Finding the Work Done in Moving a Mass on a Half-Cylinder at Constant Speed

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The discussion revolves around calculating the work done in moving a mass up a frictionless half-cylinder at constant speed. For part (a), it is established that the force exerted, F, is equal to mg cos(θ), where θ is the angle between the horizontal and the radius to the mass. In part (b), participants discuss how to set up the integral ∫F ds to find the work done, with the key insight that ds can be expressed as Rdθ. The integration leads to the conclusion that the work done is mgR. The conversation emphasizes understanding the relationship between the angle and the displacement in the context of the problem.
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Homework Statement



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.

Homework Equations



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..:redface:
 
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Ailo said:
(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.

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..:redface:

Hi Ailo! :smile:

(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.)
 
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? =)
 
Hi Ailo! :smile:

Just decide what ds is (in terms of θ), and then integrate mgcosθ ds, and you should get mgR. :wink:
 
That's the problem. I've never done a problem like this before...
 
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? :smile:
 
My best guess is to make a triangle with sides R, R and ds. Will that work?
 
Ohh! Now I get it. It's (theta)*R, right?

*palmslap
 
Yup! :biggrin:

ds = Rdθ :smile:
 
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