Homework Help: Finding the right integral

1. Feb 13, 2009

Ailo

1. The problem statement, all variables and given/known data

A small mass m is pulled to the top of a frictionless halfcylinder (of radius R) by a cord that passes over the top of the cylinder. (a) If the mass moves at a constant speed, show that $$F=mg cos(\theta)$$. The angle is between the horizontal and the radius drawn to the mass.

(b) By directly integrating
$$\int{Fds}$$
find the work done in moving the mass at constant speed from the bottom to the top of the half-cylinder. Here ds represents an incremental displacement of the small mass.

2. Relevant equations

3. The attempt at a solution

The a-part was easy when I drew a diagram. The b-part is the one I'm struggling with. With the Work-Energy theorem I get that the work done by F is mgR. But what integral should I compute and why? Have no idea whatsoever..

2. Feb 13, 2009

tiny-tim

Hi Ailo!

(I'm not sure what you mean by mgR)

The integral is given to you … ∫F ds, where s is the displacement.

(Remember, the string is always tangent to the cylinder.)

3. Feb 13, 2009

Ailo

Hi! Thx for the answer, but the problem is how to set it up. I should get an expression, integrate it, and end up with the answer mg*R. I've got the force as a function of the angle, and I don't understand how to integrate it over a distance.

Maybe I didn't explain the situation good enough. The half cylinder lies on the ground, and we pull the mass up along the quartercircle. Does anybody understand? =)

4. Feb 13, 2009

tiny-tim

Hi Ailo!

Just decide what ds is (in terms of θ), and then integrate mgcosθ ds, and you should get mgR.

5. Feb 13, 2009

Ailo

That's the problem. I've never done a problem like this before...

6. Feb 13, 2009

tiny-tim

ok …

what is ds in terms of θ?

in other words, if you increase the angle by dθ, how much do you increase the length (s) of the string by?

7. Feb 13, 2009

Ailo

My best guess is to make a triangle with sides R, R and ds. Will that work?

8. Feb 13, 2009

Ailo

Ohh! Now I get it. It's (theta)*R, right?

*palmslap

9. Feb 13, 2009

Yup!

ds = Rdθ