For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Center of Mass

- You can treat an entire complex object as if its entire mass were contained at a single point known as the object’s center of mass.
- Center of mass is the weighted average of the location of mass in an object.
- For uniform density objects, the center of mass is the geometric center.
- For objects with multiple parts, find the center of mass of each part and treat those as point particles.
- The center of gravity is the location at which the force of gravity acts upon an object as if it were a point particle with all its mass focused at that point.
- In a uniform gravitational field, center of gravity and center of mass are the same. In a non-uniform gravitational field, they may be different.

### Center of Mass

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Center of Mass
- Finding Center of Mass by Inspection
- Example I: Center of Mass by Inspection
- Calculating Center of Mass for Systems of Particles
- Example II: Center of Mass (1D)
- Example III: Center of Mass of Continuous System
- Example IV: Center of Mass (2D)
- Finding Center of Mass by Integration
- Example V: Center of Mass of a Uniform Rod
- Example VI: Center of Mass of a Non-Uniform Rod
- Center of Mass Relationships
- Center of Gravity
- Example VII: AP-C 2004 FR1

- Intro 0:00
- Objectives 0:07
- Center of Mass 0:45
- Center of Mass
- Finding Center of Mass by Inspection 1:25
- For Uniform Density Objects
- For Objects with Multiple Parts
- For Irregular Objects
- Example I: Center of Mass by Inspection 2:06
- Calculating Center of Mass for Systems of Particles 2:25
- Calculating Center of Mass for Systems of Particles
- Example II: Center of Mass (1D) 3:15
- Example III: Center of Mass of Continuous System 4:29
- Example IV: Center of Mass (2D) 6:00
- Finding Center of Mass by Integration 7:38
- Finding Center of Mass by Integration
- Example V: Center of Mass of a Uniform Rod 8:10
- Example VI: Center of Mass of a Non-Uniform Rod 11:40
- Center of Mass Relationships 14:44
- Center of Mass Relationships
- Center of Gravity 17:36
- Center of Gravity
- Uniform Gravitational Field vs. Non-uniform Gravitational Field
- Example VII: AP-C 2004 FR1 18:26
- Example VII: Part A
- Example VII: Part B
- Example VII: Part C
- Example VII: Part D
- Example VII: Part E

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: Center of Mass

*Hello, everyone, and welcome back to www.educator.com.*0000

*I’m Dan Fullerton and in this lesson we are going to talk about center of mass.*0004

*Our objectives include identifying by inspection the center of mass of a symmetrical object.*0008

*Locating the center of mass of a system consisting of two such objects.*0014

*Using integration to find the center of mass of the thin rod on non uniform density.*0018

*Applying the relation between center of mass velocity in a linear momentum*0025

*and between center of mass acceleration and net external force for system particles.*0029

*Defining center of gravity and using this concept to express the gravitational potential energy of a rigid object*0034

*in terms of the position of its center of mass.*0040

*Let us start by talking about what center of mass is.*0045

*Real objects are more complex than these theoretical particles we been dealing with.*0049

*It is never just the entire amount of mass at some tiny point and you can treat it that way.*0053

*Real objects are more irregular, they are more complicated than that.*0058

*However, what is really nice, is from a physics perspective, we can treat the entire object as*0062

*if its entire mass are concentrated in a single point that we are going to call the object center of mass.*0067

*Usually abbreviated CM or C o center of mass.*0073

*Mathematically speaking, center of mass is the weighted average of the location of mass and object.*0080

*How will we find center of mass?*0086

*Well, with some objects we can do it by inspection.*0088

*For uniform density of objects, the center of mass is going to be the geometric center of that object.*0091

*For object of multiple parts, you can find the center of mass of each part and treat it as a point and then look at their geometric center.*0097

*For the irregular objects, one way you can find it experimentally is to suspend the object from two or more points and draw a plumb line.*0105

*The lines are always going to intersect the center of mass.*0112

*If you attached it by a couple points, drop a plumb line from it, wherever they cross a right there that would be your center of mass of the object.*0115

