Trivial : Use of "negative sign" when calculating centroid

[JC2000]

Homework Statement

Find the center of mass for the following object.

Relevant Equations

For object positions along the x axis:
$\mathrm{COM}_{x}=\frac{m_{1} \cdot x_{1}+m_{2} \cdot x_{2}+m_{3} \cdot x_{3}+\ldots}{m_{1}+m_{2}+m_{3}+\ldots}$
And similarly for the y axis:
$\mathrm{COM}_{y}=\frac{m_{1} \cdot y_{1}+m_{2} \cdot y_{2}+m_{3} \cdot y_{3}+\ldots}{m_{1}+m_{2}+m_{3}+\ldots}$

I realize that this is to be solved by breaking up the object into simple objects and using their known center of mass to find the center of mass of the entire object.

1. In the solution the circular gap is also considered in the calculations with a negative center of mass, why is this done?
2. Here the entire object is in the first quadrant of the chosen frame of reference and hence the position vectors are positive. Would the position vectors be negative in the formula if the chosen frame of reference were to be in the middle of the object?

Thank you!

 

[haruspex]

Homework Statement:: Find the center of mass for the following object.
Relevant Equations:: For object positions along the x axis:
$$\mathrm{COM}_{x}=\frac{m_{1} \cdot x_{1}+m_{2} \cdot x_{2}+m_{3} \cdot x_{3}+\ldots}{m_{1}+m_{2}+m_{3}+\ldots}$$
And similarly for the y axis:
$$\mathrm{COM}_{y}=\frac{m_{1} \cdot y_{1}+m_{2} \cdot y_{2}+m_{3} \cdot y_{3}+\ldots}{m_{1}+m_{2}+m_{3}+\ldots}$$

View attachment 266606

I realize that this is to be solved by breaking up the object into simple objects and using their known center of mass to find the center of mass of the entire object.

1. In the solution the circular gap is also considered in the calculations with a negative center of mass, why is this done?
2. Here the entire object is in the first quadrant of the chosen frame of reference and hence the position vectors are positive. Would the position vectors be negative in the formula if the chosen frame of reference were to be in the middle of the object?

Thank you!

1. You can think of the hole as being created by adding a disc of negative mass to what had been a complete rectangle.
2. The formula doesn't change, but some coordinate values will go negative.

I fixed your LaTeX by doubling the dollar signs.

 

[JC2000]

1. You can think of the hole as being created by adding a disc of negative mass to what had been a complete rectangle.
2. The formula doesn't change, but some coordinate values will go negative.

I fixed your LaTeX by doubling the dollar signs.

1. Yes but why not simply ignore it?
2. So if $x_1$ is in the second quadrant then $-x_1$ is to be used in the formula?

Thanks for all the help, including the LaTex!

 

[haruspex]

1. Yes but why not simply ignore it?

How ignore it? You have a rectangular plate with a circular hole. If you ignore that there is a hole then you will find the mass centre of the rectangle, which is not what you want. What method are you suggesting?

2. So if x1 is in the second quadrant then −x1 is to be used in the formula?

Again, I am not sure what you are saying.
Consider the centre, P₁, of the lower 2x8 rectangle. Its coordinates in the given frame are (10, 2). In the formula for the y coordinate of the system, it may appear as $+y_1m_1$. Plugging in the value you get $+2m_1$.
Now switch to taking the origin as the point with coordinates (7,7) in the diagram. Now the coordinates of P₁ are (3,-5). The formula is unchanged; it still says $+y_1m_1$. But now $y_1=-5$, so we get $-5m_1$.

 

[etotheipi]

If $\mu(\vec{r})$ is the area density at position $\vec{r}$, then the position of the centre of mass, $\vec{R}$, is calculated via$$\vec{R} = \frac{1}{M}\int_S \mu(\vec{r}) \vec{r} dA$$However you can express the area density function as a sum of any number of arbitrary density functions, say $\mu(\vec{r}) = \mu_1(\vec{r}) + \mu_2(\vec{r})$, such that at any position their sum equals the actual area density. You get$$\vec{R} = \frac{1}{M}\int_S \mu_1(\vec{r}) \vec{r}dA + \frac{1}{M} \int_S \mu_2(\vec{r})\vec{r}dA$$If you have an object with a hole, like @haruspex said in #2, you can superpose the function with a positive area density of a complete object, $\mu_1(\vec{r})$, with a function with an exactly negated area density at the position of the hole, $\mu_2(\vec{r})$.

This is actually a very important technique. In electrostatics, for instance, you can use the same idea to compute things like the electric field inside charge distributions with little bits removed. Even within centres of mass, you might have something like an object made of mostly one material, say wood, but with a small pocket of metal embedded somewhere inside it. The easiest way is often to define two density functions $\rho_{\text{wood}}(\vec{r})$ and $\rho_{\text{diff}}(\vec{r}) = \rho_{\text{metal}}(\vec{r}) - \rho_{\text{wood}}(\vec{r})$. You can integrate the first over the entirety of the object and the second over the entirety of the piece embedded inside it, and add them together!

