Area defined by a rope wrapping around a circle?

[kochanskij]

How do you compute the area defined by a rope which spirals around a circle?

Here is a specific example of what I am asking:
A dog is tied to a fixed point on the outside of a circular silo with a radius of 20 feet. The rope is 50 feet long. How much total area does the dog have to roam around in?

I know that the dog has a semi-circle of radius 50 ft to walk around in front of the silo. When he goes toward the back of the silo, the rope winds around the silo and gets shorter and shorter. This winding starts when the rope is tangent to the circular silo. How do you compute the area he has to walk in as the rope is winding around and getting shorter?

I am more interested in the method you can use to solve this problem than in the exact answer.

 

[lanedance]

how about starting by writing the path of the dog at maximum limit, which bounds your area

the top section will be the seim-circle you described.

the rest will be where the rope (of length L) is around the silo for a circular section, say a, then loses contact with the silo. The remainder, b, will be a straight section, along the tangent to the circle at the last point of contact. Then L = a + b.

From there you should probably be able to set up an integral for the area

 

[kochanskij]

Can anyone offer further help on this problem?
How do you find the equation of the curve that the end of the rope makes as it wraps around the silo? I've tried it in both rectangular and polar coordinates.

 

[lanedance]

say for ease the diameter of the silo is one, set up coordinates, so the origin is at the centre of the sil & the dog is tied at the top (y axis)

the equation of the silo is:
x^2 + y^2= 1
the dog is tied at point:
(x,y) = (0,1)

now say we measure the angle, theta, from the vertical (where the dog is tied off) to the point where the rope leaves the silo.

The length of rope against the silo is just the arc length (assuming radius is 1)
l_1 = \theta
for the remaining length of rope, we assume it is straight, in the direction of the tangent to the silo

The point where the rope leaves the silo is:
\textbf{r}= (x,y) = (sin(\theta), cos(\theta))

the direction of the rope is tangent to the circle and so perpindicular to r, and the unit vector in the tangent direction, t, can be written:
\textbf{t}= (cos(\theta) , -sin(\theta))

so if L is the length of the rope, the position of the dog at maximum extent will be:
(x,y) = \textbf{r}+ (L-\theta) \textbf{t} =