Educational Innovations Blog
 Center of Mass ChallengeIn my 13 years of classroom teaching experience, I’ve learned a few tricks to keep my students from becoming overwhelmed—or bored—by their science textbooks. One of my favorite tricks is to get my kids up on their feet, doing science instead of reading about it. When I teach about gravity and center of mass, for example, I like to shake things up by turning our classroom into an impromptu biokinetics lab. I challenge my students to perform a series of seemingly simple physical tests, described below. Lift a chair? Raise your leg? Pick up a quarter? No problem! (Or so they think…) In the end, these physical experiments allow my students to grasp the basic concepts of gravity and center of mass in a far more compelling way than they’d ever learn from any textbook. The results are hilarious and unforgettable! As an added bonus, these challenges often bring my students’ families together as the students go home and challenge their siblings and parents to do them as well. Just imagine: science bringing families together, one discovery at a time! I love it! Let’s start with a quick vocabulary lesson.Gravity is a term used by scientists to describe the force of gravitation given on an object or near the surface of a celestial body such as the Earth, the moon, or another terrestrial planet. This force pulls on ALL mass. All mass—from the smallest sub-atomic particle to the largest star in the universe—exerts some amount of gravitational force. The more mass an object has, the more gravitational force is exerted. Center of Mass is a term used by scientists to describe that point in an object where the object’s mass or weight seems to be concentrated. This point could be a physical object or it could be a point outside of the object. In a round/spherical object, the equally distributed mass is the center of mass for the sphere. Finding the center of mass in an irregular object (such as the human body) takes a little more work to figure out… and a lot more fun. That’s where these challenges come into play. Center of Mass Challenge 1: The Thumb Press
WHAT HAPPENED? The reason it’s so hard to stand up is because your center of mass is located over the seat of the chair rather than over your feet, which are in front of you. Center of Mass Challenge 2: The Chair Lift
WHAT HAPPENED? Results of this challenge may vary by gender. Interestingly enough, many girls can successfully complete this challenge, while most boys cannot. Why? Again, it’s all about the center of mass. The center of mass for most girls is lower to the hips, while the center of mass in boys is much higher. Therefore, for most girls, the center of mass while bent over the chair is above their feet, while the center of mass for most boys is above the chair. Center of Mass Challenge 3: The Quarter Grab
WHAT HAPPENED? The reason this challenge is so difficult is because of our anatomy—specifically, our bottoms. As you bend over to pick up the quarter, your rear end naturally extends backward to help keep your body balanced. Since it’s pressed against the wall, your bottom has no where to extend. This causes the center of mass to shift forward, resulting in falling forward—at least for most people. Center of Mass Challenge 4: The Leg Raise
WHAT HAPPENED? This challenge is nearly IMPOSSIBLE. For the same reason as before: because the wall is in the way, your body can’t counter-balance the mass distribution. Your center of mass is over BOTH legs, not just one. Watch my students in action!If you don’t have time to perform these challenges in your own classroom—or if you’d like to get a quick look at how they can be set up—take a look at the video below, which was put together “with a little help from my (classroom) friends.” Enjoy! Educational Innovations sells many unique science demo materials. Among them, some of my favorites are – you guessed it – related to center of mass. For instance, there’s the Balancing Bird Demo , which always fascinates my students. It can balance on anything – even the tip of a pencil! Around the holidays, I like to give my students their own Small Balancing Birds as gifts. They love them! And finally, there’s the Balancing Bird Puzzle – a DIY balancing puzzle that requires students to figure out where to place the magnets in order to keep the bird balanced. This one has been a big hit among my kids. Jeremy Johnson has taught science in elementary and middle school for more than 13 years. He is also the founder of the YouTube channel “SciOnTheFly.” https://www.youtube.com/user/scionthefly This entry was posted on Friday, June 2nd, 2017 at 12:15 am and is filed under College level , Elementary level , experiments , High School level , Middle School level , Physics . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response , or trackback from your own site. 5 Responses to Center of Mass ChallengeThis is amazing and such an interesting way to conduct class. I can’t wait to try it out with my 8 graders. Thank you! Pretty cool! I tried your experiments out was found them to be very interesting. I never really thought about how my body will automatically adjust itself when I try to stand up or lift a leg. While I was trying out the chair lift exercise I found that the hard part was not lifting the chair. It was lifting the chair while leaning my head on the wall that was so hard. I tried doing the same experiment while in the middle of the room and had no issues at all with lifting the chair. After trying again but paying more attention to my body’s position, I noticed that when lifting the chair I was keeping my head closer to my body than in the first experiment. When lifting in the middle of the room I tried to lift the same as when against the wall, but my body automatically adjusted how far I extended my head without me realizing until I paid more conscious attention to my body’s position. To me that is one of the interesting things about this experiment. By leaning your head against the wall you are forcing your body to extend the head further then it naturally would if you were not leaning your head into the wall. This acts to prevent your body from adjusting your balance to counter where your center of gravity is. Awesome science! Thanks for helping me learn that by changing the center of gravity you are not simply making a task harder. Sometimes you make doing something impossible. I can also see applications to areas like engineering buildings or designing small gadgets. I did this with my siblings and it was really fun to do!! Perfect educational fun for quarantine! This was exciting to do with my kids. They got a blast out of trying to do the different things. Awesome. If we want understand this by equations and analyze the movement on XYZ , how you we can do that? Leave a Reply Cancel replyYour email address will not be published. Required fields are marked * Save my name, email, and website in this browser for the next time I comment. Notify me of follow-up comments by email. Notify me of new posts by email. This site uses Akismet to reduce spam. Learn how your comment data is processed . Sign Up for Our Newsletter!As a subscriber, you’ll receive FREE lesson ideas and exclusive sale offers – plus $5 off your next order of $50 or more! Sign up now .
