Physics Explained Through a Video Game/Systems and Center of Mass

Topic 2.1 - Systems and Center of Mass

Before looking discussing forces later in this unit, we need to consider a few definitions first.

Introduction

Collection of objects, such as the carts of a Ferris wheel, have a center of mass. Knowing this can help us examine the forces acting on a system, as introduced in the following topic.

The center of mass (sometimes abbreviated as CoM) of an object refers to the average position of all of the parts that form the object.

In classical mechanics, considering this can be useful in variety of circumstances as it oftentimes enables us to view an awkwardly-shaped object as a singular point. Later on in this article series, this will be particularly helpful in simplifying many problems and concepts.

In some cases, such as in Ferris Wheel by antidonaldtrump / Armin van buren (on the right), it may be useful to locate the center of mass of a system of objects. To note, a system in physics is just some collection of objects or parts of an object that we're interested in observing together.

Example 1: Ferris Wheel

As an example of how to calculate for a system's center of mass, we need to first consider the objects with mass that compose it. In the case of the Ferris wheel, for simplicity, we're using the assumption that only the six carts of the Ferris wheel have non-negligible mass. Thus, the rest of the Ferris wheel (such as the base or the rim) won't be considered as contributing to its center of mass.

With this out of the way, we can define that there are six objects that compose the system. For each of these objects, we need to know their own center of masses.

Using the diagram on the right hand side, which represents the locations of each of the six cart's center of mass, assume that:

• Each green dot specifies the point of a cart's center of mass and its position.
• The position values of the green dots are in meters $({\text{m}})$.
• All of the carts have the same mass, given as $10,000.\;{\text{kg}}$.

In order to calculate for where the Ferris wheel's center of mass is, we need to use the definition for the center of mass in a particular direction, as provided below. Essentially, this allows for us to separately calculate the center of mass of a system in the horizontal and vertical directions (assuming a 2-Dimensional situation). From this, we can describe where the center of mass (in a specific direction) by knowing the relative mass and coordinate of each cart.

Definition of Center of Mass in Some Direction ($x$):

$x_{\text{com}}={\frac {x_{1}m_{1}+x_{2}m_{2}+x_{3}m_{3}+\dots }{m_{1}+m_{2}+m_{3}+\dots }}$

With the formula above, $m_{1},m_{2},m_{3}\dots$ represents the mass of each object in the system. Also, $x_{1},x_{2},x_{3}\dots$ represents that respective object's center of mass in the (X) direction.

To reflect on the information that we've gathered so far, refer to the table below. In this, we're assigning each cart and its center of mass location as a part of the formula above.

Already-known information of the Ferris Wheel System

Object name	Assigned $m_{n}$	Known X-coordinate of CoM, $x_{n}$	Known Y-coordinate of CoM, $y_{n}$
Red cart	$m_{1}=10,000.\;{\text{kg}}$	$\;x_{1}=10.\;{\text{m}}$	$\;y_{1}=20.\;{\text{m}}$
Purple cart	$m_{2}=10,000.\;{\text{kg}}$	$\;x_{2}=20.\;{\text{m}}$	$\;y_{2}=30.\;{\text{m}}$
Blue cart	$m_{3}=10,000.\;{\text{kg}}$	$\;x_{3}=30.\;{\text{m}}$	$\;y_{3}=20.\;{\text{m}}$
Green cart	$m_{4}=10,000.\;{\text{kg}}$	$\;x_{4}=30.\;{\text{m}}$	$\;y_{4}=10.\;{\text{m}}$
Yellow cart	$m_{5}=10,000.\;kg$	$\;x_{5}=20.\;{\text{m}}$	$\;y_{5}=0.0\;{\text{m}}$
Orange cart	$m_{6}=10,000.\;kg$	$\;x_{6}=10.\;{\text{m}}$	$\;y_{6}=10.\;{\text{m}}$

.

With the information labeled above, we can begin calculating for the center of mass in the $x$ and $y$ directions.

We can start with calculating the $x$ direction's center of mass, $x_{\text{com}}$. To do this, we need to employ the formula specified above for the center of mass in a particular direction, $x_{\text{com}}={\frac {x_{1}m_{1}+x_{2}m_{2}+x_{3}m_{3}+\dots }{m_{1}+m_{2}+m_{3}+\dots }}$.