 

[JC2000]

How ignore it? You have a rectangular plate with a circular hole. If you ignore that there is a hole then you will find the mass centre of the rectangle, which is not what you want. What method are you suggesting?

OOF! Just realized what I was stating! Absolutely bungled it! I understand now!

But now $y_1=-5$, so we get $-5m_1$.

Yes I get it now. I was referring to this case. Thanks for the painstaking example!

 

[JC2000]

Thanks for the rigorous explanation!
I had a brain fade and missed the fact that the gap was within the rectangle! Nonetheless, I think I understand the technique much better thanks to your explanation.

Quick clarification:
$$
x_{c m}=\frac{\int_{0}^{M} x d m}{M}
$$

Here you swapped $dm$ with the area density $* dA$ (since $dm = \frac{dm}{dA} * dA$ )?

 

[JC2000]

Also, if I had to find the center of gravity where there is a non uniform gravitational field then it would be the same formula except that $m_i * r_i$ would be replaced by $r_i *m_i * g_i$ ?

 

[etotheipi]

Quick clarification:$$x_{c m}=\frac{\int_{0}^{M} x d m}{M}$$Here you swapped $dm$ with the area density $* dA$ (since $dm = \frac{dm}{dA} * dA$ )?

Yes that is correct, $dm = \mu dA$. This is done only because to actually perform the integral (in most cases, but not always - e.g. when we have symmetry) we must use $dA$. It is also perfectly fine to consider $x$ and not $\vec{r}$ in the integrand, but you must still integrate over the entire surface of the shape.

Also, if I had to find the center of gravity where there is a non uniform gravitational field then it would be the same formula except that $m_i * r_i$ would be replaced by $m_i * g_i$ ?

You must be careful. One the one hand we have the centre of mass, which only generally coincides with the centre of gravity if the body is subjected to a uniform gravitational field, and only depends on the spatial distribution of mass so will not change if the body is subjected to different fields.

If you have non-uniform gravitation, then your expression would not be right (you would just be computing the total force on the body!).

The centre of gravity is defined such that the effects (net force and net torque) of all gravitational forces on all mass elements of the rigid body can be replaced equivalently by a single force acting through the centre of gravity. You are essentially looking for the position $\vec{R}$ such that$$\int_V (\vec{r} - \vec{R}) \times \vec{g}(\vec{r}) \rho dV = \vec{0}$$If $\vec{g}(\vec{r})$ is not a constant function, $\vec{R}$ will not generally be the position of the centre of mass.

 

[JC2000]

...we must use dA. It is also perfectly fine to consider x and not r→ in the integrand, but you must still integrate over the entire surface of the shape.

I am a little confused now. In the formula you mention, to find $\vec{R}$, you used $dA$, my question is :

In order to use $dA$ shouldn't the formula for $\vec{R}$ be a double integral? Or am I missing something?!

Likewise for the formula you give for finding the center of gravity, you are using volume so that the formula can be applied to 3D objects so does this mean it has to be a triple integral?!

Also is this the formula for finding the center of mass for a 3D :

$$
\int_{V}(\vec{r}-\vec{R}) \times p(\vec{r}) d V=0
$$

Thank you for your time and patience!

 

[etotheipi]

In order to use $dA$ shouldn't the formula for $\vec{R}$ be a double integral? Or am I missing something?!

That is right, I have just used $\int_S$ instead of $\iint_S$ for the sake of laziness . In most cases the $S$ is enough to tell you that we are integrating over a surface.

Likewise for the formula you give for finding the center of gravity, you are using volume so that the formula can be applied to 3D objects so does this mean it has to be a triple integral?!

You bet!

$$\int_{V}(\vec{r}-\vec{R}) \times p(\vec{r}) d V=0$$

Nearly, but remember that the $\times$ in the torque equation I stated in the last post is the vector cross product. We don't need it here, because $\rho dV$ is just a scalar! In fact if you start with what I wrote for the centre of gravity in the case of uniform gravitation (where centre of mass is the centre of gravity), then$$\int_V (\vec{r} - \vec{R}) \times \vec{g}(\vec{r}) \rho dV = \left[\int_V (\vec{r} - \vec{R}) \rho dV \right] \times \vec{g} = \vec{0} \implies \int_V (\vec{r} - \vec{R}) \rho dV = \vec{0}$$The bit on the right is the formula for the centre of mass in 3D. Usually the integral is split in two,$$\int_V \rho\vec{r} dV = \left(\int_V \rho dV \right)\vec{R} = M\vec{R}$$and that's the form that you are probably more used to (and more likely to use in practice!).

 

[JC2000]

That is right, I have just used $\int_S$ instead of $\iint_S$ for the sake of laziness . In most cases the $S$ is enough to tell you that we are integrating over a surface.

My bad! I didn't realize they were the same!

Nearly, but remember that the × in the torque equation I stated in the last post is the vector cross product. We don't need it here!

I missed that too!

Thanks a ton! I think I understand everything very clearly now!

 

[etotheipi]

No problem . By the way, Wikipedia has a really good reference section on this in case you ever want to check some part of this: https://en.wikipedia.org/wiki/Center_of_mass#Definition