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Homework Help: Varying Force

  1. Dec 17, 2013 #1
    ImageUploadedByPhysics Forums1387274814.891516.jpg

    Okay so the answer in b) is mgR how is this possible when we integrate ?

    The work is the external force right?
    Secondly the F inside the integral is the mg sin(theta) the force of gravity?
    dr ---> pi R (semicircle)
  2. jcsd
  3. Dec 17, 2013 #2


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    The force, according to the problem, is [itex]mg cos(\theta)[/itex]. Are you saying you do not believe that?

    I don't know what you mean by "dr--> pi R". First, in this problem you are dealing with [tex]d\vec{r}[/tex], not "dr". (Since we are moving on a circle, r is constant and dr= 0!) On a circle of radius r, [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so that [itex]\vec{r}= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}[/itex] and [itex]d\vec{r}= -r sin(\theta)\vec{i}+ r cos(\theta)\vec{j} d\theta[/itex]

    No, the "work" is NOT the "external force". Work is not force. And since there is only one force here, I don't know what you mean by "external" force.
  4. Dec 17, 2013 #3
    No I believe that no problem. Force done by pulling the small particle is mgcos(θ)

    my problem is with the work done by that force to pull the particle is the W = ∫F.dr
    This is the work done by the external force in puling the particle which should be the mgcos(θ) x the distance moved

    We integrate here because the force done varies with the displacement moved due to the angle.

    In the integral I will add the which force the gravitational force which is mg j
    the dr⃗ =−rsin(θ)i⃗ +rcos(θ)j⃗ dθ

    when we integrate this it will lead us to mgr be outside the integral and cosine(θ) which integrate to sin(θ)

    Work done is mgRsin(θ) not mgR

    or did he assume that the angle is 90 which is 1 ?

    please help with that work part and correct me if I ma giving an faulty information
  5. Dec 17, 2013 #4


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    It says: "... from the bottom to the top", so what is the range for theta?
  6. Dec 18, 2013 #5
    180 sin180 is 0 so the whole thing is zero?
  7. Dec 18, 2013 #6


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    No, not 180. It's a half cylinder lying on its flat side.
  8. Dec 18, 2013 #7
    Its a semi-circle with 180 degrees
    A full circle is 360
    Quarter is 90

    Cant argue with that!
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