Using the information from the table above and then substituting the known variables, we will find that: $x_{\text{com}}={\frac {(10.)(10,000.)+(20.)(10,000.)+(30.)(10,000.)+(30.)(10,000.)+(20.)(10,000.)+(10.)(10,000.)}{(10,000.)+(10,000.)+(10,000.)+(10,000.)+(10,000.)+(10,000.)}}{\frac {{\text{m}}\cdot {\text{kg}}}{\text{kg}}}$.
This can be algebraically simplified such that:
$\implies x_{\text{com}}={\frac {1,200,000}{60,000}}{\text{m}}\cdot {\cancelto {\mathbf {1} }{\frac {\text{kg}}{\text{kg}}}}$

$\implies x_{\text{com}}=20.\;{\text{m}}\;{\text{[2}}\;{\text{sigfigs]}}$

Practice Question: Using the provided table and the process shown for solving for the center of mass in a specific direction, solve for $y_{\text{com}}$. Using this information, where is the center of mass of the Ferris wheel system, ($x_{\text{com}}$, $y_{\text{com}}$)?

Object name	Assigned $m_{n}$	Known Y-coordinate of CoM, $y_{n}$
Red cart	$m_{1}=10,000.\;{\text{kg}}$	$\;y_{1}=20.\;{\text{m}}$
Purple cart	$m_{2}=10,000.\;{\text{kg}}$	$\;y_{2}=30.\;{\text{m}}$
Blue cart	$m_{3}=10,000.\;{\text{kg}}$	$\;y_{3}=20.\;{\text{m}}$
Green cart	$m_{4}=10,000.\;{\text{kg}}$	$\;y_{4}=10.\;{\text{m}}$
Yellow cart	$m_{5}=10,000.\;kg$	$\;y_{5}=0.0\;{\text{m}}$
Orange cart	$m_{6}=10,000.\;kg$	$\;y_{6}=10.\;{\text{m}}$

Using the table on the right, we can consider the columns dedicated for the assigned mass ($m_{\text{n}}$) and the Y-coordinate ($y_{\text{n}}$) of the parts of the system (the Ferris wheel).

Also, since the center of mass in the Y-direction is being solved for, the related equation can have the particular direction be specified as $y$ such that:

$y_{\text{com}}={\frac {y_{1}m_{1}+y_{2}m_{2}+y_{3}m_{3}+\dots }{m_{1}+m_{2}+m_{3}+\dots }}$

Similar to how $x_{\text{com}}$ was calculated, we can then use substitution of variables and algebraically simplify to find $y_{\text{com}}$. This derivation results in finding $y_{\text{com}}$, as shown below.

$y_{\text{com}}=15.\;{\text{m}}\;{\text{[2}}\;{\text{sigfigs]}}$

With knowing both the values for the $x_{\text{com}}$ and $y_{\text{com}}$, we can now specify the center of mass of the Ferris wheel's as being located at $(20.,15.)\;{\text{m}}$. For more context, we can again consider the reference image of the Ferris wheel system. By using the surrounding labeled coordinates, we could plot where the center of mass as being just below the center shaft (where the large grey rods in the image meet together).

See the Ferris wheel image below with a superimposed bold yellow star indicating the approximated center of mass's location.

Example 2: Blocky Snakes

Section I: Content Primer

In the previous example about a Ferris wheel, a specific moment was considered for calculating for the system's center of mass at that instant. However, it required a considerable amount of effort despite the system only having a few considered parts.

A video showing a simple snake moving around a rectangular grid.

This raises a couple of questions:

1. Are there sometimes shortcuts towards for the center of mass?
2. Can we expand the idea of an object's center of mass to objects that are in motion?

In this example, we'll address these questions, providing more of an insight towards the intuition and applications behind a system's center of mass.

Consider the map "Strange Snake" by 11vanya11. In this, a snake made out of seven blocks circles around a rectangular perimeter, as seen on the right. Suppose that all of these seven blocks with a uniform density* each with the same mass.

*An important sidenote about uniform density:

When objects have a uniform density, this means that the mass is evenly distributed within it. In other words, if we were to consider one of the snake's blocks as multiple chunks, any two chunks will have the same mass if they take up the same amount of space.

A block of the snake is broken into three different chunks. Because the three chunks all have the same size and the snake blocks have a uniform density, the chunks have the same mass.

Taking this idea one step further, suppose we were to find the snake block's center of mass by removing some small part of the snake's block away from consideration. Then, a portion of the same size on the snake block's opposite side is also removed. This process can be repeated until we approach the geometric center of the block, which is where the snake block's center of mass is.

