Let's talk about yoga. Have you ever been doing yoga, and the instructor said something along the lines of, "Okay, everyone, get centered?" Or maybe you haven't done yoga, but you've played Wii Fit, and you've done the Zen Level, where you had to sit on the Fit Board and be perfectly balanced not to make the candle burn out. So, what's up with all this talk about centering? Is being centered really that important?
I can't say much about fitness and meditation but being centered is extremely important for physics. In this article, we'll explore the implications of the center of gravity and how being "centered" can have a massive impact on the outcome of a physics problem.
Center of Gravity Background Knowledge
To talk about the center of gravity, it is helpful to first review the concept of center of mass.
The center of mass of a system is the mean position of an object's or system's mass distribution.
At the point of the center of mass, the system can be thought of as a single particle. This concept has astronomical value to physicists because it simplifies many problems with complex systems. The center of mass can also be implied as the place where an outside force acts on the entirety of the system or object, whichever we are dealing with.
Gravitational Force
To understand the center of gravity, we must comprehend the gravitational force.
The force of gravity is a pull (gravity is always attraction) due to the massiveness of an object.
The magnitude of the force of gravity can be calculated with this formula:
$$F_g = G\frac{m_1 m_2}{r^2},$$
where \(F_g\) is the magnitude of the force of gravity, \(G\) is the gravitational constant which is equal to \(6.674×10^{−11}\,\mathrm{\frac{ m^3}{kg\,s^2}\\}\), \(m_1\) is your first mass, \(m_2\) is your second mass, and \(r\) is the distance between the center of mass of the two objects.
In general, the force of gravity on two objects is always felt on the line connecting the center of masses of the two objects.
Fig. 1 - Note how the two masses are exerting an equal and opposite force on each other along the line connecting their centers of mass.
In the section explaining the center of mass, we learned that forces acting on a particular object or system are thought of as acting on the center of mass of that object or system. This principle also applies to the force of gravity. Gravity will always act along an object's or system's center of mass in the presence of a uniform gravitational field.
Gravity is a force that acts without the need for physical contact; this means that the force can be felt in dispersed regions of space and we can think of space as covered by a uniform gravitational field.
We can calculate the strength of the gravitational field, \( g \) by using the formula for gravitational force. $$\mid\vec g\mid=\frac{\mid\vec F_g\mid}{m}\\=G\frac{M}{r^2}\\{,}$$
whose variables only differ from the one above in the notation of the masses. Here \(M\) is the source of the gravitational field, and \(m\) is the mass of the object subject to that field. The notation used for \( \mid\vec g\mid \) and \( \mid\vec F_g\mid \) means that we are just using the magnitude of those vectors.
If we use the values for the mass and radius of the Earth we find that
\begin{aligned} g & = G\frac{M}{r^2}\\ g & = \frac{\left(6.67 \times 10^{-11}\;\mathrm{\frac{N m^2}{kg^2}}\right)5.97\times10^{24}\;\mathrm{kg}}{ (6.37\times 10^6\;\mathrm{m})^2}\\ g & \approx 10\,\mathrm{\frac{N}{kg}}. \end{aligned}
It is important to note that, for an object at a certain height \( h \) from the surface of Earth, the value of \( r \) we would need to use would be the sum of Earth's radius, \( r_\text{E} \) plus that height, \( r=r_\text{E} +h. \) However, normally, this height is too small in comparison to the Earth's radius,
$$r = r_\text{E} +h \approx r_\text{E}.$$
Therefore, it doesn't make a noticeable difference to include this height in our calculations. This is why we can model the gravitational field strength on Earth to be constant and simply calculate the weight of any object as $$w = mg,$$
without caring about the height of the object with respect to the surface of the Earth. However this is an approximation, and if the height is big enough we need to consider it in our calculations.
For any application where we can model Earth's gravitational field as constant, we can consider that it exerts the total force on the object's center of mass.
Center of Gravity Definition
Now with a proper understanding of the center of mass and the gravitational force, we are equipped to comprehend what the center of gravity means.
The center of gravity is similar to the center of mass, with one key difference.
The center of gravity of a system is the mean position of an object's or system's weight distribution.
