# Center of mass

• Mar 5th 2006, 01:24 PM
Jameson
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.
• Mar 5th 2006, 01:34 PM
CaptainBlack
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

RonL
• Mar 5th 2006, 01:47 PM
Jameson
Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).
• Mar 5th 2006, 02:02 PM
CaptainBlack
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
• Mar 5th 2006, 02:02 PM
ThePerfectHacker
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.
• Mar 5th 2006, 02:03 PM
ThePerfectHacker
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.
• Mar 5th 2006, 02:07 PM
Jameson
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.
• Mar 5th 2006, 08:29 PM
CaptainBlack
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
• Mar 5th 2006, 08:35 PM
CaptainBlack
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