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Center of mass problem

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    A baseball bat of length L has a peculiar linear density given by [tex]\lambda[/tex] = [tex]\lambda[/tex]0 * (1+X2/L2)

    Find the x coordinate of the center of mass in terms of L

    2. Relevant equations

    Mxcm= mr

    3. The attempt at a solution

    So I use integration

    The integrand I have is x*[tex]\lambda[/tex] dx and substitue whatever on the right side of the [tex]\lambda[/tex] equation in. Then I just took normal integral.

    However I got wrong answer. The right answer does not contain [tex]\lambda[/tex]0 but mine does

    Can you guys help me ??
     
  2. jcsd
  3. Dec 15, 2008 #2
    You need to also do an integration to find the total mass, M. This will have a factor lambda_o that will cancel the one you have on the right side.
     
  4. Dec 15, 2008 #3
    How do I integrate to find mass M ??

    Do I plug in lamda formula*L for M or do I have to integrate lamda*L ??
     
  5. Dec 15, 2008 #4
    M = int(lambda * dx) will do the job where you take the integral over the length 0 to L.
     
  6. Dec 15, 2008 #5
    I got (L^2/2 +L^3/4)/(1+L^2/3)

    But it's still not the answer in the book.

    Did I do something wrong ??
     
  7. Dec 15, 2008 #6

    LowlyPion

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    You've almost got it looks like to me.
    But the numerator should have been multiplied by x

    yielding as an integrand x + x3/L2

    that gives

    x2/2 + x4/4 | from 0 to L or ... 3L2/4

    The total Mass looks integrated a little off.

    Shouldn't that be (L + L/3) = 4L/3 ?

    Then dividing denominator into numerator

    (3/4L2)/(4/3*L) = 9L/16
     
    Last edited: Dec 15, 2008
  8. Dec 16, 2008 #7
    How can you get 3L^2/4 and 4L/3

    I thought the integral of the total mass will yield L+ L^3/3 ???
     
  9. Dec 16, 2008 #8

    LowlyPion

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    The integrand for the volume of the mass is 1 +X2/L2 |evaluated between 0 and L

    That yields the result X + X3/3L2 The 0 terms are of no account leaving L + L3/3L2 = L + L/3 = 4L/3

    The integrand for the incremental moments is as I outlined previously.
     
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