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Center of Mass: Definition, Equation, How to Find (w/ Examples)Have you ever seen one of those toy birds that is able to balance on your fingertip by its beak without tipping over, as if by magic? It isn’t magic that allows the bird to balance at all, but the simple physics associated with center of mass. Understanding the physics behind the center of mass allows you not only to understand conservation of momentum and other related physics, but can also inform stability and dynamics in the sports you play, as well as allow you to perform some creative balancing acts. Definition of Center of MassAn object’s center of mass , sometimes also called the center of gravity, can be thought of as the point where the total mass of an object or a system can be treated as a point mass. In certain situations, external forces can be treated as though they are acting on the center of mass of the object. For the toy bird balancing on your fingertip, the center of mass is at its beak. This might seem wrong at first, which is why the act of balancing appears magical. Indeed, for a bird sitting on a branch, its center of mass is somewhere in its body. But the balancing bird toy often has weighted wings that span outward and forward, causing it to balance differently. The center of mass can be determined for a single object – such as the balancing bird – or it can be computed for a system of several objects, as you will see in a later section. Center of Mass for a Single ObjectThere will always be a single point on a rigid body that is the location of that body’s center of mass. The position of the center of mass of an object depends on the distribution of mass. If an object is of uniform density, its center of mass is easier to determine. For example, in a circle of uniform density, the center of mass is the center of the circle. (This would not be the case, however, if the circle was denser on one side than the other). In fact, the center of mass will always be at the geometric center of the object when density is uniform. (This geometric center is called the centroid .) If the density is not uniform, there are other ways to determine the center of mass. Some of these methods involve the use of calculus, which is beyond the scope of this article. But one simple way to determine the center of mass of a rigid object is to simply try to balance it on your fingertip. The center of mass will be at the balancing point. Another method, useful for planar objects, is as follows:
For some objects, however, it is possible for the balance point to be outside the bounds of the object itself. Think of a ring, for example. The center of mass for a ring shape is in the center, where no part of the ring exists at all. Center of Mass of a System of ParticlesThe position of the center of mass for a system of particles can be thought of as their average mass position. The same idea can be used as for a rigid object if you imagine this system of particles are all connected by rigid, massless plane. The center of mass would then be the balance point of that system. To determine the center of mass of a system of particles mathematically, the following simple formula can be used: Where M is the total mass of the system, m i are the individual masses and r i are their position vectors. In one dimension (for masses distributed along a straight line) you can replace r with x . In two dimensions, you can find the x -coordinate and y -coordinate of the center of mass separately as: Examples of Calculating the Center of Mass Example 1: Find the coordinates of the center of mass of the following system of particles: particle of mass 0.1 kg located at (1, 2), particle of mass 0.05 kg located at (2, 4) and particle of mass 0.075 kg located at (2, 1). Solution 1: Apply the formula for the x -coordinate of the center of mass as follows: Then apply the formula for the y -coordinate of the center of mass as follows: So the location of the center of mass is (0.079, 2.11). Example 2: Find the location of the center of mass of a uniform density equilateral triangle whose vertices lie at points (0, 0), (1, 0) and (1/2, √3/2). Solution 2: You need to find the geometric center of this equilateral triangle with side length 1. The x -coordinate of the geometric center is straightforward – it is simply 1/2. The y -coordinate is a little trickier. It will occur at the location that a line from the top of the triangle to the point (0, 1/2) intersects with a line from any of the other vertices to the midpoint of one of the opposite side. If you sketch such an arrangement, you will find yourself with a 30-60-90 right triangle whose long leg is 0.5 and short leg is the y -coordinate. The relationship between these sides is √3y = 1/2, hence y = √3/6, and the coordinates of the center of mass are (1/2, √3/6). Motion of the Center of MassThe location of the center of mass of an object or system of objects can be used as a reference point in many physics calculations. When working with a system of interacting particles, for example, finding the center of mass of the system allows for an understanding of linear momentum. When linear momentum is conserved, the center of mass of the system will move with a constant velocity even as the objects themselves bounce off one another. For a falling rigid object, gravity can be treated as acting on that object’s center of mass, even if that object is rotating. The same is true of projectiles. Imagine tossing a hammer, and as it flies through an arc in the air, it rotates end over end. This might seem like complex motion to model at first, but it turns out that the center of mass of the hammer moves in a nice smooth parabolic path. A simple experiment can be performed which demonstrates this by taping a small piece of glow tape to the hammer’s center of mass, and then tossing the hammer as described in a dark room. The glow tape will appear to move in a smooth arc, like a tossed ball. A Simple Experiment: Find the Center of Mass of a BroomA fun center-of-mass experiment that you can perform at home involves using a simple technique for finding the center of mass of a broom. All you need for this experiment is one broom and two hands. With your hands relatively far apart, hold up the broom on the end of two pointer fingers. Then, slowly bring your hands closer together, sliding them underneath the broom. As you move your hands closer together you may notice one hand wants to slide along the underside of the broom handle while the other one stays put for a while before sliding. The entire time your hands move, the broom remains balanced. Eventually, when your two hands meet, they will meet at the location of the broom’s center of mass. Center of Mass of the Human BodyThe center of mass of the human body is located somewhere near the navel (belly button). In men, the center of mass tends to be a little higher since they carry more body mass in their upper body, and in women, the center of mass is lower because they carry more mass in their hips. If you stand on one foot, your center of mass will shift toward the side of the foot you are standing on. You may notice yourself leaning more toward that side. This is because in order to stay balanced, your center of mass needs to stay over the foot you are balancing on, or else you will tip over. If you stand with one leg and hip against a wall and try to lift your other leg, you will likely find it impossible because the wall prevents your weight from shifting over the balance leg. Another thing to try is standing with your back to the wall and your heels touching the wall. Then try to bend forward and touch the floor without bending your legs. Women may be more successful at this task than men because their center of mass is lower in their body and may end up still being over their toes as they lean forward. Center of Mass and StabilityThe location of the center of mass relative to an object’s base determines its stability. Something is considered stably balanced if, when tipped slightly and then released, it then returns back to its original position instead of tipping further and falling over. Consider a three-dimensional pyramid shape. If balanced on its base, it is stable. If you lift one end slightly and let it go, it falls back down. But if you try to balance the pyramid on its tip, then any deviations from perfect balance will cause it to fall over. You can determine if an object will fall back to its original position or tip over by looking at the location of the center of mass relative to the base. Once the center of mass moves past the base, the object will tip over. If you play sports, you might be familiar with the ready position where you stand with a wide stance and knees bent. This keeps your center of mass low, and the wide base makes you more stable. Consider how hard someone would have to push you to tip you over if you are in the ready position vs. when you are standing up straight with your feet together. Some cars have problems with rolling over when they take sharp turns. This is because of the location of their center of mass. If the center of mass of a vehicle is too high and the base is not wide enough, then it doesn’t take much to cause it to tip over. It’s always best for the stability of a vehicle to have most of the weight as low as possible. Related ArticlesScience project on how the mass of a paper airplane..., what are the three principles of gravity that affect..., levers used in everyday life, how to build a knee school project, how to use a single and double pulley system, principles of pulleys & levers, how to calculate cg, how to calculate center of mass, how to make a model of the pivot joint, how to make a clay model of the urinary system, how to make a compass at home for geometry, tips to make a strong marshmallow tower, how to make a bridge out of straws, how to find a hermit crab, how to build a model of the leaning tower of pisa, how to build a catenary curve arch, how to build an easy catapult for kids, 6 types of freely movable joints.