*We can also figure this out by inspection especially if we have a highly symmetric object.*0125

*If we have something like this with the uniform density even though it is got a complex object it is pretty easy to see*0130

*that the center of mass is going to be right in the geometric center of our object.*0136

*That was pretty straightforward.*0141

*If we have a system of particles, however, we need to get a little bit more detail.*0144

*The position vector to the center of mass is the sum of all the little objects there mass × position ÷ total mass of your system.*0149

*Or if you want to look at coordinates, an x center of mass coordinates that would be your first mass × its x position + your second mass × x position and so on.*0159

*Divided by the sum of the masses.*0170

*In a similar fashion, if you wanted the center of mass for Y, that would be M1Y1 + M2Y2 and so on divided by the total mass.*0173

*Or m + m2 + 3 or total mass typically written capital M.*0187

*Let us do a quick example here.*0194

*Find the center of mass of an object model as two separate masses on the x axis.*0196

*Our first mass is2kg in the x coordinate of 2 and the second mass is 6kg here in the x coordinate of 8.*0201

*To find that, we can go to our function, that x coordinate of center of mass is M1X1 + M2X2 / M*0210

*which would be M1 2kg × its x position 2m + our second mass 6kg × its x position 8m divided by their total mass 8 kg.*0224

*4 + 48 ÷ 8 is just going to give us 6.5m.*0244

*We could treat this system as if its entire mass was one object here at about 6 1/2 with a mass of 8kg.*0250

*What if it is a continuous system?*0268

*Find the center of mass of the combination object here below.*0271

*The density of the object is uniform.*0274

*What we can do by inspection if this is a 3 kg block, we can see that its center of mass*0278

*is right in the center at what the position we are calling 00.*0283

*This object, we can find its center of mass here at 0, 3 on the number line, the center of that object.*0286

*What we can do is add these up as if the particles.*0293

*We can pretty easily see that the x center of mass is going to be right on that to x =0 line by symmetry.*0297

*All we need to figure out is the y center of mass which will be M1Y1 + M2Y2 divided by our total mass M*0304

*which will be 3kg × its y position 0 + 6kg × this objects y position 3 divided by our total mass 9 kg.*0317

*Our y position, center of mass is going to be 6 × 3 is 18 ÷ 9 are going to be 2.*0333

*The center of mass of our system would be at 0, 2 or right about there.*0344

*Let us do a two dimensional problem, find the coordinates of the center of mass for the system shown below,*0358

*where we got 3 kg mass at 1, 2, a 4kg mass 5, 3, and a 1kg mass at 7, 1.*0364

*We can do the x first, the x coordinate for the center of mass is going to be M1X1 3 × 1 + M2X2 4 × 5 + M3X3 1 × 7 divided by our total mass 3 + 4 + 1 or 8.*0374

*I come up with the value of 3.75.*0399

*Our y coordinate for the center of mass, we find the same way.*0405

*We will have 3 × 2 + 4 × 3 + 1 × 1 divided by their total mass 8.*0409

*6 + 12 =18 + 1= 19/8 or 2.38.*0421

*Our total, our center of mass of the system of object is going to be at 3.75 up to .38 is going to be right around there.*0428

*We can treat the entire system as if we had one particle with the mass of 8kg here at 3.75, 2. 38.*0440

*For more complex objects, you can find the center of mass by summing up all the little pieces a position vectors multiplied by that differential,*0459

*that little tiny piece of mass and then dividing by the total mass.*0466

*Just doing the calculus version, the integration version, is to make all of those mass is smaller and smaller.*0470

*The position vector to the center of mass is equal to the integral of the position vector × all those little tiny small masses divided by the total mass.*0477

*Let us put that in the play so you can see how that works.*0488

*We will start by finding the center of mass of a uniform rod of length LMSM.*0491

*The first thing I'm going to do is I'm going to do a couple definitions here.*0496