Weight is the total force acting on an object due to the gravitational force exerted by an astronomical body.
This means that the center of gravity is the point in an object or system where the average of all the weight lies. Therefore, the center of gravity could appropriately be called the center of weight.
Center of Gravity vs. Center of Mass
The center of mass and the center of gravity are very similar. To understand better how they differ, we need to understand the differences between mass and weight.
Mass is the amount of "stuff" or matter inside an object. It is a scalar quantity that does not change in a system. Weight is a force caused by an interaction with a gravitational field. It is a vector quantity that can change because the force of gravity is different depending on the strength of the gravitational field. Therefore, if you were to go to the Moon, you would have a different weight than on Earth because the gravitational field of the Moon is weaker.
The center of mass is always the same because it depends on the mass distribution. However, the center of gravity can change because it depends on the gravitational field.
In a uniform gravitational field, the center of mass and the center of gravity are located at the same point!
Center of Gravity Calculation
As we mentioned before, finding the center of mass is equivalent to finding the center of gravity in a uniform gravitational field. The equation for the center of mass of a system of particles is:
$$x_{\mathrm{cm}} =\frac{\sum m_i x_i}{\sum m_i}\\$$
where \(x_{\mathrm{cm}}\) is the location of the center of mass of the system, \(m_i\) is the mass value for one of the particles comprising it, and \(x_i\) is the particle's position.
Notice that the center of mass equation is simply finding the mean position of all the mass (it is adding up all the constituent parts, each mass times its distance, and dividing by the whole mass of the system, leaving us with only a value of position).
If a function of mass density is given for a solid, the total mass can be determined by integrating the mass density over the length (1-dimension), area (2-dimensions), or volume (3-dimensions) of the solid. For example, we would be able to use the equation \(\int M_\text{total} =p(r) \mathrm{d}V\), which takes a density function \(p(r)\) as an integrated, to solve for the total mass of a solid!
For systems or objects with symmetrical mass distribution, the center of mass lies on the lines of symmetry.
Therefore, to find the center of mass of a system composed of infinitely many differential masses \(\mathrm{d}m\) (this is the mathematical term for "very tiny bits") we can use the following formula:
$$\vec{r}_{\mathrm{cm}}=\frac{\int \vec{r} \mathrm{d}m}{\int \mathrm{d}m}.$$
This equation is similar to the equation for a system of particles with position \(x\). Still, it is more general and powerful because we integrate the differential of a mass function that tells us how is mass distributed in the system and uses a position vector \(\vec r\). Thus, this formula works for any system describing a finite-sized body.
Center of Gravity Formula
However, sometimes, we do not have a uniform gravitational field. In that case, we use the same formulas as above but replace all the masses with their weight! We can do this because the center of mass and the center of gravity are the same conceptually; one simply has to do with mass and the other with weight! Therefore, our above equations become
$$x_{\mathrm{cw}} =\frac{\sum w_i x_i}{\sum w_i},$$
and
$$\vec r_{\mathrm{cw}} =\frac{\int r \mathrm{d}w}{\int \mathrm{d}w},$$
respectively.
Center of Gravity and Center of Mass in a Uniform Gravitational Field
Consider a constant gravity field such as the one we use to model Earth's. Then, for a system of particles, we have:
$$\begin{aligned}x_{\mathrm{cm}} &=\frac{\sum w_i x_i}{\sum w_i}\\ x_{\mathrm{cm}}&= \frac{w_1x_1+w_2x_2+ \ldots}{w_1+w_2+ \ldots}\\x_{\mathrm{cm}}&= \frac{gm_1x_1+gm_2x_2+ \ldots}{gm_1+gm_2+ \ldots}\\x_{\mathrm{cm}} &= \frac{\bcancel{g}(m_1x_1+m_2x_2+ \ldots)}{\bcancel{g}( m_1+m_2+ \ldots)}\\ x_{\mathrm{cm}} &= \frac{m_1x_1+m_2x_2+ \ldots}{m_1+m_2+ \ldots}\\ x_{\mathrm{cw}}&= x_{\mathrm{cm}.}\end{aligned}$$
Although you do not need to know how to derive the equivalence of the center of mass and gravity as we do above, you do need to remember the essential concept that the center of mass equals the center of gravity in a uniform gravitational field.