About the Author Gayle Towell is a freelance writer and editor living in Oregon. She earned masters degrees in both mathematics and physics from the University of Oregon after completing a double major at Smith College, and has spent over a decade teaching these subjects to college students. Also a prolific writer of fiction, and founder of Microfiction Monday Magazine, you can learn more about Gayle at gtowell.com. Find Your Next Great Science Fair Project! GO9.6 Center of MassLearning objectives. By the end of this section, you will be able to:
We have been avoiding an important issue up to now: When we say that an object moves (more correctly, accelerates) in a way that obeys Newton’s second law, we have been ignoring the fact that all objects are actually made of many constituent particles. A car has an engine, steering wheel, seats, passengers; a football is leather and rubber surrounding air; a brick is made of atoms. There are many different types of particles, and they are generally not distributed uniformly in the object. How do we include these facts into our calculations? Then too, an extended object might change shape as it moves, such as a water balloon or a cat falling ( Figure 9.26 ). This implies that the constituent particles are applying internal forces on each other, in addition to the external force that is acting on the object as a whole. We want to be able to handle this, as well. The problem before us, then, is to determine what part of an extended object is obeying Newton’s second law when an external force is applied and to determine how the motion of the object as a whole is affected by both the internal and external forces. Be warned: To treat this new situation correctly, we must be rigorous and completely general. We won’t make any assumptions about the nature of the object, or of its constituent particles, or either the internal or external forces. Thus, the arguments will be complex. Internal and External ForcesSuppose we have an extended object of mass M , made of N interacting particles. Let’s label their masses as m j m j , where j = 1 , 2 , 3 , … , N j = 1 , 2 , 3 , … , N . Note that If we apply some net external force F → ext F → ext on the object, every particle experiences some “share” or some fraction of that external force. Let: Notice that these fractions of the total force are not necessarily equal; indeed, they virtually never are. (They can be, but they usually aren’t.) In general, therefore, Next, we assume that each of the particles making up our object can interact with (apply forces on) every other particle of the object. We won’t try to guess what kind of forces they are; but since these forces are the result of particles of the object acting on other particles of the same object, we refer to them as internal force s f → j int f → j int ; thus: f → j int = f → j int = the net internal force that the j th particle experiences from all the other particles that make up the object. Now, the net force, internal plus external, on the j th particle is the vector sum of these: where again, this is for all N particles; j = 1 , 2 , 3 , … , N j = 1 , 2 , 3 , … , N . As a result of this fractional force, the momentum of each particle gets changed: The net force F → F → on the object is the vector sum of these forces: This net force changes the momentum of the object as a whole, and the net change of momentum of the object must be the vector sum of all the individual changes of momentum of all of the particles: Combining Equation 9.22 and Equation 9.23 gives Let’s now think about these summations. First consider the internal forces term; remember that each f → j int f → j int is the force on the j th particle from the other particles in the object. But by Newton’s third law, for every one of these forces, there must be another force that has the same magnitude, but the opposite sign (points in the opposite direction). These forces do not cancel; however, that’s not what we’re doing in the summation. Rather, we’re simply mathematically adding up all the internal force vectors. That is, in general, the internal forces for any individual part of the object won’t cancel, but when all the internal forces are added up, the internal forces must cancel in pairs. It follows, therefore, that the sum of all the internal forces must be zero: (This argument is subtle, but crucial; take plenty of time to completely understand it.) For the external forces, this summation is simply the total external force that was applied to the whole object: As a result, This is an important result. Equation 9.25 tells us that the total change of momentum of the entire object (all N particles) is due only to the external forces; the internal forces do not change the momentum of the object as a whole. This is why you can’t lift yourself in the air by standing in a basket and pulling up on the handles: For the system of you + basket, your upward pulling force is an internal force. Force and MomentumRemember that our actual goal is to determine the equation of motion for the entire object (the entire system of particles). To that end, let’s define: p → CM = p → CM = the total momentum of the system of N particles (the reason for the subscript will become clear shortly) Then we have and therefore Equation 9.25 can be written simply as Since this change of momentum is caused by only the net external force, we have dropped the “ext” subscript. This is Newton’s second law, but now for the entire extended object. If this feels a bit anticlimactic, remember what is hiding inside it: p → CM p → CM is the vector sum of the momentum of (in principle) hundreds of thousands of billions of billions of particles ( 6.02 × 10 23 ) ( 6.02 × 10 23 ) , all caused by one simple net external force—a force that you can calculate. Center of MassOur next task is to determine what part of the extended object, if any, is obeying Equation 9.26 . It’s tempting to take the next step; does the following equation mean anything? If it does mean something (acceleration of what, exactly?), then we could write which follows because the derivative of a sum is equal to the sum of the derivatives. Now, p → j p → j is the momentum of the j th particle. Defining the positions of the constituent particles (relative to some coordinate system) as r → j = ( x j , y j , z j ) r → j = ( x j , y j , z j ) , we thus have Substituting back, we obtain Dividing both sides by M (the total mass of the extended object) gives us Thus, the point in the object that traces out the trajectory dictated by the applied force in Equation 9.27 is inside the parentheses in Equation 9.28 . Looking at this calculation, notice that (inside the parentheses) we are calculating the product of each particle’s mass with its position, adding all N of these up, and dividing this sum by the total mass of particles we summed. This is reminiscent of an average; inspired by this, we’ll (loosely) interpret it to be the weighted average position of the mass of the extended object. It’s