Results 1 to 5 of 5

Math Help - Centre of Mass problem

  1. #1
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1

    Centre of Mass problem

    Got a problem here dealing with finding the centre of mass which I'm a bit stuck on.

    A uniform lamina of mass m is bounded by the curve with equation

    y = \frac{x^2}{a^2}(a-x)

    and that part of the x-axis between x = 0 and x = a, where a is a positive constant. You are given that the area of the lamina is \frac{1}{12}a^2. By considering thin strips of thickness \delta x parallel to Oy find the position of the centre of mass G of the lamina.

    For the x coordinate:

    \bar{x} = \frac{\int^a_0 x(\frac{x^2}{a} - \frac{x^3}{a^2}) dx}{\frac{1}{12}a^2}

    \frac{1}{12}a^2 \bar{x} = \int^a_0 x(\frac{x^2}{a} - \frac{x^3}{a^2}) dx

    This is because the mass of the strip would be \delta x multiplied by the height, y. The distance from the origin would then be x, giving us the above equation.

    So after working out the integral we're left with:

    \bar{x} = \frac{12}{a^2} \left( \frac{a^3}{20}\right) = \frac{3a}{5} which I believe is right?

    However when I come to work out the y coordinate it get's a bit confusing.

    As far as I'm aware, the right hand side of the equation should be the integral of the mass multiplied by the distance from your origin, over the integral of the range of masses? This is how I proceeded with the first equation. This is what I get:

    \frac{1}{12}a^2 \bar{y} = \int^a_0 (\frac{x^2}{a} - \frac{x^3}{a^2}).(\frac{x^2}{a} - \frac{x^3}{a^2}) dx

    Which simplifies to:

    \frac{1}{12}a^2 \bar{y} = \int^a_0 (\frac{x^2}{a} - \frac{x^3}{a^2})^2 dx

    This is because the mass of the strip would again be (\delta x)y, and the distance from the origin in the y direction would be y, giving us the above equation.

    However in the answers provided, they have the integral as:

    \frac{1}{12}a^2 \bar{y} = \int^a_0 \frac{1}{2} \left(\frac{x^2}{a} - \frac{x^3}{a^2}\right)^2 dx

    I'm just not sure where that \frac{1}{2} has come from?

    Thanks in advance for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Anyone got any ideas on this?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Consider a thin strip like you have there, with a dx width, and y = f(x) height. Where is its center of mass?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    The COM would be halfway up the strip, hence the half y...

    Thank you very much for pointing out the obvious for me there, sometimes you spend so much time looking for a complicated reason you just overlook the obvious!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Yeah. Been there, done that. You're welcome, and have a good one!
    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
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 7th 2010, 12:00 PM
  3. Centre of mass problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 1st 2009, 04:23 AM
  4. Volume and centre of mass
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 13th 2008, 01:08 PM
  5. Centre of mass
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 17th 2008, 10:31 AM

Search Tags


/mathhelpforum @mathhelpforum