Page 1 of 2 12 LastLast
Results 1 to 15 of 29

Math Help - Centre of Mass by integration

  1. #1
    Newbie superphysics's Avatar
    Joined
    Oct 2007
    Posts
    15

    Centre of Mass by integration

    I've come across this question (which looks more like physics than maths, to be honest), and I really need some help...

    Q. Find the centre of mass of a homogenous semi-circular plate. Given that x_{cm}=\frac{1}{M}\int{x}\ dm and
    y_{cm}=\frac{1}{M}\int{y}\ dm

    where x_{cm} and y_{cm} refer to the centre of mass along the respective axis, M is the mass of the plate, and dm is the small change in mass.

    Oh, by the way, if there's a type of solution apart from calculus (geometric or otherwise), I'd be glad to know of it...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by superphysics View Post
    I've come across this question (which looks more like physics than maths, to be honest), and I really need some help...

    Q. Find the centre of mass of a homogenous semi-circular plate. Given that x_{cm}=\frac{1}{M}\int{x}\ dm and
    y_{cm}=\frac{1}{M}\int{y}\ dm

    where x_{cm} and y_{cm} refer to the centre of mass along the respective axis, M is the mass of the plate, and dm is the small change in mass.

    Oh, by the way, if there's a type of solution apart from calculus (geometric or otherwise), I'd be glad to know of it...
    Hello,

    I don't know how to get the formula but I can give you possible way to get a valid solution:

    The "centre of mass" of a plane area of homogenous material which is bounded by the x-axis, the curve of y = f(x), the straight lines x = x_1 and x = x_2 is the centroid of this area. The coordinates of the centroid are calculated by:

    x_c=\frac{\int_{x_1}^{x_2}x \cdot y dx}{\int_{x_1}^{x_2}y dx} and y_c=\frac{\int_{x_1}^{x_2}y^2 dx}{2 \int_{x_1}^{x_2}ydx}

    With your problem:

    y = \sqrt{r^2-x^2} , that menas the center of the semi-circle is (0, 0), x_1 = -r, x_2 = r

    Obviously \int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2

    Plug in all values you know:

    x_c=\frac{\int_{-r}^{r}x \cdot \sqrt{r^2-x^2} dx}{\frac12 \cdot \pi \cdot r^2} = 0 Use integration by substitution

    y_c=\frac{\int_{-r}^{r}(r^2-x^2)dx}{\pi \cdot r^2}=\frac{\frac43 r^3}{\pi \cdot r^2}=\frac{4}{3\pi}r

    Therefore the centre of mass has the coordinates \left(0, \frac{4}{3\pi}r \right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    29

    Centre of mass by integration semi-circle with thickness

    Supposing that you wanted to find the centre of mass of a semi-circle plate with radius A and this semi circle had a smaller semi-circle plate with radius a removed then would the centre of mass be equal to

    y_c=\frac{\int_{-A}^{A}(A^2-x^2)dx}{\pi \cdot A^2}-\frac{\int_{-a}^{a}(a^2-x^2)dx}{\pi \cdot a^2}=\frac{\frac43 a^3}{\pi \cdot a^2}=\frac{4}{3\pi}(A-a)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Quote Originally Posted by earboth View Post
    Hello,

    I don't know how to get the formula but I can give you possible way to get a valid solution:

    The "centre of mass" of a plane area of homogenous material which is bounded by the x-axis, the curve of y = f(x), the straight lines x = x_1 and x = x_2 is the centroid of this area. The coordinates of the centroid are calculated by:

    x_c=\frac{\int_{x_1}^{x_2}x \cdot y dx}{\int_{x_1}^{x_2}y dx} and y_c=\frac{\int_{x_1}^{x_2}y^2 dx}{2 \int_{x_1}^{x_2}ydx}

    With your problem:

    y = \sqrt{r^2-x^2} , that menas the center of the semi-circle is (0, 0), x_1 = -r, x_2 = r

    Obviously \int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2

    Plug in all values you know:

    x_c=\frac{\int_{-r}^{r}x \cdot \sqrt{r^2-x^2} dx}{\frac12 \cdot \pi \cdot r^2} = 0 Use integration by substitution

    y_c=\frac{\int_{-r}^{r}(r^2-x^2)dx}{\pi \cdot r^2}=\frac{\frac43 r^3}{\pi \cdot r^2}=\frac{4}{3\pi}r

    Therefore the centre of mass has the coordinates \left(0, \frac{4}{3\pi}r \right)
    Hello ,I understand that the r^2 cancels with the r^3 to give r on the top, but where does the number 4 come from please?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    someone please
    where does the number 4 come from?????????
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1
    Quote Originally Posted by wolfhound View Post
    someone please
    where does the number 4 come from?????????