actually called the center of mass of the object. Notice that the position of the center of mass has units of meters; that suggests a definition: So, the point that obeys Equation 9.26 (and therefore Equation 9.27 as well) is the center of mass of the object, which is located at the position vector r → CM r → CM . It may surprise you to learn that there does not have to be any actual mass at the center of mass of an object. For example, a hollow steel sphere with a vacuum inside it is spherically symmetrical (meaning its mass is uniformly distributed about the center of the sphere); all of the sphere’s mass is out on its surface, with no mass inside. But it can be shown that the center of mass of the sphere is at its geometric center, which seems reasonable. Thus, there is no mass at the position of the center of mass of the sphere. (Another example is a doughnut.) The procedure to find the center of mass is illustrated in Figure 9.27 . Since r → j = x j i ^ + y j j ^ + z j k ^ r → j = x j i ^ + y j j ^ + z j k ^ , it follows that: Therefore, you can calculate the components of the center of mass vector individually. Finally, to complete the kinematics, the instantaneous velocity of the center of mass is calculated exactly as you might suspect: and this, like the position, has x -, y -, and z -components. To calculate the center of mass in actual situations, we recommend the following procedure: Problem-Solving StrategyCalculating the center of mass. The center of mass of an object is a position vector. Thus, to calculate it, do these steps:
Here are two examples that will give you a feel for what the center of mass is. Example 9.16Center of mass of the earth-moon system. From Appendix D , We defined the center of Earth as the origin, so r e = 0 m r e = 0 m . Inserting these into the equation for R gives SignificanceCheck your understanding 9.11. Suppose we included the sun in the system. Approximately where would the center of mass of the Earth-moon-sun system be located? (Feel free to actually calculate it.) Example 9.17Center of mass of a salt crystal. There are eight ions in this crystal, so N = 8: The mass of each of the chloride ions is For the sodium ions, The total mass of the unit cell is therefore From the geometry, the locations are Substituting: Similar calculations give r CM, y = r CM, z = 1.18 × 10 −10 m r CM, y = r CM, z = 1.18 × 10 −10 m (you could argue that this must be true, by symmetry, but it’s a good idea to check). Check Your Understanding 9.12Suppose you have a macroscopic salt crystal (that is, a crystal that is large enough to be visible with your unaided eye). It is made up of a huge number of unit cells. Is the center of mass of this crystal necessarily at the geometric center of the crystal? Two crucial concepts come out of these examples:
Center of Mass of Continuous ObjectsIf the object in question has its mass distributed uniformly in space, rather than as a collection of discrete particles, then m j → d m m j → d m , and the summation becomes an integral: In this context, r is a characteristic dimension of the object (the radius of a sphere, the length of a long rod). To generate an integrand that can actually be calculated, you need to express the differential mass element dm as a function of the mass density of the continuous object, and the dimension r . An example will clarify this. Example 9.18Cm of a uniform thin hoop. We replace dm with an expression involving the density of the hoop and the radius of the hoop. We then have an expression we can actually integrate. Since the hoop is described as “thin,” we treat it as a one-dimensional object, neglecting the thickness of the hoop. Therefore, its density is expressed as the number of kilograms of material per meter. Such a density is called a linear mass density , and is given the symbol λ λ ; this is the Greek letter “lambda,” which is the equivalent of the English letter “l” (for “linear”). Since the hoop is described as uniform, this means that the linear mass density λ λ is constant. Thus, to get our expression for the differential mass element dm , we multiply λ λ by a differential length of the hoop, substitute, and integrate (with appropriate limits for the definite integral). The center of mass is calculated with Equation 9.34 : We have to determine the limits of integration a and b . Expressing r → r → in component form gives us In the diagram, we highlighted a piece of the hoop that is of differential length ds ; it therefore has a differential mass d m = λ d s d m = λ d s . Substituting: However, the arc length ds subtends a differential angle d θ d θ , so we have One more step: Since λ λ is the linear mass density, it is computed by dividing the total mass by the length of the hoop: Notice that the variable of integration is now the angle θ θ . This tells us that the limits of integration (around the circular hoop) are θ = 0 to θ = 2 π θ = 0 to θ = 2 π , so a = 0 a = 0 and b = 2 π b = 2 π . Also, for convenience, we separate the integral into the x - and y -components of r → CM r → CM . The final integral expression is as expected. Center of Mass and Conservation of MomentumHow does all this connect to conservation of momentum? Suppose you have N objects with masses m 1 , m 2 , m 3 , ... m N m 1 , m 2 , m 3 , ... m N and initial velocities v → 1 , v → 2 , v → 3 , ... , v → N v → 1 , v → 2 , v → 3 , ... , v → N . The center of mass of the objects is Its velocity is and thus the initial momentum of the center of mass is After these masses move and interact with each other, the momentum of the center of mass is But conservation of momentum tells us that the right-hand side of both equations must be equal, which says This result implies that conservation of momentum is expressed in terms of the center of mass of the system. Notice that as an object moves through space with no net external force acting on it, an individual particle of the object may accelerate in various directions, with various magnitudes, depending on the net internal force acting on that object at any time. (Remember, it is only the vector sum of all the internal forces that vanishes, not the internal force on a single particle.) Thus, such a particle’s momentum will not be constant—but the momentum of the entire extended object will be, in accord with Equation 9.36 . Equation 9.36 implies another important result: Since M represents the mass of the entire system of particles, it is necessarily constant. (If it isn’t, we don’t have a closed system, so we can’t expect the system’s momentum to be conserved.) As a result, Equation 9.36 implies that, for a closed system, That is to say, in the absence of an external force , the velocity of the center of mass never changes . You might be tempted to shrug and say, “Well yes, that’s just Newton’s first law,” but remember that Newton’s first law discusses the constant velocity of a particle, whereas Equation 9.37 applies to the center of mass of a (possibly vast) collection of interacting particles, and