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Algebraic operation

  1. Aug 12, 2015 #1
    1. The problem statement, all variables and given/known data

    In a rotation of rigid bodies-problem there's a question regarding the speed of a falling block, attached to a solid cylinder which can rotate frictionless around its axis. I have a question regarding the algebraic operations.


    2. Relevant equations

    So I have this expression: ##mgh=1/2mv^2+1/2(1/2MR^2)(v^2/R^2)##.

    3. The attempt at a solution

    I can operate on the expression to solve for v. So I eventually get that ##2mgh=v^2(m+1/2M)##. And from here I just want to divide both sides by ##(m+1/2M)## and take the square root of both sides.

    Which would leave me with ##v=sqrt((2mgh)/(m+(1/2)M)##

    My text book expresses this result as ##v=sqrt((2gh)/(1+(M/2m))##

    Are those the same results? If so, what operations were made to end up with that result? If not, what did I do wrong?
     
  2. jcsd
  3. Aug 12, 2015 #2

    SteamKing

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    Divide both sides of ##2mgh=v^2(m+1/2M)## by m and then solve for v.

    You should be careful to distinguish (1/2)M from 1/(2M) though.
     
  4. Aug 12, 2015 #3
    Thanks. I get that if I divide ##(m+1/2M)## by ##m## I get ##(1+M/2m)##. But doesn't dividing the whole right-hand side with ##m## affect ##v^2## and I'll end up with ##v^2/m##? I hope you can get what's confusing me. :)
     
  5. Aug 12, 2015 #4

    SteamKing

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    Why should dividing by m affect v2? What you are doing is dividing ##(m+1/2M)## by m.

    It's like saying m * y = (1 + k) * x2. If you divide both sides of the equation by the same quantity, what's left must be equal. That's not really algebra even; it's arithmetic.

    m * y = (1 + k) * x2 will be the same as y = [(1 + k)/m] * x2, as long as m ≠ 0.

    The value of x is not going to be affected by dividing by m.
     
  6. Aug 13, 2015 #5

    RUber

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    Test the relationship:
    ## \sqrt{ \frac{2mgh}{m+M/2} } = \sqrt{ \frac{(m)2gh}{(m)(1+M/(2m)} } = \sqrt{ \frac{2gh}{(1+M/(2m)} }##
    They look the same to me.
     
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