Center of Gravity Examples
How could we get away with writing a center of gravity article without talking about jungle gyms?
Bob and Jimmy are sitting in a jungle gym. Jane is sliding downward on a diagonal part of the jungle gym inclined at \(30^\circ\). The normal force acting on her by the incline is \(500 \,\mathrm{N}\). Bob has a mass of \(60\,\mathrm{kg}\), and Jimmy has a mass of \(70\,\mathrm{kg}\). From the leftmost point of the jungle gym, Bob is \(1.5\,\mathrm{m}\) away, Jimmy is \(2.75\,\mathrm{m}\) away, and Jane is \(5\,\mathrm{m}\) away. Where is the center of gravity of Bob, Jimmy, and Jane, as a system, with respect to the leftmost point of the jungle gym?
Fig. 2 - Apparently, there are wooden jungle gyms. You learn something new every day.
Remember our formula for the center of gravity:
$$x_{cm} =\frac{\sum w_i x_i}{\sum w_i}\\\mathrm{.}$$
First, let's find all of the weights. Remember that we have to multiply \(60\,\mathrm{kg}\) by \(10\,\mathrm{\frac{N}{kg}\\}\) to get Bob's weight, which is the force on Bob due to gravity. Therefore,
$$60\,\mathrm{kg}\times 10\,\mathrm{\frac{N}{kg}\\}=600\,\mathrm{N},$$
is Bob's weight.
Now, we'll move on to Jimmy:
$$70\,\mathrm{kg}\times 10\,\mathrm{\frac{N}{kg}\\}=700\,\mathrm{N.}$$
To find Jane's weight, we need to calculate the upward normal force exerted on her. To do this, we treat her position like a right triangle with the forces acting on her as side lengths.
Fig. 3 - The normal force acting on Jane equals her weight \(mg\) times the cosine of the incline's angle.
The purple vector in the image above labeled \(\vec{F}_\mathrm{N}\) is the normal force acting on Jane, whose magnitude is given to us in the problem as \(500 \,\mathrm{N}\). To find the magnitude of Jane's weight, recall that the normal force is equal to the perpendicular component of the weight, as depicted in figure 4. From this relationship, we can solve for the weight.
$$\begin{align*} |\vec F_{\mathrm{N}}| &= F_\mathrm{N} \\ \vec |w| &= w \\[20pt] w &= mg \\ F_\mathrm{N} &= mg\cos{30.0^\circ } \\ F_\mathrm{N} &= w\cos{30.0^\circ } \\ w&= \frac{F_\mathrm{N}}{\cos{30.0^\circ .}} \\ \end{align*}$$
Let's now calculate her weight using the above result:
$$w=\frac{5.00\times 10^2 \,\mathrm{N}}{\cos{30.0^\circ }}=577\,\mathrm{N.}$$
Now that we know all the weights, we are ready to use the formula for the center of gravity.
$$\begin{aligned} x_{\text{cw}} &= \sum\frac{w_ix_i}{w_i}\\[8pt] x_{\text{cw}} &= \frac{(6.00\times10^2\,\mathrm{N})1.50\;\mathrm{m}+ (7.00\times10^2\,\mathrm{N}) 2.75\;\mathrm{m} + (577\,\mathrm{N}) 5\;\mathrm{m}}{600\,\mathrm{N} + 700\,\mathrm{N} + 577\,\mathrm{N}}\\[8pt]x_{\text{cw}} &= 3.04\,\mathrm{m}\end{aligned}$$
Therefore, the system's center of gravity is \(3.04\,\mathrm{m}\) from the leftmost part of the gym. Now that was a physics workout!
Let's now move on to an example that will dip into a little of your calculus knowledge.
Imagine that we wanted to design a megastructure and require to determine the center of gravity of a \(3\,000\,\mathrm{km}\) long rod. We can model the rod as one-dimensional, so we can describe its mass distribution with a linear mass density function \( \lambda (y) \). The rod needs to be partially buried so only \(2\,000\,\mathrm{km}\) will be visible. For your calculations, use Earth's center as the origin and consider the radius of Earth as \(6\,371\,\mathrm{km}\)). At what height would the center of gravity for this rod be?