that there may not be any particle at the center of mass at all! So, this really is a remarkable result. Example 9.19Fireworks display. The picture shows radial symmetry about the central points of the explosions; this suggests the idea of center of mass. We can also see the parabolic motion of the glowing particles; this brings to mind projectile motion ideas. At the instant of the explosion, the thousands of glowing fragments fly outward in a radially symmetrical pattern. The symmetry of the explosion is the result of all the internal forces summing to zero ( ∑ j f → j int = 0 ) ; ( ∑ j f → j int = 0 ) ; for every internal force, there is another that is equal in magnitude and opposite in direction. However, as we learned above, these internal forces cannot change the momentum of the center of mass of the (now exploded) shell. Since the rocket force has now vanished, the center of mass of the shell is now a projectile (the only force on it is gravity), so its trajectory does become parabolic. The two red explosions on the left show the path of their centers of mass at a slightly longer time after explosion compared to the yellow explosion on the upper right. In fact, if you look carefully at all three explosions, you can see that the glowing trails are not truly radially symmetric; rather, they are somewhat denser on one side than the other. Specifically, the yellow explosion and the lower middle explosion are slightly denser on their right sides, and the upper-left explosion is denser on its left side. This is because of the momentum of their centers of mass; the differing trail densities are due to the momentum each piece of the shell had at the moment of its explosion. The fragment for the explosion on the upper left of the picture had a momentum that pointed upward and to the left; the middle fragment’s momentum pointed upward and slightly to the right; and the right-side explosion clearly upward and to the right (as evidenced by the white rocket exhaust trail visible below the yellow explosion). Finally, each fragment is a projectile on its own, thus tracing out thousands of glowing parabolas. Check Your Understanding 9.13How would the firework display change in deep space, far away from any source of gravity? You may sometimes hear someone describe an explosion by saying something like, “the fragments of the exploded object always move in a way that makes sure that the center of mass continues to move on its original trajectory.” This makes it sound as if the process is somewhat magical: how can it be that, in every explosion, it always works out that the fragments move in just the right way so that the center of mass’ motion is unchanged? Phrased this way, it would be hard to believe no explosion ever does anything differently. The explanation of this apparently astonishing coincidence is: We defined the center of mass precisely so this is exactly what we would get. Recall that first we defined the momentum of the system: We then concluded that the net external force on the system (if any) changed this momentum: and then—and here’s the point—we defined an acceleration that would obey Newton’s second law. That is, we demanded that we should be able to write which requires that where the quantity inside the parentheses is the center of mass of our system. So, it’s not astonishing that the center of mass obeys Newton’s second law; we defined it so that it would. This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
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Center of MassLaunch interactive. August 11, 2016 "X" Marks the Spot: Finding the Center of MassA centering science activity By Ben Finio & Science Buddies  Can you find the center of a shape? You'll be able to--even for the oddest oblong creation--with this simple science activity. No strings attached (okay, maybe one)! George Retseck Key concepts Physics Geometry Gravity Center of mass Introduction With a little time, you can probably find the center of simple shapes such as circles and squares pretty easily. But how do you find the “middle” of an irregular shape such as a drawing of a dog or a cat? This project will show you how to do it using nothing but string and paper clips! Background How do you define the exact center of an object? One way to do this is to find the object's center of mass. The center of mass is the point about which an object will balance if you try to rest it on your fingertip. Or if you hang an object, for example a picture frame from a nail, the center of mass will hang directly below the nail. On supporting science journalismIf you're enjoying this article, consider supporting our award-winning journalism by subscribing . By purchasing a subscription you are helping to ensure the future of impactful stories about the discoveries and ideas shaping our world today. For symmetrical objects, finding the center of mass is relatively easy. For example, for a rectangular picture frame, you know the center of mass is in the middle of the rectangle and you can find that with a ruler. When you hang the picture frame, you will make sure it is centered on the nail—otherwise it will tip to one side and will be off-center. The same applies to other symmetrical objects such as a spherical basketball; you know the center of mass is in the middle of the sphere. What about irregularly shaped objects such as a dog or cat or person? Now finding the center of mass is not so easy! This activity will show you how to find the center of mass for any two-dimensional shape you cut out of paper using a trick that has to do with the hanging picture frame mentioned above. If you hang a shape from a single point, you know the center of mass will always rest directly below that point. So, if you hang a shape from two different points (one at a time) and draw a line straight down from each point, the center of mass is where those lines intersect. This technique can be used for any irregular two-dimensional shape. Don't believe it? Try this activity to find out! Paper (Heavier paper, such as construction paper, card stock or thin cardboard from the side of a cereal box will work best.) Scissors (Have an adult help with cutting if necessary—especially on thicker materials.) Two paper clips or a pushpin and another small, relatively heavy object you can tie to the string (such as a metal washer) Preparation Cut a piece of string about one foot long and tie a paper clip to each end. (Alternatively, you can use any other small object such as a metal washer on one end—this will serve as a weight—and any other small, pointy object like a needle or pushpin on the other end—this will be used to puncture the paper.) Start with an easy shape: Cut out a rectangular piece of paper or cardboard. Can you guess where the center of mass of the rectangle is? If so, use a ruler to measure where you think it will be and mark this spot with your pencil. Punch several small holes around the edge of the paper. Make them as close to the edge as possible without