Center of Gravity / Mass

By Brad Rodgers (P1930) on Saturday, May 12, 2001 - 10:18 pm :

Why does the formula for the center of gravity of mass end up working?

Thanks,

Brad

By Dan Goodman (Dfmg2) on Saturday, May 12, 2001 - 10:21 pm :

Which formula? End up working in what sense?

By Brad Rodgers (P1930) on Saturday, May 12, 2001 - 11:42 pm :

"For a plane area, the center of gravity is given by:

x_{-} = (\int \int_{A} x dS) / A

$y_{-} = (\int \int_{A} y dS) / A$
where A is the area of the given region, and dS is the area of an infinitesimal portion of the surface."

To be honest, I'm really not even sure what this means, not to mention how to prove that it gives the point where no matter how the body is turned the same resultant force of gravity is given.

Brad

By Dan Goodman (Dfmg2) on Saturday, May 12, 2001 - 11:55 pm :

Brad, the

dS

is the same as

dx dy

if you are integrating in those coordinates, or it might be

r dr d θ

in plane polars.

I can't remember how to prove it works. Hopefully someone who's doing the first year Dynamics course will chip in and give the proof. If I remember rightly, you divide the object into infinitessimal particles and prove it by considering the integrals as sums and thinking about how gravity acts on each individual particle. It's been too long since I did an applied maths course I'm afraid...

By Sean Hartnoll (Sah40) on Sunday, May 13, 2001 - 12:20 am :

Brad - of the top of my head I can't remember rigorously what you do with this. But if you do the integral for a few easy shapes (circle, square, rectangle etc.) you should see that it gives you a logical answer, i.e. the centre. If you think about what the integral actually means for a bit you should see why it makes sense. Perhaps you can also see how to generalise to higher dimensions? and to surfaces of varying density?

It turns out to be very important in dynamics that you can reduce the whole motion of a rigid body to the motion of its centre of mass and rotation about the centre of mass.

Sean

By Brad Rodgers (P1930) on Sunday, May 13, 2001 - 12:49 am :

What does the subbed A mean on the integral?

Brad

By Sean Hartnoll (Sah40) on Sunday, May 13, 2001 - 01:14 am :

It is the region you are meant to do the integral over. So for example if the shape is a square with sides length

a

the integral for

x_{-}

$\int_{0}^{a} x dx \int_{0}^{a} dy / (a^{2}) = a^{2} / 2 \times a / (a^{2}) = a / 2$ .

And for a circle radius $a$ (using $x = r \cos (t)$ , use $t$ for theta)

$\int_{0}^{a} r^{2} dr \int_{0}^{2 π} \cos (t) dt / (π a^{2}) = 0$ .

Sean

By Brad Rodgers (P1930) on Sunday, May 13, 2001 - 02:32 am :

I'm still not sure I entirely understand how to use the subbed A. For example say we were to find the center of mass for a 45,45,90 triangle with legs measuring a, it would seemingly be the same as a 45,45,90 triangle facing the opposite way. Perhaps if I could see how to do this for a triangle, as it doesn't show the sort of symmetry of the others, I would be less confused.

Thanks,

Brad

By Sean Hartnoll (Sah40) on Sunday, May 13, 2001 - 11:43 am :

Okay, that's a good question. Integrating over non-symmetric shapes can be a bit tricky. However, the triangle is not too bad:

The 45, 45, 90 triangle with legs $a$ has base $2^{1 / 2} a$ by Pythagoras. And height $a / 2^{1 / 2}$ , also by Pythagoras.

So, to integrate over the triangle, we will let $y$ go from 0 to $a^{1 / 2}$ , the height, but we need to restrict the range of $x$ , so that $(x, y)$ stays on the triangle.

The lines bounding the triangle (assuming the centre of the base is at the origin) are, perhaps you can check this,

$x = \pm (y - a / 2^{1 / 2})$

$\pm$ is plus or minus for the left and right legs respectively.

So the integral is (note how the limits of the $x$ integral depend on the $y$ position, this is because the width of the triangle gets smaller as we go up

$x_{-} = \int_{0}^{a / 2^{1 / 2}} (\int_{y - a / 2^{1 / 2}}^{- y + a / 2^{1 / 2}} x dx) dy / A = \int_{0}^{a / 2^{1 / 2}} dy 0 / A = 0$

$y_{-} = \int_{0}^{a / 2^{1 / 2}} (\int_{y - a / 2^{1 / 2}}^{- y + a / 2^{1 / 2}} dx) y dy / A = \int_{0}^{a / 2^{1 / 2}} (2 y - 2^{1 / 2} a) y dy / A = a^{3} / (6 2^{1 / 2}) / A$

but $A = a^{2} / 2$

So $y_{-} = a / (3 2^{1 / 2})$

Sean

By Sean Hartnoll (Sah40) on Sunday, May 13, 2001 - 11:44 am :

Hope all those calculations were right, to do the inverted triangle you just change the lines that were limiting the x integration. Perhaps you can try to do this?

Sean

By Brad Rodgers (P1930) on Monday, May 14, 2001 - 04:12 am :

Alright, thanks. I think I see how to do that. I've gone ahead and attempted to do the work for the equations of a right triangle refelcted across the x axis, and intuitionally and mathematically, I've gotten (0,-a/[2^1/2 3]). To do this mathematically, you just integrate for y parameters from 0 to -a/2^1/2 , and for y parameters from +-(y+a/2^1/2 ), right?

