Results 1 to 6 of 6

Math Help - Z-coordinate of centroid , centre of mass prob

  1. #1
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    Z-coordinate of centroid , centre of mass prob

    Hi!

    Problem: Determine the z-coordinate of the centroid of the solid generated by revolving the quarter-circular area thorugh 90 degress about the z-axis.

    Solution attempt:

    I tried to calculate the volume of the solid. I got it to be  \frac{\pi a^{3}}{24}(4+3\pi)

    I have some trouble setting up the sum of moments integral, that is  \int z_{c} \; dz = V\bar{z}

    I tried to use a horizontal strip as differential element:
    Witdh:  \sqrt{a^{2}-z^{2}}

    Height: dz

    Moment arm: z

    But I also need thickness, right? expressed in terms of z

    Thx!
    Attached Thumbnails Attached Thumbnails Z-coordinate of centroid , centre of mass prob-pic_08.jpg  
    Last edited by Twig; July 8th 2009 at 02:39 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    First , calculate the numinator :  I = {\int\int\int}_{D} z dV

    Since it is uniform , we may neglect the density .

     I = \int_0^{\frac{\pi}{2}} \int_a^{2a} \frac{ 4a^2 - r^2 }{2}  rdr d\theta

     = \frac{\pi}{4} [ 2a^2r^2 - \frac{r^4}{4} ]_{a}^{2a}

     = \frac{\pi}{4} \frac{9a^4}{4}

     = \frac{9a^4 \pi }{16}


    Oh i made a mistake , i thought that is a part of a sphere ...

    The function should be  z = \sqrt{ a^2 -  ( r-a)^2}

    so  I = \frac{\pi}{4}[ \int_a^{2a} (a^2 r - r(r-a)^2) dr ]

     = \frac{\pi}{4} (a^4) [ \frac{11}{12} ]

     = \frac{ 11 \pi a^4 }{48}

    and  V = \frac{\pi}{4} [\int_a^{2a} \sqrt{ a^2 - (r-a)^2} rdr

    Sub  r-a = a\sin{\theta}  \implies

     V = \frac{\pi}{2} [ \int_0^{\frac{\pi}{2}} a^3 ( 1 + \sin{\theta}) \cos^2{\theta} d\theta ]

     = \frac{\pi}{2} (a^3) [\frac{\pi}{4} + \frac{1}{3} ]

     = \frac{a^3 \pi}{2} \frac{3\pi + 4}{12}

     = \frac{ a^3 \pi ( 3\pi + 4) }{24}

    So bar-z  = \frac{11a}{2(3\pi + 4)}
    Last edited by simplependulum; July 8th 2009 at 03:41 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    I dont get it, did you change to polar coordinates? Why do yo multiply by another r?

    What is the moment arm?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,993
    Thanks
    1653
    Quote Originally Posted by Twig View Post
    I dont get it, did you change to polar coordinates?
    Well, yes. The " r dr d\theta is kind of a tip off, don't you think? There is plenty of circular symmetry here so using polar coordinates is kind of obvious.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    Quote Originally Posted by HallsofIvy View Post
    Well, yes. The " r dr d\theta is kind of a tip off, don't you think? There is plenty of circular symmetry here so using polar coordinates is kind of obvious.

    Yes, true.

    z^{2}+(r-a)^{2}=a^{2} , it feels like he just changes x to r.

    I am with on the moment arm:  \frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}

    Shoudlnt it work just integrating "as usual" without polar coordinates, even if polar will make it easier?

    What is wrong below?

    \int_{a}^{2a}\underbrace{\frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}}_{z_{c}}\cdot \underbrace{\sqrt{a^{2}-(x-a)^{2}}}_{Height} \; dx

    I thought polar was x=r\cdot cos(\theta)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by Twig View Post
    Yes, true.

    z^{2}+(r-a)^{2}=a^{2} , it feels like he just changes x to r.

    I am with on the moment arm:  \frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}

    Shoudlnt it work just integrating "as usual" without polar coordinates, even if polar will make it easier?

    What is wrong below?

    \int_{a}^{2a}\underbrace{\frac{1}{2}\sqrt{a^{2}-(x-a)^{2}}}_{z_{c}}\cdot \underbrace{\sqrt{a^{2}-(x-a)^{2}}}_{Height} \; dx

    I thought polar was x=r\cdot cos(\theta)
    I think  \int_a^{2a} \sqrt{ a^2 - (x-a)^2 } ^2 ~dx is equal to the moment sum of circular plate with radius a and center (a,0)

     \int\int z dzdL ~~ and  \int\int\int z dz dA
    are actually different .

    The moment sum of the required solid is
     {\int \int \int}_D  z dz dA
    The reason using polar co. is for convenient calculation  dA = rdrd\theta

    So the moment sum is also equal to

     {\int \int \int}_D z dz (rdrd\theta)  D = [ (\theta,r,z) \in {\mathbb{R}}^3 |0 \leq z \leq \sqrt{a^2 - (r-a)^2 } , a \leq r \leq  2a , 0 \leq \theta \leq \frac{\pi}{2}]
    Last edited by simplependulum; July 8th 2009 at 08:50 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. centre of mass
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: October 25th 2010, 05:45 AM
  2. centre of mass questions
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: August 9th 2010, 10:52 AM
  3. Centre of mass
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2010, 12:00 PM
  4. Centre of mass problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 1st 2009, 04:23 AM
  5. Centre of mass
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 17th 2008, 10:31 AM

Search Tags


/mathhelpforum @mathhelpforum