ripping the paper. (This is important for the accuracy of this technique). The exact location of the holes does not matter but this technique will work best if you space them all the way around the edge (not just put two holes right next to each other). Now poke one end of one paper clip (or pushpin) through one of the holes to act like a hanging hook. Make sure the paper can swing easily from the hook and does not get stuck (Rotate it back and forth a few times to loosen the hole if necessary). Hold on to your “hook” and hold the paper up against the wall. Let the paper swing freely and make sure the string can hang straight down and does not get stuck. Use a pencil and ruler to draw a straight line on the paper along the string. Does this line go through the center of mass you predicted earlier? Now, hang your paper from a different hole and repeat the process. Where does this line intersect the first line? Repeat the process several more times with different holes. Do all the lines intersect at the same point? Now cut out an irregular shape. You can cut out a “blob” or draw something like a dog or cat and then cut out the outline. Make sure the shape you cut out remains stiff and flat. (That is, do not cut very thin sections that might be floppy.) Can you use a ruler to predict where the center of mass of your irregular shape will be? This is much harder! Punch holes around the edge of your irregular shape and repeat the activity. One at a time, hang the shape and the string from one of the holes and draw a line along the string. Where do the lines intersect? Does this match up with what you predicted? Extra: If you use a stiff enough material to cut out your shape (such as cardboard), can you try balancing it on your fingertip at the center of mass? What happens if you try to balance it about another point? Observations and results You should have found that the center of mass of the rectangle is right in the middle of the piece—halfway along the width and halfway along the height. You can easily locate this spot with a ruler. Then, when you hang the rectangle from a hole on its edge, the string should always pass through this point, regardless of which hole you use. Whereas it is much harder to predict the center of mass for an irregular shape, the same principle holds true. Regardless of what point you hang the irregular shape from, the string will always pass through the center of mass. So, if you hang it from two or more points (one at a time), you can find the intersection of these lines—and that is the center of mass. Note that due to small variables in the activity (such as friction on the hook that prevents the paper from rotating perfectly or the holes not being close enough to the edge of the paper), if you draw multiple lines, they might not all intersect in exactly the same place but they should still be fairly close to one another. More to explore Fun with Gravity and Center of Mass ( pdf ), from Gravity Recovery and Climate Experiment Circus Science: How to Balance Anything , from Scientific American Swinging with a Pendulum , from Scientific American Science Activities for All Ages! from Science Buddies This activity brought to you in partnership with Science Buddies  VanCleave's Science Fun Your Guide to Science Projects, Fun Experiments, and Science Research Center of Gravity: Balancing Soda CanBy Janice VanCleave  The center of gravity (center of mass) is the point on an object that when supported on that spot, the object will balance. A line through the center of gravity of an object is always vertical. Note that the leaning soda can in the diagram is balanced where the center of gravity of the can touches the table. The volume of liquid in the can is such that a line through the center of gravity (center of mass) of the can passes though the bottom of the can as shown. A tilted object will not fall over as long as the line through its center of gravity does not fall past the base of the object. In the following video, a balancing soda can is used as a game. The object being to balance three soda cans on their bottom edge. I suggest that you change the rules and use soda cans filled with water. Then instead of drinking the liquid, kids can pour out or add water to the cans as needed. The second video shows a more engaging Challenge. Shop Experiment Center of Mass Motions ExperimentsCenter of mass motions. Experiment #15 from Physics with Video Analysis IntroductionWhen analyzing collisions between several objects, physicists define a point in space known as the center of mass of the system of objects in order to simplify the analysis of motions before and after collisions. In particular, the center of mass of a colliding system that experiences no net outside force is defined as a point that moves at a constant velocity before, during and after the collisions. In this activity, you will
Sensors and EquipmentThis experiment features the following sensors and equipment. Additional equipment may be required.  Ready to Experiment?Ask an expert. Get answers to your questions about how to teach this experiment with our support team.
Purchase the Lab BookThis experiment is #15 of Physics with Video Analysis . The experiment in the book includes student instructions as well as instructor information for set up, helpful hints, and sample graphs and data.  Stack Exchange NetworkStack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Experiment to determine the center of massConsider an axially symmetric rigid body with asymmetric mass distribution. Generally, the center of mass (and therefore moments of inertia) can be determined by knowing the center of mass of the consituent components of the of the body. However this data may not always be available. Is there an experiment which can determine the center of mass of an axially symmetric rigid body with asymmetric mass distribution in the absence apriori knowledge of the component masses and their individual centers of mass? Please provide any new ideas or insights on extending the ideas towards center of mass determination to determination of (at least some) moments of inertia.
If you hang the object from a point, the center of mass will hang directly below the point. This gives you a line that contains the center of mass. If you do that using two or three different points, you have 2 or 3 lines that all pass through the center of mass. Find where the lines intersect. You can also hang the object from two points. The center of mass will hang directly below the line connecting the points. That gives you a plane containing the center of mass. If the body is difficult to hang or balance, put it on a table with three or more legs and place a scale under each leg. From the distance and weights, you can infer the point under the center of mass. Rotate the object and repeat. You can find two or three planes containing the center of mass. If the object is easy to slide, you can avoid calculation. Slide it until the weight under each leg is the same. 