I do have one more question though. I think I understand why the formula holds assuming there is a center of gravity in the first place (is it because we're taking the average of all the points contained?). But why should we assume that there is a point such that all lines drawn through that point have equal areas(i.e. the figure would balance on this point)? I constructed the following argument to show that there is a center of gravity, in which we find a curve centered at (0,0), make sure that its areas on the right and left of the y-axis are equal and the areas on the top and bottom of the x-axis are equal, then if a line is drawn through the origin then the area above this line is equal to the area below it.

Intuitionally, one would think that this would have to be untrue, but I constructed the following mathematical argument in case anyone can think of a way to prove/disprove the above conjecture (which as I understand it is the principle behind center of mass.

For f₁ above the x axis, and f₂ below it, and for x_i and x_I that satisfy f₁ (x_i )=m*x_i ;f₂ (x_I )=m*x_I

\int_{a}^{b} f_{1} (x) dx = - \int_{a}^{b} f_{2} (x) dx

and

\int_{a}^{0} f_{1} (x) - f_{2} (x) dx = \int_{0}^{b} f_{1} (x) - f_{2} (x) dx

then

\int_{x_{i}}^{b} f_{1} (x) dx + \frac{m (x_{i})^{2}}{2} - [\int_{x_{I}}^{b} f_{2} (x) dx + \frac{m (x_{I})^{2}}{2}

For any given f(x)'s.

I don't think that will be very easy to prove or even disprove; any ideas as to how to go about showing that for all regions, there exists some center of mass?

Thanks,

Brad

By Sean Hartnoll (Sah40) on Monday, May 14, 2001 - 01:51 pm :

Good. That is the right answer (obviously!). And your reasoning as to why it is the centre of mass is fine. I'm not sure I completely follow what you're trying to do after that, the point is that we have a formula for the centre of mass for any object, therefore it must exist (by construction).

I'm not sure what you mean Brad by saying that "all lines through the c.o.m. have equal areas".

Sean

By Brad Rodgers (P1930) on Monday, May 14, 2001 - 06:08 pm :

It's rather hard to explain without some pictures: maybe this will help.
Diagrams
And so on for all lines. One would think that this would have to be true for an object to balance on 1 point, but perhaps there is another way to explain why objects can indeed be balanced on a single point.

Thanks,

Brad

By Michael Doré (Md285) on Thursday, May 17, 2001 - 11:45 pm :

Brad,

I'm not quite sure I follow. Is your conjecture the following?

Given a uniform lamina bounded by a simple curve, any straight line through the centre of mass divides the lamina into two halves of equal area.

This is false. For a counter-example, imagine the lamina is two squares ABCD and PQRS joined together with ABCD above PQRS such that C,P,Q,D appear in that order along a straight line.

By Tom Hardcastle (P2477) on Friday, May 18, 2001 - 01:01 am :

One can also disprove the more general conjecture: 'for any lamina there exists some point such that all lines through that point divide the lamina into two halves of equal area'.
The most obvious counterexample I can think of involves a squat T shape:

T shape

By Michael Doré (Md285) on Friday, May 18, 2001 - 09:20 am :

Or if you want a convex shape, take a square ABCD of side 2 then draw an isoceles triangle ABP exterior to the square so that the square and triangle share side AB and AP = BP = sqrt(17). The perpendicular from P to AB is 4, i.e. the square and triangle have equal area. Then if there existed a point so that each line through it divided the shape into two equal areas, this point would certainly have to be the midpoint of AB. However the line through the midpoint of AB which is at 45 degrees to AB clearly doesn't divide the shape into equal areas, contradiction.
Diagram

By Brad Rodgers (P1930) on Friday, May 18, 2001 - 04:53 pm :

Indeed, I didn't think that all shapes would be able to be divided like that. How, then, does one explain the way that all shapes can be balanced on a single point?

Brad

By Sean Hartnoll (Sah40) on Friday, May 18, 2001 - 05:17 pm :

It's probably because what the body actually feels is a TORQUE not a force. If you don't know what a torque is then imagine two people, one trying to keep a door open and the other trying to shut it. The person who is further away from the hinge will probably win because he has to exert less force to achieve the same torque.

Basically, torque is to rotational motion what force is to linear motion. And if we are keeping a point fixed, then torque is what is appropriate.

If we are exerting the force perpendicularly to the door then the torque is defined as

T=Fr

where r is the distance from the hinge to the point where I am applying the force.

Now in the case of the body, we are balancing the body at the centre of mass and the force of gravity is acting downwards, so it is perpendicular to the body. So the total torque acting on the body is

T² = (int (x - x_) dx dy)² + (int (y - y_) dx dy)² = 0

by definition of the centre of mass. (note that you sum before squaring, this is because the torque is in fact a vectorial quantity, defined as T = rxF, of which I have then taken the modulus squared).

Sean

By Sean Hartnoll (Sah40) on Friday, May 25, 2001 - 05:58 pm :

Another thing that may clarify the idea of Torque: suppose you want to balance two weights of 1kg and 2kgs on a rod, with the rod supported at a point. Would you put them at the same distance from the support? Or suppose you had two weights of 1kg and put them at different distances from the support. Would it be stable? The point in each case is that what you need to balance is the rF, the distance from the support to the weight times the force, not just the force.

Sean