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Object is dragged across a semicircle's perimeter

  1. Dec 16, 2016 #1
    1. The problem statement, all variables and given/known data

    A small object with mass m is dragged without friction to the highest point of a semicircle with a radius of R, by a weightless rope.

    a) If the magnitude of his velocity is constant, then his acceleration that is parallel to the semicircle, is zero. Prove that F = mgcosθ

    b) Integrate the following swiftly W = ∫ F*dr, to calculate the work that is produced when the object is moved from the highest point, to the lowest.

    Uz8qH6M.jpg


    2. Relevant equations


    3. The attempt at a solution

    (a) So, first up I set the direction of the force F as the Y axis, and then created the net force (I don't know the proper term yet, but you get the gist of it from the pic). Like this:

    XF1pHbE.jpg

    Thus: ΣFy = mat = 0 <=> F = Fgy = Fgcosθ = mgcosθ

    (b) Now's the part where I can't quite figure it out. I didn't do much intergals at school, so it's pretty new to me, and I haven't reached that part in my math book. I glanced at the back and found the basic ones (eg ∫cosθdθ = sinθ + c), but I've never come across an intergal with both x & y coordinates. I took a look at other examples from the solution manual, but only one other problem was intergals, and it's quite different.

    The book's answer is "mgR", and I kinda sorta found the same, but from lowest to highest, not the opposite, like the book's asking. If anyone could help me with the intergal, I'd appeciate it!
     
  2. jcsd
  3. Dec 16, 2016 #2

    haruspex

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    I'm not sure what the problem setter had in mind. From the instruction to integrate F, I would say highest to lowest is a typo. Must have meant lowest to highest. There is no information about F in a highest to lowest phase.
    The "swiftly" is suspicious. Might mean that you are not supposed to do the integral at all, just use work conservation and write down mgr straight away. Hard to tell.
     
  4. Dec 18, 2016 #3
    Yeah, I just took it as a typo as well. From lowest to highest I get the book's result, so that's probably it.
     
  5. Dec 18, 2016 #4
    You shouldn't convert to Cartesian coordinates. It is easiest to answer both questions in terms of theta. That integral solution you found is exactly what you need. Now you just need the initial and final theta.
     
  6. Dec 18, 2016 #5
    I'm not sure how though. As I've said, I'm not that familiar with intergals, so I'm not sure how to get it to work. I figure that at the highest point, theta is 90 degree, and at the lowest, it's 0 degree. But I'm not exactly sure how to handle the intergal.
     
  7. Dec 18, 2016 #6
  8. Dec 18, 2016 #7
    I know how the basic version works (only one kind of variable, say, x), and I have some experience with more complex types, but I don't know how to connect the angle (theta) and the displacement (r). The best I've got is:

    W = ∫Fdr = ∫HighestLowestmgcosθdy = ∫900mgcosθRdθ = mgR∫900(sinθ)'dθ = mgR[sinθ]900 = mgR(sin90 -sin0) = mgR

    I'm not sure if that's correct though.
     
  9. Dec 18, 2016 #8

    haruspex

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    That all looks fine, but it is a little confusing to use r and y as variable names with such non-standard meanings. You could use ds as a generic displacement along a path, and write ds=Rdθ.
     
  10. Dec 18, 2016 #9
    I had the y axis in my mind, hence why I used that symbol. But yeah, your way is better.
     
  11. Dec 18, 2016 #10
    Yes, that's it. And better with Haruspex's suggested labeling.
     
  12. Dec 18, 2016 #11
    Well, that settles it then.

    Thanks for the help everybody!
     
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