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 Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 2.6 Measures of CenterLet’s keep working through the acronym SOCS for describing key aspects of our data, this time focusing on the center. The “center” is a way of describing the “central tendency” or “typical value” of a dataset. The two most widely used measures of the center of the data are the mean (average) and the median . Most people are familiar with the ideas of these two: (1) to calculate the mean weight of 50 people, add the 50 weights together and divide by 50, and (2) to find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. However, some datasets may be better summarized by one or the other. The most “appropriate” measure of center depends on the shape of the distribution and the presence of extreme values or potential outliers. The mean is the most common measure of the center. The words “mean” and “average” are often used interchangeably. The technical term is “arithmetic mean,” and “average” technically refers to a center location. However, in practice among non-statisticians, “average” is commonly accepted for “arithmetic mean.”  The Greek letter μ (pronounced “mew”) represents the population mean . We will often use the sample mean to estimate the population mean. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample to be taken truly at random. Calculate the mean of the sample: 1, 1, 1, 2, 2, 3, 4, 4, 4, 4, 4. Calculate the mean of the sample: 7, 10, 14, 14, 15, 21, 38, 38, 38, 56. The median is generally a better measure of the center when there are extreme values or outliers because it is more robust , or not affected by the precise numerical values of those outliers. Especially for larger datasets, you may choose to use the following location function over the traditional counting method to find the median:  Remember that this function simply tells you where to look for the median, not the actual value itself, and n is the total number of data values in the sample (sample size).  AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest): 3, 4, 8, 8, 10, 11, 12, 13, 14, 15, 15, 16, 16, 17, 17, 18, 21, 22, 22, 24, 24, 25, 26, 26, 27, 27, 29, 29, 31, 32, 33, 33, 34, 34, 35, 37, 40, 44, 44, 47. Calculate the median. Calculate the median of the sample: 7, 10, 14, 14, 15, 21, 38, 38, 38, 56. Another measure of the center is the mode . The mode is the most frequent value. There can be more than one mode in a dataset as long as those values have the same frequency and that frequency is the highest. A dataset with two modes is called bimodal. For example, if five real estate exam scores are 430, 430, 480, 480, 495, then the dataset is bimodal because the scores 430 and 480 each occur twice. When is the mode the best measure of the “center”? Consider a weight loss program that advertises a mean weight loss of six pounds during the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing. The mode can be calculated for categorical data as well as quantitative data but has different uses and interpretations for each. For example:
Statistics exam scores for 20 students are as follows: 50, 53, 59, 59, 63, 63, 72, 72, 72, 72, 72, 76, 78, 81, 83, 84, 84, 84, 90, 93 Find the mode. Order Relationship of Measures of CenterConsider the following dataset: 4, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 9, 10. This dataset can be represented by the histogram in Figure 2.44. Each interval has a width of one, and each value is located in the middle of an interval.  The histogram displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shapes to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each 7 for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median. The histogram for the data {4, 5, 6, 6, 6, 7, 7, 7, 7, 8} is not symmetrical. The right-hand side seems “chopped off” compared to the left side. A distribution of this type is said to be skewed to the left because it is pulled out to the left.  The mean is 6.3, the median is 6.5, and the mode is 7. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so. The histogram for the data {6, 7, 7, 7, 7, 8, 8, 8, 9, 10} is also not symmetrical. It is skewed to the right.  The mean is 7.7, the median is 7.5, and the mode is 7. Of the three statistics, the mean is the largest, while the mode is the smallest. Again, the mean reflects the skewing the most. To summarize, if the distribution of data is skewed to the left, the mean is often less than the median, which is often less than the mode. If the distribution of data is skewed to the right, the mode is often less than the median, which is less than the mean. Skewness and symmetry become important when we discuss probability distributions in later chapters. Statistics are used to compare and sometimes identify authors. The following lists shows a simple random sample that compares the letter counts for three authors. Darnell: 7, 9, 3, 3, 3, 4, 1, 3, 2, 2 Mary: 3, 3, 3, 4, 1, 4, 3, 2, 3, 1 Lee: 2, 3, 4, 4, 4, 6, 6, 6, 8, 3 a. Make a dot plot for the three authors and compare the shapes. Darnell’s distribution has a right (positive) skew:  Mary’s distribution has a left (negative) skew:  Lee’s distribution is symmetrically shaped:  b. Calculate the mean for each. c. Calculate the median for each. d. Describe any pattern you notice between the shape and the measures of center. Suppose that in a small town of 50 people, one person earns $5,000,000 per year, and the other 49 each earn $30,000. Which is the better measure of the “center,” the mean or the median? Calculating the Mean of Grouped Frequency TablesWhen only grouped data is available, we do not know the individual data values (only the intervals and interval frequencies); therefore, we cannot compute an exact mean for the dataset. What we must do is estimate the actual mean by calculating the mean of a frequency table (a data representation in which grouped data is displayed along with the corresponding frequencies). To calculate the mean from a grouped frequency table, we can apply the basic definition of mean:  We simply need to modify the definition to fit within the restrictions of a frequency table. Since we do not know the individual data values, we can instead find the midpoint of each interval. The midpoint is:  We can now modify the mean definition to be:  where f = the frequency of the interval and m = the midpoint of the interval. A frequency table displaying professor Blount’s last statistic test is shown.
Find the midpoints for all intervals. Calculate the sum of the product of each interval frequency. Calculate the midpoint. Maris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data:
What is the best estimate for the mean number of hours spent playing video games? Click here for more multimedia resources, including podcasts, videos, lecture notes, and worked examples. Figure References Figure 2.45: Kindred Grey (2020). Symmetrical distribution. CC BY-SA 4.0. Figure 2.46: Kindred Grey (2020). Skewed left. CC BY-SA 4.0. Figure 2.47: Kindred Grey (2020). Skewed right. CC BY-SA 4.0. Figure 2.48: Kindred Grey (2020). Darnell’s letter count. CC BY-SA 4.0. Adaptation of Figure 2.21 from OpenStax Introductory Statistics (2013) (CC BY 4.0). Retrieved from https://openstax.org/books/statistics/pages/2-6-skewness-and-the-mean-median-and-mode Figure 2.49: Kindred Grey (2020). Mary’s letter count. CC BY-SA 4.0. Adaptation of Figure 2.22 from OpenStax Introductory Statistics (2013) (CC BY 4.0). Retrieved from https://openstax.org/books/statistics/pages/2-6-skewness-and-the-mean-median-and-mode Figure 2.50: Kindred Grey (2020). Lee’s letter count. CC BY-SA 4.0. Adaptation of Figure 2.23 from OpenStax Introductory Statistics (2013) (CC BY 4.0). Retrieved from https://openstax.org/books/statistics/pages/2-6-skewness-and-the-mean-median-and-mode A number that measures the central tendency of the data The middle number in a sorted list The arithmetic mean, or average, of a dataset The arithmetic mean, or average, of a population Not affected by violations of assumptions such as outliers The most frequently occurring value How many peaks or clusters there appear to be in a quantitative distribution Significant Statistics Copyright © 2024 by John Morgan Russell, OpenStaxCollege, OpenIntro is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted. Share This BookAelita Otdykh Na Vode
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Observations and Results. You should have found that the center of mass of the rectangle is right in the middle of the page - halfway along the width, and halfway along the height. You can easily locate this spot with a ruler. Then, when you hang the rectangle from a hole on its edge, the string should always pass through this point ...
