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Help with center of mass question

  1. Feb 3, 2014 #1
    1. The problem statement, all variables and given/known data

    The problem is attached in this post.

    2. Relevant equations

    x=My/M=(∫xf(x)dx)/(∫f(x)dx) from a to b

    3. The attempt at a solution

    My=4∫xtan(x^2)dx from 0 to √π/2 = 2ln(2/√2)

    M=4∫tan(x^2)dx from 0 to √π/2 =2xln(2/√2)

    My/M= 2ln(2√2)/2xln(2/√2) = 1/x or 1/A

    The answer however is (1/A)ln2
     

    Attached Files:

  2. jcsd
  3. Feb 3, 2014 #2
    Wrong.

    You have calculated the area but ∫xdm or 'My' (as you have put it) is
    [itex] σ*∫xf(x)dx [/itex] {σ = Mass per unit Area}
    [itex]∴ My = σ*2ln√2

    [/itex]

    Wrong.
    First of all, you have calculated the area incorrectly. This integration is different from the previous one- there is no 'x' over here.
    Moreover,
    M = ∫dm = ∫σdA (Where σ= Mass per unit Area)=σ*Area.


    Hope you got it!
    Adithyan.
     
  4. Feb 3, 2014 #3
    For "My", I'm still getting -2ln(√2/2) as the value of my integral. How did you get 2ln√2? Could you please clarify how you did the integration etc.?
     
  5. Feb 3, 2014 #4
    [itex] -2ln(√2/2) = -2ln(1/√2) = -2ln(√2^{-1}) = (-2)*(-1)ln(√2)= 2ln√2. [/itex]

    I guess, you are okay with my answer now. By the way, I did the first integration (ie [itex] ∫xtan(x^{2})dx [/itex]) by putting [itex] x^{2} = t[/itex]. (I hope you are well versed with integration by substitution.)
     
  6. Feb 3, 2014 #5
    Yeah I also used substitution, also could you also please clarify the integration for "M", I keep getting -2xln(√2/2), which is equal to 2xln(ln√2) (Is this correct?). Also what exactly do I do with the "x" in my answer? (I got the "x" because I used substitution to solve for this integral).

    Also the answer to the question is (ln2)/A, where "A" is equal to Area, however I don't know how to apply the variable "A" in regards to this question.
     
  7. Feb 3, 2014 #6
    Let
    [itex]
    σ =\frac{Mass}{Area }. [/itex]


    [itex]
    M = σ*A
    [/itex]

    [itex]
    ∴dm = σdA
    [/itex]
    [itex]
    M = ∫dm = ∫σdA =σ∫dA = σ*A
    [/itex]

    Here [itex]
    A =∫f(x)dx = ∫tan(x^{2})dx [/itex]
    and in this integration you cannot use substitution method since, there is no xdx term in it.
     
  8. Feb 3, 2014 #7
    Is it possible to integrate 4tan(x^2) then? How exactly do you go about integrating a function such as 4tan(x^2) etc.?
     
  9. Feb 3, 2014 #8
    I tried some of the common methods of integration, but they get messy later on. I don't think I can integrate this term at my current level of understanding of integration. Moreover, the question didn't ask for the value, they themselves substituted the value of this integral with A.
     
  10. Feb 3, 2014 #9
    Ok that makes sense since I also had a lot of trouble trying to integrate 4tan(x^2).

    Also after following the steps you provided, I ended up getting my answer to be (2ln2)/A instead of the actual answer (ln2)/A

    I set x=My/M=((p)(2ln2))/((p)(A))

    My=(p)(2ln2)
    M=(p)(A)

    p=density
    A=Area

    Did I make an error in my setup?
     
  11. Feb 3, 2014 #10

    You made an error here.
    You should get this term after integration: [itex] 2∫^{\frac{√\pi}{2}}_{0}ln(sec(x^{2})) [/itex]
     
  12. Feb 3, 2014 #11
    I don't understand how you got that as the integral, after substitution, don't the limits of integration become π/4 and 0?

    My=(p)4∫xtan(x^2)dx from 0 to √π/2

    My=(p)4∫xtan(x^2)dx from 0 to √π/2

    u=x^2 -> 0 to π/4 become the new limits of integration
    du=2xdx

    My=(p)(4/2)∫tan(u)du from 0 to π/4

    My=(p)(2)(-ln(cos(u)) from 0 to π/4

    My=(p)((-2ln(√2/2)-(-2ln(1))

    My=(p)(-2ln(√2/2))=2ln(2)
     
  13. Feb 3, 2014 #12
    I think you need to get your logarithmic knowledge correct.
    [itex] -2(ln(√2/2)) = -2(ln(\frac{1}{√2}) = 2ln(√2)= ln(√2^{2}) = ln(2) [/itex]

    Here, go through these rules and learn them to avoid such calculation errors in future.

    http://en.wikipedia.org/wiki/List_of_logarithmic_identities
     
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