Results 1 to 9 of 9

Math Help - Center of mass

  1. #1
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599

    Center of mass

    What is the center of mass of the region bounded by y=\sqrt{3}|x| and the line y=3?

    A) (0,0)
    B) (0,1)
    c) (0,2)
    d) (0,3)

    Ok, I know the x-coordinate of the center of mass is 0 because this is an even function. The y-coordinate of the center of mass can be found by this correct? \frac{\int\int_{R}xdydx}{\int\int_{R}dydx}?

    If so I first must find the bounds. I set \sqrt{3}|x|=3 and get that the two graphs intersect at +- \frac{3}{\sqrt{3}} or \sqrt{3}

    So if this correct then the y-coordinate of the center of mass is \frac{\int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}xdydx}{\  int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}dydx}

    I hope someone can check me up to here. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jameson
    What is the center of mass of the region bounded by y=\sqrt{3}|x| and the line y=3?

    A) (0,0)
    B) (0,1)
    c) (0,2)
    d) (0,3)

    Ok, I know the x-coordinate of the center of mass is 0 because this is an even function. The y-coordinate of the center of mass can be found by this correct? \frac{\int\int_{R}xdydx}{\int\int_{R}dydx}?

    If so I first must find the bounds. I set \sqrt{3}|x|=3 and get that the two graphs intersect at +- \frac{3}{\sqrt{3}} or \sqrt{3}

    So if this correct then the y-coordinate of the center of mass is \frac{\int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}xdydx}{\  int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}dydx}

    I hope someone can check me up to here. Thanks.
    You seem to be making a bit of a meal out of this one.

    The region is a triangle symmetric about the y-axis, with a vertex at (0,0)
    and the two other vertices on the line y=3.

    So the centre of mass is on the axis of symmetry x=0, also it cannot be at
    either (0,0) or (0,3), as these are a vertex and on a side respectively.

    Also it cannot be at (0,1), as this leaves much more area above than below
    the line y=1.

    So we conclude that the centre of mass must be at (0,2) if it is at any of the
    given candidate answers.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jameson
    Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).
    Actually a simple dissection of the triangle into 9 congruent triangles all similar to
    the big triangle will show that the centre of mass divides the axis of symmetry
    in the ratio of 2:1, again showing that (c) is the answer.

    RonL
    Last edited by CaptainBlack; March 5th 2006 at 03:04 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    This is an elementary way to do this problem but it works here. No need to use the formula for finding a lamina just use a compass and straightedge. But that I mean the centroid of this isoseles triangle can easiliy be found by algebra.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CaptainBlack
    Actualy a simple disection of the triangle into congruent triangles all similar to
    the big triangle will show that the centre of mass divides the axis of symetry
    in the ratio of 2:1, again showing that (c) is the answer.

    RonL
    Sorry, CaptainBlack you posted that before me while I was posting.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Thanks guys. Am I correct in my setup though? Usually these problems don't make regular polygons where geometry can be used like this one.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jameson
    What is the center of mass of the region bounded by y=\sqrt{3}|x| and the line y=3?

    A) (0,0)
    B) (0,1)
    c) (0,2)
    d) (0,3)

    Ok, I know the x-coordinate of the center of mass is 0 because this is an even function. The y-coordinate of the center of mass can be found by this correct? \frac{\int\int_{R}xdydx}{\int\int_{R}dydx}?
    This is correct.

    [quote]
    If so I first must find the bounds. I set \sqrt{3}|x|=3 and get that the two graphs intersect at +- \frac{3}{\sqrt{3}} or \sqrt{3}

    So if this correct then the y-coordinate of the center of mass is \frac{\int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}xdydx}{\  int_{-\sqrt{3}}^{\sqrt{3}}\int_{\sqrt{3}|x|}^{3}dydx}[QUOTE]

    The limits of integration also look OK.

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jameson
    Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).
    Another reason for choosing (c) now that I think of it is that the centre
    of mass of a triangle is the point of intersection of the medians, and that
    the medians divide one another into segments in the ratio 2:1.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 15th 2009, 06:04 PM
  2. Replies: 1
    Last Post: November 11th 2009, 05:39 PM
  3. center of mass
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 17th 2009, 08:20 AM
  4. Center of Mass
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 10th 2009, 02:48 PM
  5. center of mass of a rod
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 27th 2007, 01:15 PM

Search Tags


/mathhelpforum @mathhelpforum