Center of Mass Challenge 2: The Chair Lift. Place a chair sideways along a wall. Stand next to the chair. Make sure your feet are beside the chair, not under the chair. Bend over the chair, creating a 90-degree angle with your body. Place your head against the wall. Now try to lift the chair while keeping your head pressed against the wall.
This week's Science Sunday episode is geared toward Junior High and High School students with a fascinating lesson in Newton's first law of Motion. Dr. Timot...
Once an object is balanced, wherever you are holding up that object is where it's COM is. You could probably guess where the center of mass is for a lot of symmetric objects. For example, a ruler's COM is in the middle. But if you add some extra weight to the end, it's COM will shift! A fork is pretty asymmetic, and that shows in it's COM.
During our center of mass experiment, we wrote down how long it took for each spinner to stop. We did this by watching the time-lapse on the video, and noting when the fidget spinner started and stopped spinning. As you can see, when the fidget spinner had three inserts in, it spun for over a minute. With two inserts in the spinner, it only ...
Example 1: Find the coordinates of the center of mass of the following system of particles: particle of mass 0.1 kg located at (1, 2), particle of mass 0.05 kg located at (2, 4) and particle of mass 0.075 kg located at (2, 1). Solution 1: Apply the formula for the x -coordinate of the center of mass as follows:
Find the center of mass of a uniform thin hoop (or ring) of mass M and radius r. Strategy First, the hoop's symmetry suggests the center of mass should be at its geometric center. If we define our coordinate system such that the origin is located at the center of the hoop, the integral should evaluate to zero.
The Center of Mass Interactive allows a learner to explore the effect of object shape upon the center of mass. Draw out a shape on the workspace and watch the center of mass change its location as you change the shape. Then pin the shape on the corkboard and see how it hangs. Can you find the pattern between the pivot point from which is hung and the location of the center of mass?
Observations and results. You should have found that the center of mass of the rectangle is right in the middle of the piece—halfway along the width and halfway along the height. You can easily ...
NCSSM Online Physics Collection: This video deals with locating the center of mass of an object and it's effect on stability. http://www.dlt.ncssm.eduPlease ...
In this experiment, you will. Use video analysis techniques to obtain position, velocity, and time data for two carts undergoing a variety of collisions. Analyze the position-time graphs for the individual carts and compare these to the position-time graph for the center of mass of the system. Compare the momentum of the system before and after ...
The center of gravity (center of mass) is the point on an object that when supported on that spot, the object will balance. A line through the center of gravity of an object is always vertical. Note that the leaning soda can in the diagram is balanced where the center of gravity of the can touches the table. The volume of liquid in the can is ...
Heres a fun experiment for kids involving the center of mass. All you need is a popsicle stick (or craft stick) a pipe cleaner and two binder clips. Or, if ...
Center of mass
It depends on two things: the location of the object's center of mass*, and where the object is in contact with the ground. Imagine looking down on a chair from directly above, and drawing imaginary lines connecting the chair's four legs, forming a square on the ground. The chair's center of mass will be inside this square. Gravity pulls down ...
Introduction. When analyzing collisions between several objects, physicists define a point in space known as the center of mass of the system of objects in order to simplify the analysis of motions before and after collisions. In particular, the center of mass of a colliding system that experiences no net outside force is defined as a point ...
From the distance and weights, you can infer the point under the center of mass. Rotate the object and repeat. You can find two or three planes containing the center of mass. If the object is easy to slide, you can avoid calculation. Slide it until the weight under each leg is the same.
Definition of Centre of Mass. The Centre of mass of a body is defined as a single point at which the whole mass of the body or system is imagined to be concentrated and all external forces are applied there. It is the point where if a force is applied it moves in the direction of the force without rotating. x, y, & z coordinates of the centre ...
The two most widely used measures of the center of the data are the mean (average) and the median. Most people are familiar with the ideas of these two: (1) to calculate the mean weight of 50 people, add the 50 weights together and divide by 50, and (2) to find the median weight of the 50 people, order the data and find the number that splits ...
The largest of all the scientific institutions of the center is the Institute of Chemical Physics in Chernogolovka ... and mass spectroscopy. ... to produce equipment for scientific investigations and tools that automatically control experiments. Founded in 1972, the factory is now one of the best producers of scientific equipment in Russia and ...
A 100 hp motor, a cabin, a bio-toilet, a music center. Maximum number of passengers 5 + captain Captain Natalia will tell you about local attractions, places of recreation on the shore, show you beautiful places. We will go for a ride on a water tubing, wakeboard (for a fee), stop for swim. ... Outdoor Activities in Boltino.
A 100 hp motor, a cabin, a bio-toilet, a music center. Maximum number of passengers 5 + captain Captain Natalia will tell you about local attractions, places of recreation on the shore, show you beautiful places. We will go for a ride on a water tubing, wakeboard (for a fee), stop for swim. Boltino, Moscow Oblast, Russia.
Aelita Otdykh Na Vode. We offer boat trips on Aelita! We have everything you need for an unforgettable vacation on the water! A 100 hp motor, a cabin, a bio-toilet, a music center. Maximum number of passengers 5 + captain Captain Natalia will tell you about local attractions, places of recreation on the shore, show you beautiful places.