# Problems with work and F*dr

1. Jan 25, 2006

### dimensionless

A semi-circle is given by the function $$y=R cos(\frac{x\pi}{2R})$$. An attractive force F at $$x = R$$ and $$y = 0$$ such that $$F(x, y) = F(R, 0) = 0$$ and $$F(x, y) = F(-R, 0) = F_{0}$$. This force along the semi-circle, relative to its source, is given by $$F(\theta) = F_{0} sin(\theta/2)$$. What amount of work is done against the force when a particle moves along the circle from $$\theta = 0$$ to $$\theta = \pi$$.
To find the work I have the equation
$$W = - \int^{\pi}_{0} F dr$$
For the force I have the equation
$$F= F_{0} sin(\theta/2)$$
For the distance from the particle to the orgin of the force I have
$$2 R sin(\theta/2)$$
so for dr I have
$$2 R sin(\theta/2) d\theta$$
Putting this all tegether I get
$$W = - \int^{\pi}_{0} F_{0} sin(\theta/2) 2 R sin(\theta/2) d\theta$$
This rearanges to
$$W = -F_{0} R \int^{\pi}_{0} \left[1-cos(\theta) \right] d\theta$$
Integrating I get
$$W = -F_{0} R \left[ \theta-sin(\theta) \right]^{\pi}_{0}$$
This works out to
$$W = -F_{0} R \pi$$
The answer I am suppose to get is
$$W = -F_{0} R$$
What did I do wrong?

Last edited: Jan 26, 2006
2. Jan 25, 2006

### Gokul43201

Staff Emeritus
The question doesn't make very much sense. Could you please reproduce the question exactly as it appears in the text, and also provide the name of the text ?

3. Jan 26, 2006

### dimensionless

I rewrote the problem a little bit. I'm afraid it's still a little unclear with out the diagram though. This question is not from a text book.

4. Jan 26, 2006

### Gokul43201

Staff Emeritus
I think the mistake you are making is in interpreting what dr refers to. This is an elemental distance traveled by the particle (along the semicircular path) - it is not an element of the distance of the particle from the origin. Also keep in mind that the angle between F and dr changes with the motion of the particle.