*I'm going to say the linear of mass density which we are going to call λ.*0500

*The mass in the length λ is M / L.*0504

*When I go and break this up into little pieces and when you break it up into little pieces along the X axis Dx and V.*0510

*Linear mass density there is just going to be that little bit of mass DM / dx, the mass in length again.*0519

*Which implies that just a little tiny piece of mass, DM is going to be our linear mass density × DX.*0528

*I can go to my formula for the position vector to the center of mass.*0539

*Our center of mass = the integral of our DM / our total mass which implies then that the vector to the center of mass is equal to the integral.*0547

*Our r in this case is just going to be our x coordinate.*0562

*X DM / our total mass but what is this the DM?*0567

*For that, we have to go up to our definition here, the little bit of mass contain a bit in that piece Dx is λ.*0572

*The linear mass density × DX.*0579

*That is going to be equal to the integral of x × differential of mass λ DX all divided by M.*0583

*We can pull out our constants because it is a uniform rod, λ is constant.*0597

*Our total mass already is constant.*0601

*We would have the λ divided by M integral of XDX.*0604

*We are going to integrate from x = 0 to X = some final value L the length of the rod.*0611

*This implies then that the vector r CM = we have a λ / M integral of X is x² /2 evaluated from 0 to L which is going to be λ/M L² /2.*0619

*We also know now that if λ = M/L well M = L λ.*0646

*We can replace M with L λ then we have λ L² /M which is L λ 2.*0657

*Our linear mass density cancels out and we end up with just L /2.*0667

*It is right in the middle.*0678

*That is common sense right.*0678

*If it is a uniform mass density rod the center of mass is going to be right in the center.*0681

*We proved that using calculus, going step by step as an exercise to see if we could do it, to show how you would go doing integration like this.*0686

*How you would determine the center of mass which is going to be useful when we do our next example problem.*0694

*Let us find the center of mass, now the non uniform rod.*0701

*This non uniform rod has length L and mass M, and its density is given by λ = KX.*0705

*You get the larger and larger values as x gets more and more dense.*0711

*Right away, just common sense tells me I would expect that we are going to have the center of mass somewhere to the right of center of this object.*0716

*We will use that as a check when we are all done.*0724

*To find the total mass that is going to be helpful here.*0726

*The total mass is going to be the integral from X = 0 to L of our linear mass density DX which is going to be our integral from 0 to L.*0729

*λ is KXDX, this is going to be KL² /2.*0746

*Let us go to our function again.*0756

*The position vector to the center of mass is the integral of r DM divided by our total mass.*0758

*We know that our R is just x coordinate.*0768

*We are dealing with one dimension.*0773

*Now, λ = KX which means that our differential of mass is going to be λ DX but now that is KX DX.*0776

*Our position vector to the center of mass = the integral of X × DM which is KXDX all divided by our total mass M.*0787

*Or position vector to the center of mass is we can pull K and M out of there, they are constants K /M integral for x=0 to x=L of we have got an x² in here now DX.*0806

*That is going to be K/M integral of x² is x³/3 evaluated from 0 to L.*0821

*We are going to have KL² / 3M.*0829

*But remember, we said M was KL² / 2*0836

*this implies then that our position vector to the center of mass is going to be, we got KL² / 3 and our M KL² / 2.*0844

*A little bit of simplification here, L² we got an L there, we have got a K and I end up with 2/3 L.*0860

*It is somewhere over in this area, just like we predicted, a little bit to the right of center.*0875

*Let us take a look at some relationships here, if the position vector to the center of mass is 1/the total mass × the sum*0883

*over all the little tiny pieces of the mass of those pieces × their position vector.*0894

*That implies that the velocity of the center of mass is 1/ that mass × the sum over all i for all those little pieces of Mi × there individual little velocities.*0903

*We also know that momentum is mass × velocity so we can write this as the velocity of the center of mass is 1/ M sum*0920

*overall I, all of those individual little momentum of each of those pieces.*0933