Solution
The linear mass density of a straight rod is the derivative of the rod's mass with respect to the position(in this case, it is constant):
$$\lambda = \frac{\mathrm{d}m}{\mathrm{d}y}.$$
This means that
$$\mathrm{d}m = \lambda\mathrm{d}y$$
Recall that the gravitational force exerted by the Earth on a particle of mass \( m \) is the particle's weight. And we can calculate it with the formula
$$ w = F_g =G\frac{M_{\text{Earth}}m}{r^2} $$
In this case, we have to use this formula, and we can't consider the gravitational strength constant because the length of our system (\( 3\,000\;\mathrm{km} \)) is not negligible when compared to the radius of the Earth (\(6\,371\,\mathrm{km}\)). Now, we consider the rod as comprised of infinitely many pieces of mass, \( dm, \) located at a distance \( y \) from the center of the Earth. Therefore the equation for the weight of that piece of mass becomes
$$ \mathrm{d}w =G\frac{M_{\text{Earth}}\mathrm{d}m}{y^2}.$$
We are almost ready to use our formula for the center of mass:
$$\vec y_{\mathrm{cw}} =\frac{\int y \mathrm{d}w}{\int \mathrm{d}w}\\\mathrm{.}$$
Note that here, we just consider one dimension so we are using \( y \) instead of \( \vec{r} .\) Since we are measuring the distance with respect to the Earth's center, the distance to the beginning of the rod is \[6\,371\,\mathrm{km}-1000\;\;\mathrm{km}=5371\;\mathrm{km},\]
since the rod's first \( 1\,000\;\mathrm{km} \) are buried. Similarly, the distance of the top of the rod from the center of Earth is
\[6\,371\,\mathrm{km}+2\,000\;\mathrm{km}=8371\;\mathrm{km}.\]
This is because the remaining \( 2\,000\;\mathrm{km} \) protrude from Earth's surface. These values define our limits of integration!
Fig. 4 - This image helps us to visualize the scenario.
Now we have all we need, and we can substitute all our data in the formula for the center of gravity.
$$\begin{aligned}y_{\text{cw}} &= \frac{\int y \textcolor{#00b695}{\mathrm{d}w}}{\int \textcolor{#00b695}{\mathrm{d}w}}\\[8pt]y_{\text{cw}} &= \frac{\displaystyle\int y \textcolor{#00b695}{ \frac{GM_{\text{Earth}}\mathrm{d}m}{y^2}}}{\displaystyle\int \textcolor{#00b695}{\frac{ GM_{\text{Earth}}\mathrm{d}m}{y^2} }}\\[8pt]y_{\text{cw}} &= \frac{\textcolor{#00b695}{\bcancel{\textcolor{black}{G}}} \textcolor{#1478c8}{\bcancel{\textcolor{black}{M_{\text{Earth}}}}} \displaystyle\int y \frac{\mathrm{d}m}{y^2}}{\textcolor{#00b695}{\bcancel{\textcolor{black}{G}}} \textcolor{#1478c8}{\bcancel{\textcolor{black}{M_{\text{Earth}}}}} \displaystyle\int \frac{\mathrm{d}m}{y^2}}\\[8pt]y_{\text{cw}} &= \frac{\displaystyle\int \frac{\mathrm{d}m}{y}}{\displaystyle\int \frac{\mathrm{d}m}{y^2}}\end{aligned}$$
Notice that the \(G\) and the \(M_{\text{Earth}}\) cancel out. Giving us the mass of the Earth was only an attempt to throw us off!