    It comes from the integral in the numerator:

    \int_{-r}^r (r^2 - x^2)dx = [r^2x]_{-r}^r - \left[\frac{1}{3}x^3\right]_{-r}^r

    = (r^3 + r^3) - \left(\frac{1}{3}r^3 + \frac{1}{3}r^3\right)

     = 2r^3 - \frac{2}{3}r^3

     = \frac{4}{3}r^3
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Quote Originally Posted by Aryth View Post
    It comes from the integral in the numerator:

    \int_{-r}^r (r^2 - x^2)dx = [r^2x]_{-r}^r - \left[\frac{1}{3}x^3\right]_{-r}^r

    = (r^3 + r^3) - \left(\frac{1}{3}r^3 + \frac{1}{3}r^3\right)

     = 2r^3 - \frac{2}{3}r^3

     = \frac{4}{3}r^3
    but why don't you get r^3/3 (on first line) when you integrate it?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1
    Quote Originally Posted by wolfhound View Post
    but why don't you get r^3/3 (on first line) when you integrate it?
    No, the integral is done with respect to x not respect to r. So r remains a constant in the integral.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Quote Originally Posted by Aryth View Post
    No, the integral is done with respect to x not respect to r. So r remains a constant in the integral.
    Ok, could you explain the 2nd line please, when you evaluate r and -r, how did you get this bit on the 2nd line. please
    thanks
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1
    Quote Originally Posted by wolfhound View Post
    Ok, could you explain the 2nd line please, when you evaluate r and -r, how did you get this bit on the 2nd line. please
    thanks
    Alright.

    [r^2x]_{-r}^r - \left[\frac{1}{3}x^3\right]_{-r}^r

    We know that:

    [f(x)]_a^b = [f(b) - f(a)]

    So that:

    [r^2x]_{-r}^r = [(r^2)(r) - (r^2)(-r)] = (r^3 - (-r^3)) = 2r^3

    And:

    \left[\frac{1}{3}x^3\right]_{-r}^r = \left[\frac{1}{3}r^3 - \frac{1}{3}(-r)^3\right] = \left(\frac{1}{3}r^3 - \frac{1}{3}(-r)^3\right) = \frac{2}{3}r^3
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Sep 2009
    Posts
    300

    Smile

    Thanks for that
    I understand now
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Sep 2009
    Posts
    300

    one more question

    y = \sqrt{r^2-x^2} , that menas the center of the semi-circle is (0, 0), x_1 = -r, x_2 = r

    Obviously \int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2


    if y = sqroot of r^2 - x^2
    when its integrated how does it become 3.14xr^2
    where does pie come from?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    Quote Originally Posted by wolfhound View Post
    y = \sqrt{r^2-x^2} , that menas the center of the semi-circle is (0, 0), x_1 = -r, x_2 = r

    Obviously \int_{-r}^{r}ydx = \frac12 \cdot \pi \cdot r^2


    if y = sqroot of r^2 - x^2
    when its integrated how does it become 3.14xr^2
    where does pie come from?
    first of all ...

    this is the lower case Greek letter pi :



    this is pie (something to eat) :






    this is where \pi comes from in calculating the integral of the semicircle ...


    \int_{-r}^r \sqrt{r^2-x^2} \, dx

    2\int_0^r \sqrt{r^2-x^2} \, dx

    let x = r\sin{\theta}

    dx = r\cos{\theta} \, d\theta

    2\int_0^{\frac{\pi}{2}} \sqrt{r^2 - r^2\sin^2{\theta}} \cdot r\cos{\theta} \, d\theta

    2r^2\int_0^{\frac{\pi}{2}} \sqrt{1 - \sin^2{\theta}} \cdot \cos{\theta} \, d\theta

    2r^2\int_0^{\frac{\pi}{2}} \sqrt{\cos^2{\theta}} \cdot \cos{\theta} \, d\theta

    2r^2\int_0^{\frac{\pi}{2}} \cos^2{\theta} \, d\theta

    2r^2\int_0^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta

    r^2\int_0^{\frac{\pi}{2}} 1 + \cos(2\theta) \, d\theta

    r^2\left[\theta +\frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}}

    r^2\left[\frac{\pi}{2} - 0 \right] = \frac{\pi r^2}{2}
    Last edited by skeeter; February 21st 2010 at 12:34 PM. Reason: fix sign error
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    I dont understand this
    is this the easiest way?
    how can I pluck values of sin and cos out of the sky?
    I like apple pie but pi is making me angry............
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    Quote Originally Posted by wolfhound View Post
    I dont understand this
    is this the easiest way?
    how can I pluck values of sin and cos out of the sky?
    I like apple pie but pi is making me angry............
    other than recognizing that the area of a semicircle is \frac{\pi r^2}{2} , no.

    the method of integration is called trig substitution ... not "plucking" values of sine and cosine out of the sky.

    Trigonometric substitution - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

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. [SOLVED] Centre of mass
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: February 25th 2009, 05:13 PM
  5. Centre of mass
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 17th 2008, 10:31 AM

Search Tags


/mathhelpforum @mathhelpforum