*Which implies then that the total momentum which is what this is the total momentum = the total mass × the velocity of the center of mass.*0941

*The total momentum is mass × the velocity of the center of mass.*0954

*You can find the total momentum by just using that center of mass piece as well.*0957

*Or total momentum, if that is equal to M × velocity of the center of mass*0963

*and we know from previous work that force is that derivative of momentum with respect to time.*0971

*Let us take the derivative of our both sides here.*0982

*For the left hand side, we have a derivative of the total momentum with respect to T.*0986

*Which we know is must be our total force and that must be equal to the derivative of the right hand side.*0993

*The derivative with respect to T of mass × the velocity of the center of mass.*1004

*Which implies then that our total force must be equal to, the derivative of mass × velocity*1011

*mass is a constant the derivative of the velocity of the center of mass must be the acceleration of the center of mass.*1020

*Newton’s second law, the total force is equal to our total mass × the acceleration of the center of mass.*1026

*With Newton’s second law again, we do not have to worry about individual point particles or irregularly shaped object.*1033

*We treat the whole thing as if it had its entire mass right at that point the center of mass and we are allowed to do that.*1046

*Simplifies this up tremendously.*1053

*Another term you might have heard is called center of gravity.*1056

*Center of gravity refers to the location at which the force of gravity acts upon the object is if it were point particle with all of its mass at that point.*1060

*In a uniform gravitational field, the center of gravity and the center of mass are the same.*1068

*But if you happen to be the non uniform gravitational field they could be different.*1074

*If you have a humongous object that is so big, that parts that are in different gravitational field strengths where you have to bring that into account.*1078

*The center of mass and the center of gravity can be different.*1088

*For the most part, when we are in uniform gravitational fields where*1091

*we make the estimation like here on the surface of earth with relatively small objects, they are the same.*1094

*Technically speaking now, they are different quantities.*1101

*Let us finish up by looking in an old free response problem.*1105

*We will take the 2004 Mechanics exam free response number 1, you can find the link up here or google it.*1108

*Take a minute, print it out, and give it a try, then come back here and we will see how you do.*1117

*Taking a look at this question.*1126

*We have got someone swinging from a rope, right when they get to the vertical position on the rope they are grabbing that rock,*1130

*that mass, and then they are flying off as a projectile.*1136

*Find the speed of a person just before the collision with the object.*1140

*I would do that by conservation of energy.*1143

*The gravitational potential energy at position A must be converted into kinetic energy at position B.*1146

*Which implies that MG L the length of the rope must equal ½ M1 V².*1154

*I suppose that is M1 as well.*1164

*Which implies then the V² = 2 GL or V is going to be equal to the √2 GL right before the collision.*1167

*It also asks us to find the tension in the rope just before the collision with the object.*1181

*Our free body diagram, we have tension up and we have M1 G down.*1186

*We know that the net force, this case we will talk about a centripetal force, must be T - M1 G toward the center of the circle the positive direction.*1192

*And that has to be equal to mass × centripetal acceleration.*1204

*In this case, that will be M1 V² /L.*1208

*Therefore, solving for tension T will be M1 V² /L + M1 G.*1215

*But we just found out that V² = 2 GL right up there.*1225

*We can write this as T = M1 2 GL ÷ L + M1 G.*1233

*We got an L so I have 2M1 G + M1 G is just going to be 3 M1 G.*1249

*Let us move on to part C, find the speed of the person and object right after the collision.*1263

*That is a conservation of momentum problem.*1271

*Our initial momentum must equal our final momentum.*1273

*You can make a momentum table here if you want to.*1277

*Which implies then the D mass 1 × initial velocity must equal the combined mass after they collide × their final velocity.*1280

*Or V final is going to be equal to M1/M1 + M2 × the initial.*1293

*But V initial is √2 GL so that is just going to be M1/M1 + M2 √2 GL.*1301

*Moving on the part D, find the ratio of the kinetic energy of the person object system before the collision so the kinetic energy after the collision.*1325

