# Work Done By A Force of Constant Magnitude in Moving an Object in a Circle

1. Dec 15, 2011

### Zach Knight

1. The problem statement, all variables and given/known data
A wagon is drawn by a student pulling with a constant force of F newtons applied at an angle of θ° to the horizontal. If the wagon is drawn in a circle with radius r meters, how much work is done on the wagon? (I don't remember the actual numbers)

2. Relevant equations
W=F*d*cos(θ)
$W=\int_{C}\vec{F}\cdot d\vec{r}$
3. The attempt at a solution
The teacher's solution manual proceeded as follows.
First, multiply the magnitude of the force by the cosine of the angle to the horizontal to obtain the force in the direction of motion. Then find the circumference of the circle to find the distance the wagon travels. Finally, find the product of these two quantities to find the work done on the wagon. This yields W=2$\pi$rFcos(θ). I felt uneasy with this answer, though, so I began pondering the question on my own.

I considered the point on the circle where the wagon began and the point directly opposite it. I then imagined the wagon moved a small distance along the circle counterclockwise from both points. Since the magnitude of the force remains constant, and the displacements are in opposite directions, the work along the two paths should cancel. I continued this process until the whole circle was traversed and concluded that the work done must be zero.

I explained this line of reasoning to my teacher, but he denied its validity. I too was a bit uncertain because of its lack of rigor, so I approached the problem using line integrals.
I began by writing the displacement and force as functions of time. This yielded $\vec{r}=r\cdot cos(\omega t)\hat{i}+r\cdot sin(\omega t)\hat{j}$ and $\vec{F}=F\cdot cos(\theta)\cdot cos(\omega t)\hat{i}+F\cdot cos(\theta)\cdot sin(\omega t)\hat{j}$, where $\omega$ is the angular speed of the wagon. I then found the dot product of $\vec{F}$ and $\frac{d\vec{r}}{dt}$, which was zero. The integral of this w.r.t. time from 0 to $\frac{360}{\omega}$ was also zero.

I showed this to my teacher, but he doesn't know calculus; he just denied my argument again. Is the book correct, or am I?

2. Dec 15, 2011

### Staff: Mentor

The magnitude of the force may remain constant, but its direction changes all around the circle. In fact, it always points in the instantaneous direction of the motion. So there can be no cancellation -- the work is always positive when the force is in the same direction as the motion. The book is correct.

3. Dec 15, 2011

### Zach Knight

Okay. I think I see where I went wrong with the line integral approach too. Because the force is applied perpendicular to the motion, $\vec{F}$ should equal $-F\cdot cos(\theta) \cdot sin(\omega t)\hat{i}+F\cdot cos(\theta) \cdot cos(\omega t)\hat{j}$.