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Centroid of half circle

  1. Oct 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the location of the centroid (See PDF)

    ----------------------------------------
    I am stuck on how to find the centroid of a half circle, item #5. I don't how to find the distance of that centroid on the x-axis. Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Oct 19, 2008 #2
    No one knows?
     
  4. Oct 20, 2008 #3

    cristo

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    Firstly, you should learn some patience: the homework helpers here give up their time voluntarily, and so you cannot demand that they answer your question within a couple of hours of posting, especially if you do not conform to the homework rules. Where is your work? What have you tried for the problem? How do you calculate a centroid, in general?
     
  5. Oct 20, 2008 #4
    I didn't demand nothing, I simply asked a question and was trying to bump the thread because my .pdf was pending approval for awhile. Second, I didn't ask how to calculate a general centroid. Finally, I asked how to calculate the centroid of a half circle. There was no work to be shown because I had no idea how to calculate the centroid of a half circle. I didn't try anything on this problem.
     
    Last edited by a moderator: Oct 20, 2008
  6. Oct 20, 2008 #5

    mgb_phys

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    I was waiting for the attachment to approve to see what work you had already done.
    Presumably you have done calculus to be asked this question?

    You should look up a semicircular lamina, there is a difference between the CofG of a semicircle (ie a wire forming half a rim) and a half a disc.

    You can find an explanation here:
    http://books.google.com/books?id=4w...&hl=en&sa=X&oi=book_result&resnum=6&ct=result
     
  7. Oct 20, 2008 #6
    Thank you for answering my question in a kind manner unlike some other people... Anyways I figured it out, but again, thank you!
     
  8. Oct 20, 2008 #7
    Would you mind posting your solution?
     
  9. Oct 20, 2008 #8
    Sure, the equation i used is 4*R/3pi, in my case it was (4*4in)/(3*Pi) which gives me 1.69765in, which is the distance from the center towards x-direction.
     
  10. Oct 20, 2008 #9

    mgb_phys

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    Looking at your attachment this is an engineering question about the CofG of a complex shape that involves a semicircle. So in this case it's probably perfectly reasoanble to just lookup the equation and use it.
    If this had been a maths/calculus question asking you to prove the CofG of a semicircle then us telling you that the answer is 4*R/3pi does no good - that was cristo's point.

    Attachments have to be individually checked / approved by one of the site mentors, this takes time but is need to stop people just posting junk/porn images. There was a big server move this weekend and the mentors are busy fixing the problems that that caused.
     
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