car is constantly changing its kinetic energy.

4. How do we calculate work and energy in circular motion?

In circular motion, the work done by a force is calculated by multiplying the magnitude of the force by the distance traveled in the direction of the force. The change in kinetic energy is calculated by subtracting the initial kinetic energy from the final kinetic energy.

5. What are the key factors that affect work and energy in circular motion?

The key factors that affect work and energy in circular motion are the magnitude and direction of the external forces acting on the object, the mass of the object, and the radius of the circular path. These factors determine the amount of work done and the change in energy of the object.

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