Rotation vs lift distance I'm working on a homework problem and I think I found an answer, but not positive. Here is the question: A winch with a $6$-inch radius is used to lift a container. The winch is designed so that as it is rotated, the cable stays in contact with the surface of the winch. That is, the cable does not wrap on top of itself. a. Find the distance the container is lifted as the winch is rotated through an angle of $5\pi/6$ radians. b. (I haven't started this yet. Don't know where to start.) Determine the angle, in radians, through which the winch must be rotated to lift the container a distance of $2$ feet. This is what I did: $((5\pi/6)/2\pi) * 12\pi$ =$10\pi/6 * 12\pi$ = $120\pi/6$ =$20\pi$ inches

The idea was right, there was a slip of arithmetic. The distance the container is raised is $$\frac{\frac{5\pi}{6}}{2\pi}\times 12\pi.$$ This is $5\pi$.

For the second problem, the idea is much the same. Let $\theta$ be the angle of rotation. Then by the reasoning of the first problem, we have $$\frac{\theta}{2\pi}\times 12\pi=24.$$ We want to solve for $\theta$.