Now let's rewrite the differential of mass using the linear mass density. Recall that we found that \( \mathrm{d}m = \lambda dy \). Using this result allows us to integrate with respect to \( y \) and use the limits of integration that we determined earlier.
$$\begin{aligned} y_{\text{cm}} &= \frac{\displaystyle \int \frac{\textcolor{#00b695}{\mathrm{d}m}}{y}}{\displaystyle \int \frac{\textcolor{#00b695}{\mathrm{d}m}}{y^2}}\\[10pt]y_{\text{cm}} &= \frac{\displaystyle\int \frac{\textcolor{#00b695}{\lambda \mathrm{d}y}}{y}}{\displaystyle \int \frac{\textcolor{#00b695}{\lambda \mathrm{d}y}}{y^2}}\\[10pt]y_{\text{cm}} &= \frac{\bcancel{\lambda} \displaystyle\int \frac{ \mathrm{d}y}{y}}{\bcancel{\lambda} \displaystyle \int \frac{ \mathrm{d}y}{y^2}}\end{aligned}$$
We can take the linear density out of the integral because it is constant. Finally, let's integrate using the limits we found before and simplify.
$$\begin{aligned} y_{\text{cm}} &= \frac{\displaystyle\int_{5\,371}^{8\,371} \frac{ \mathrm{d}y}{y}}{\displaystyle\int_{5\,371}^{8\,371} \frac{ \mathrm{d}y}{y^2}}\\[10pt]y_{\text{cm}} &= \frac{\ln(y)\Big |_{5\,371}^{8\,371}}{\left. -\frac{1}{y}\right|_{5\,371}^{8\,371}}\\[10pt]y_{\text{cm}} &= \frac{\ln(8\,371)-\ln(5\,371)}{\frac{1}{5\,371}-\frac{1}{8\,371}}\\[10pt]y_{\text{cm}} &= 6\, 651\end{aligned}$$
Therefore, the center of gravity of the rod is \(6\, 651\,\mathrm{km}\) from the center of the Earth, which means that the center is
$$6\,651\;\mathrm{km}- 6\,371\,\mathrm{km} = 280\,\mathrm{km}$$
above the Earth's surface. Remember that one half of the rod's length is \( 1\,500\;\mathrm{km} \), but since \( 1\,000\;\mathrm{km} \) are buried down, the geometric center would \( 500\,\mathrm{km} \) above the surface. Therefore, the center of gravity of the rod is not in its geometric center!
Fig. 5 - Notice that the center of mass is not in the same place as the center of gravity.
This is because the gravitational field of Earth changes with distance, and it is stronger near the bottom of the rod, closer to Earth's center of mass.
Now go try that Zen Level on Wii Fit again, and see if your knowledge of physics turns you into a master.
Center of Gravity - Key takeaways
The center of mass of a system is the mean position of an object's or system's mass distribution.
The center of gravity of a system is the mean position of an object's or system's weight distribution.
Forces acting on a particular object or system are thought of as acting on the center of mass of that object or system. In general, the force of gravity exerted by two objects is always felt on the line connecting the center of masses of the two objects.
Weight is the amount of force acting on an object due to the gravitational force of an astronomical body.
The center of gravity is located at the center of mass for a uniform gravity field.
On Earth, the gravitational field strength can be considered to be constant, with an approximate value of \( 10\;\mathrm{\frac{N}{kg}}. \) This allows us to calculate the weight of an object as \( w=mg \). However, if the object's dimensions or its height with respect to the surface of Earth are considerable, this approximation won't be accurate.
For systems or objects with symmetrical mass distribution, the center of mass lies on the line of symmetry.
If the gravitational field is not uniform, we can calculate it using the formula $$x_{cm} =\frac{\sum W_i x_i}{\sum W_i},$$ if it is a system of particles, or we can use $$\vec r_{\mathrm{cw}} =\frac{\int r \mathrm{d}w}{\int \mathrm{d}w},$$ if it is a finite-sized body.
References
- Fig. 1 - Gravitational force, StudySmarter Originals.
- Fig. 2 - Wooden Jungle Gym with Slide (https://www.publicdomainpictures.net/en/view-image.php?image=101134&picture=wooden-jungle-gym-with-slide) by Lynn Greyling (http://www.artseasons.co.za/) is licensed by CC0 1.0 (https://creativecommons.org/publicdomain/zero/1.0/)
- Fig. 3 - Person on an inclined plane, StudySmarter Originals
- Fig. 4 - Rod on Earth, StudySmarter Originals
- Fig. 5 - Rod on Earth with center of mass and center of gravity, StudySmarter Originals
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