By gradually removing opposite pairs of the block with uniform density, we can find that the block has a center of mass located at its geometric center.

Because the snake's block has a uniform density, the removed portions will have the same mass. Also, because they are on opposite sides, they won't change where the snake block's center of mass is. This is because the center of mass in other words is the average point of where an object's mass is. To note, with many basic flat geometric shapes, such as a circle, square, equilateral triangle, etc., if it has a uniform density, then the center of mass is where the geometric center is.

Section II: Helpful Techniques for Finding the CoM

A blue snake created out of more blocks capable of moving around in a more complicated path than in the previous video.

When considering the information of the snake's blocks (and the sidenote's information above), in many cases, we can simplify the process of finding the center of an object's mass. To explain this concept, let's consider a still image of "Block Snake" by O_o O_o O_o, a similar map as shown on the right. Suppose that each block of the snake has a mass of $1\;{\text{kg}}$ and is $0.2\;{\text{m}}$ in length and width.

We now know that if (1) a system or system part creates a simple geometric shape and (2) that system or system part has a uniform density, then it has a center of mass where the geometric center is.

When looking at the image on the right, it is clear that the snake as a whole does not create a simple geometric shape. However, there are parts of the snake that each create a rectangular shape. Because the entire snake has a uniform density, each of these snake parts (shown below) has its own center of mass at its geometric center.

The blue snake is broken into four subsections, each with a center of mass specified at that subsection's geometric center. The mass of each subsection is also labeled.

Through using the method shown above, if we were to use the formulas for $x_{\text{com}}$ and $y_{\text{com}}$ (provided a coordinate plane), the computation is much simpler. Instead of having to consider the $(x,y)$ coordinates of 16 different blocks, we only have to consider four distinct groups to find the entire snake's center of mass at that moment.

Continue this example:

It is possible to find the relative position of each of the center of masses because we know the size of each block.

If we were aware of the coordinate position of each of center of masses or otherwise were able solve for them, we would be able to figure out where the snake's entire center of mass is.

Recall that the snake's blocks each have a length and width of $0.2m$. If we allow for the green segment's center of mass to be located at the $(x,y)$ coordinate $(0.0,0.0)\;{\text{m}}$, we can manually figure out where the other center of masses are located as shown in the diagram on the right.

From this point, we could use the formulas for $x_{\text{com}}$ and $y_{\text{com}}$, as introduced in Example 1.

$x_{\text{com}}={\frac {x_{1}m_{1}+x_{2}m_{2}+x_{3}m_{3}+\dots }{m_{1}+m_{2}+m_{3}+\dots }}$

$x_{\text{com}}={\frac {x_{1}m_{1}+x_{2}m_{2}+x_{3}m_{3}+x_{4}m_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}}$

$x_{\text{com}}={\frac {(0.00)(4.0)+(0.40)(3.0)+(1.1)(7.0)+(1.8)(2.0)}{(4.0)+(3.0)+(7.0)+(2.0)}}{\frac {{\text{m}}\cdot {\text{kg}}}{\text{kg}}}$ $x_{\text{com}}=0.78\;{\text{m}}\;{\text{[2}}\;{\text{sigfigs]}}$

$y_{\text{com}}={\frac {y_{1}m_{1}+y_{2}m_{2}+y_{3}m_{3}+\dots }{m_{1}+m_{2}+m_{3}+\dots }}$

$y_{\text{com}}={\frac {(0.00)(4.0)+(0.30)(3.0)+(0.60)(7.0)+(0.40)(2.0)}{(4.0)+(3.0)+(7.0)+(2.0)}}{\frac {{\text{m}}\cdot {\text{kg}}}{\text{kg}}}$

$y_{\text{com}}=0.37\;{\text{m}}\;{\text{[2}}\;{\text{sigfigs]}}$

With both $x_{\text{com}}$ and $y_{\text{com}}$, we can define that the center of the snake's mass is approximately at $(0.78,0.37)\;{\text{m}}$, as labeled by the image now with a superimposed star on the right.

To clarify, it is okay for the center of an object's mass to be not on the object itself. For instance, consider a doughnut that (1) has a hole in the middle of it and (2) has uniform density. Although there isn't mass from the doughnut itself inside of the hole, the doughnut mass surrounding the hole averages out such that it's center of mass is in the hole.

