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Equation re-arrangement

  1. Feb 29, 2012 #1
    Hi newbie question,

    I'm trying to work through an example of a equation where circular motion can be expressed . I'm having difficultly understanding where the next line in the equation comes from. I have shown them below. My goal is to express 'V' although currently i am a few steps away from this.

    Line 1

    mgr + 0.5m(2[itex]\sqrt{}rg[/itex])2 = 0.5mv2

    Line 2

    gr + (2[itex]\sqrt{}rg[/itex])2 / 2 = 0.5v2

    I can follow the equation such that i understand both 0.5m cancels out from both sides but i get lost when i try to figure out how i lose the m from the mgr expression in line 1 from line 2?
     
  2. jcsd
  3. Mar 1, 2012 #2
    1. original ....... mgr + 0.5m(2√rg)2 = 0.5mv2

    2. divide both sides by 0.5m........ (mgr / 0.5m) + (0.5m(2√rg)2 / 0.5m) = v2

    3. result ........... (gr / 0.5) + (2√rg)2 = v2

    4. rearrange ....... gr + (2√rg)2 = 0.5v2

    I haven't touched physics & maths for 11 years since i left university with a masters but i think that is how you get the above.

    :cool:
     
  4. Mar 4, 2012 #3
    Thanks for the reply. It now makes sense i was forgetting about the 0.5m on the right hand side which would cancel a 0.5m on the left hand side. Thanks for the explanation. :D
     
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