# Math Help - Center of mass

1. ## 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.

2. 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

3. Good reasoning. I should have also added that on these tests there is always an option E. NOTA (None of the above).

4. 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

5. 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.

6. 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.

7. 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.

8. 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

9. 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