*Let us start off with a kinetic energy before, that is going to be ½ M1 Vb 4² which is ½ M1 V² was 2 GL.*1338

*That is going to be M1 GL.*1353

*To find the kinetic energy after, the kinetic energy after will be 1/2 and their combined mass is M1 + M2 VF²*1359

*which is ½ F M1 + M2 × (M1 / M1) + M2 × √2 GL² our velocity we found in part C.*1370

*That implies then with a little bit of algebra, kinetic energy after is going to be ½ M1 + M2.*1393

*We will have M1² / M1 + M2² × 2 GL.*1402

*We can do little bit of simplification, M1 + M2 ÷ M1 + M2 to give us ½ M1 + M2 that is going to be equal to M1² / M1 + M2 GL.*1415

*When we take the final ratio, kinetic energy of B / kinetic energy at A, we are going to have kinetic energy at B/A.*1447

*We will have M1 GL / A.*1457

*We have got M1² GL M1 + M2 left over.*1463

*A little bit of simplification GGLL.*1472

*We got that M1 vs. M1².*1477

*I come up with M1 + M2 all divided by M 1.*1480

*We have got a part E in this question here as well.*1493

*Let us go give ourselves a little more room again.*1496

*For part E, find the total horizontal displacement x of the person from position A and tell the person object when the water at D.*1499

*That second part looks like it is a projectile motion problem, something launch horizontally.*1510

*Let us do that first.*1515

*We will find out how long they are in the air.*1517

*We will take a look at the vertical motion, we will call down the positive y direction.*1520

*V initial vertically is 0.*1525

*Our δ y is going to be L.*1528

*Our acceleration is G, the acceleration due to gravity.*1531

*δ Y = V initial T + ½ AY T².*1536

*Our V initial is 0 so that term goes away.*1543

*We could then write that L = ½ GT² or T is going to be equal to √2 L/G.*1547

*If we want the total horizontal displacement from B to D, δ x from B to D is just going to be the velocity in x × the time it is in the air.*1560

*We figured out the velocity in the x direction as they went the edge of the cliff previously, that was M1 / M1 + M2 × √2GL.*1576

*We need to multiply all of that by the time they are in the air √2L/G.*1591

*√2L √2L = 2L √G √G is a little bit of simplification here then I come up with 2 LM1 / M1 + M2.*1598

*We are not asked for just that distance, we are asked for the total distance.*1617

*The total horizontal displacement from position A.*1621

*Our total displacement is going to be L + what we just found there 2 LM1 / M1 + M2.*1625

*Or if you want to do a little bit of algebra, I think which you have there perfectly acceptable.*1642

*But you could go distribute that through a little bit and have L × get a common denominator M1 + M2 / M1 + M2 + 2 L M1 / M1 + M2,*1647

*which would be L × M1 + M2 + 2 M1 / M1 + M2.*1666

*Which is equivalent to, we would have L × 3 M1 + M2 / M1 + M2.*1678

*I'm not sure that is a whole lot prettier than what we have right there but either one of those should work.*1690

*Hopefully, that gets you a good feel for center of mass and center of gravity.*1698

*Thanks for watching www.educator.com.*1701

*We will see you again soon and make it a great day everybody.*1703

2 answers

Last reply by: Professor Dan Fullerton

Tue Jan 20, 2015 6:35 AM

Post by Caleb Lear on January 20, 2015

Curious, I thought that your calculation only covered the distance from point B to point D. You calculated the time it would take for an object to fall from B, then checked how far it could go in that time period given its velocity at B. Isn't that just the distance from B to D? The problem asked for the distance from A to D, so don't we need to add the distance from A to B?

1 answer

Last reply by: Professor Dan Fullerton

Sun Dec 14, 2014 7:10 AM

Post by Thadeus McNamara on December 13, 2014

for that ap question at the end... was there any center of mass application? i didnt use any but it was still good practice