As such, as also seen by the example just shown with the snake, the center of an object's (or a system's) mass does not necessarily have to be inside of the object itself.

Section III: Velocity of the CoM

Players are climbing up a handful of small green platforms that are all falling downwards.

As seen in the previous sections, we've idealized finding the center of mass of a system by object considering still frames. However, oftentimes, the center of mass is undergoing motion, making it necessary to account for this in some situations.

To explore this concept, consider the map Climb by Fantao.

Assume that:

• The platforms are all falling a constant velocity of $-0.5\;{\text{m/s}}$.
• The smaller platforms each have a mass of $1.0kg$.
• The larger top platform has a mass of $2.5kg$.

What is the velocity of the center of mass of the system of visible platforms?

Approaching this problem may be intuitive. To explain, if an entire system (such as a car, or in this case: a set of green platforms) is moving at the same velocity, then it makes sense for the center of mass of that system to be moving at that velocity. This would be regardless of where that center of mass is positioned.

More formally, we can also solve this problem by considering the formula for the velocity of the center of mass:

Definition of Center of Mass's Velocity in Some Direction ($x$):

$v_{\text{com,x}}={\frac {v_{1,x}m_{1}+v_{2,x}m_{2}+v_{3,x}m_{3}+\dots }{m_{1}+m_{2}+m_{3}+\dots }}$

$v_{\text{com,x}}={\frac {v_{1,x}m_{1}+v_{2,x}m_{2}+v_{3,x}m_{3}+\dots }{m_{1}+m_{2}+m_{3}+\dots }}$
$\implies v_{\text{com,x}}={\frac {(-0.5)(1.0)+(-0.5)(1.0)+(-0.5)(1.0)+(-0.5)(2.5)}{(1.0)+(1.0)_{(}1.0)+(2.5)}}{\frac {\text{m}}{\text{s}}}\cdot {\frac {\text{kg}}{\text{kg}}}$
$\implies v_{\text{com,x}}=-0.5{\frac {\text{m}}{\text{s}}}{\text{[1}}{\text{ sigfig]}}$

Alongside this, because the platforms aren't moving in the horizontal direction, there is a $v_{\text{com,y}}$ that equals $0.0\;{\frac {\text{m}}{\text{s}}}\;{\text{[1}}{\text{ sigfig]}}$.

We can add these vectors together similar to how displacement was calculated for in Topic 1.2. As such, we can specify that $v_{\text{com}}=\pm {\sqrt {(v_{\text{com,x}})^{2}+(v_{\text{com,y}})^{2}}}$.

Thus, $v_{\text{com}}=\pm {\sqrt {(-0.5)^{2}+(0.0)^{2}}}$
$v_{\text{com}}=\pm {\sqrt {(-0.5)^{2}+(0.0)^{2}}}$

$\implies v_{\text{com}}=-0.5{\frac {\text{m}}{\text{s}}}\;{\text{[1}}{\text{ sigfig, Cartesian coordinate convention]}}$.

Question 1: Icebergs

An visual of the three icebergs, their respective above-water mass and their respective center of mass* (labeled in pink).

*For their respective iceberg portions that are above water.

Unit conversion: $1{,}000\;{\text{kg}}=1\;{\text{Mg}}$

Consider the map Breaking Ice by Semi_Cow124, as pictured on the right.
Part (a):

Calculate the location of the center of mass of parts of the three icebergs that are above water.

Assume that 90% of each of the iceberg's total mass is below water. It's been discovered that the center of mass for the three iceberg's portions that are underwater is at $(22.0,-15.0)\;{\text{m}}$.
Part (b):

(i): How much mass do three icebergs combined have underwater?
(ii): Calculate the location of the center of mass of the three entire icebergs.

Several players are pictured leaping and bouncing off of the largest iceberg until it beings to collapse.

The falling iceberg with labels providing the vertical velocities of specified portions of it.

Suppose a new in-game situation (above) where players are actively jumping on the icebergs. Because of their impacts, the largest iceberg in the center of the map begins to fracture and fall apart. At a certain moment, 70% of the largest iceberg that is above water has its vertical velocity recorded, as diagrammed on the right. Assume that the remainder of the iceberg remains stationary.
Part (c):

How quickly is the center of mass of the center iceberg's portion above water falling vertically?

Consider discussing your solutions on this article's Talk Page, where you find help from others.

References