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Lecture Comments (17)

1 answer

Last reply by: Professor Dan Fullerton
Fri Mar 11, 2016 12:41 PM

Post by David Löfqvist on March 11 at 10:56:48 AM

Doesn't the thickness of the rod play a roll? Or is that just because of air resistance?

1 answer

Last reply by: Professor Dan Fullerton
Tue Mar 1, 2016 6:17 AM

Post by Ore Okusanya on February 29 at 09:35:02 PM

hey dan,
in mechanics doing this topic moments, i find it hard to distinguish which moments is going either clockwise or anticlockwise.

1 answer

Last reply by: Professor Dan Fullerton
Mon Jan 4, 2016 6:33 AM

Post by Shehryar Khursheed on January 2 at 03:27:47 PM

Hello Mr. Fullerton,

I have a question about the moment of inertia of the cylinder. I noticed that you used an integral from 0 to R. However, I was wondering whether or not you can go the other way, and keep R constant while changing L; that is, going from the top to the bottom instead of inside out as you did. I will explain the process I went through:

1) (rho)=M/(pi*R^2*L)
2) Now, i'm going to divide the cylinder into tiny cylinders (can be considered disks), with a thickness dl that will go from the top (L=0) to the bottom of the cylinder (L=L)
3) (rho)dv= dm
4) dv= pi*R^2*dl
5) integral(r^2*dm)= inegral((R^2)(rho*pi*R^2*dl))
6) now taking the constants out:  rho*pi*R^4*integral(dl) from 0 to L
7) after solving the integral, I got I=MR^2, which is the moment of inertia of a disk, not a cylinder

My question is did I do something wrong while solving for the moment of inertia this way? Or are you not allowed to do it the way I did it, keeping R constant but changing L? If so, why?

1 answer

Last reply by: Professor Dan Fullerton
Sun Dec 20, 2015 8:50 AM

Post by Jim Tang on December 19, 2015

hey dan,

I'm confused about how you set up dm.

for the rod problem, how are you incorporating the area of the rod if you don't use pi r^2? i can't see how dx is the volume here?

also, for the cylinder, you lost when you said area but you said 2 pi r, which is circumference.

i think I'm missing the intuitive understand of dm, but i can't see how the area is incorporated here in both these scenarios.

1 answer

Last reply by: Professor Dan Fullerton
Tue May 5, 2015 7:26 PM

Post by Xinyuan Xing on May 5, 2015

In example you add an excessive /3 under λ,which may lead to confusion. Btw it's a nice video for students to understand the mathematic procedure for the derivation of physics formula, helps me a lot, too.

1 answer

Last reply by: Professor Dan Fullerton
Mon Apr 6, 2015 6:18 AM

Post by Jason Kim on April 5, 2015

On the last example in this video how is it that you do not have to account for moment of inertia of two spheres that are attached to the rod?

2 answers

Last reply by: Thadeus McNamara
Sat Jan 3, 2015 5:21 PM

Post by Thadeus McNamara on December 31, 2014

I dont understand the step that starts with dm= at 14:19

1 answer

Last reply by: Thadeus McNamara
Wed Dec 31, 2014 4:55 PM

Post by Thadeus McNamara on December 31, 2014

at around 11:25 you are finding the moment of inertia for a rod spinning about its center. i understand that your bounds of integration are from L/2 to -L/2. The first time I did it, i used bound of integration from L/2 to 0. I then multiplied that answer by 2. How come my way does not work?

Moment of Inertia

  • Inertial mass is an object’s ability to resist a linear acceleration. Moment of Inertia (or rotational inertia) is an object’s resistance to a rotational acceleration.
  • Objects that have most of their mass near their axis of rotation have smaller rotational inertias than objects with more mass farther from their axis of rotation.
  • The moment of inertia can be found by summing the product of the mass of each object in the system and the square of its distance from the axis of rotation for all the objects in the system.

Moment of Inertia

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 0:00
  • Objectives 0:07
  • Types of Inertia 0:34
    • Inertial Mass
    • Moment of Inertia
  • Kinetic Energy of a Rotating Disc 1:25
    • Kinetic Energy of a Rotating Disc
  • Calculating Moment of Inertia (I) 5:32
    • Calculating Moment of Inertia (I)
  • Moment of Inertia for Common Objects 5:49
    • Moment of Inertia for Common Objects
  • Example I: Point Masses 6:46
  • Example II: Uniform Rod 9:09
  • Example III: Solid Cylinder 13:07
  • Parallel Axis Theorem (PAT) 17:33
    • Parallel Axis Theorem (PAT)
  • Example IV: Calculating I Using the Parallel Axis Theorem 18:39
  • Example V: Hollow Sphere 20:18
  • Example VI: Long Thin Rod 20:55
  • Example VII: Ranking Moment of Inertia 21:50
  • Example VIII: Adjusting Moment of Inertia 22:39

Transcription: Moment of Inertia

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

I’m Dan Fullerton and in this lesson we are going to talk about moment of inertia.0004

Our objectives include determining by inspection which set of symmetrical objects has the greatest moment of inertia.0008

Determining by what factor an object’s moment of inertia changes if its dimensions are increased by a consistent factor.0015

Calculating the moment of inertia for various objects 0022

and stating applying the parallel axis theorem which we use to find a moment of inertia about a different point for an object.0026

Let us talk about types of inertia for a moment.0033

An inertial mass or translational inertia is an object's ability to resist a linear acceleration.0037

When we talk about things rotating, you also know that if you try and rotate something it has some resistance to being rotated as well.0044

It has a resistance to the angular acceleration, we will call that the moment of inertia or rotational inertia.0051

It is an object's resistance to rotational acceleration.0058

Objects that have most of their mass near their axis of rotation have a smaller rotational inertia, 0063

an object that have most of their mass farther from their axis of rotation.0067

Now inertial mass, we are going to give the symbol M, we have been using that.0072

For moment of inertia, we are going to call that capital I or rotational inertia.0075

The rotational analogue is linear inertia.0080

Let us start our exploration by looking at the kinetic energy of a rotating disk.0083

If we look at this disk and if it starts to spin, it is pretty obvious that it is in motion.0089

Therefore, it must have some kinetic energy but we do not know how to deal with that yet.0094

All we have talked about is the kinetic energy of objects moving translationally at different points in space.0098

Now parts of this are rotating, parts are maintaining their position.0105

The way we are going to do this is, let us start by defining the direction for angular velocity ω.0110

Let us assume that is spinning about its center point with some angular velocity and we will draw a point for a center here.0116

Let us take and let us start by finding just the kinetic energy of a little tiny piece of our uniform disk here.0125

As we do this, let us figure out that the entire disk has a radius capital R, we have a vector RI, 0135

to our little piece of mass of our disk MI, that is moving with some velocity at this point in time VI.0146

With that, I think we can start to look at the kinetic energy of just that little tiny piece of disk.0156

The kinetic energy of that little piece I is going to be ½ × its mass MI × its speed Vi².0163

We also know for objects rotating that linear velocity is its angular velocity × its radius.0174

We can write this as the kinetic energy of that little piece I is ½ × its mass × the square root of its angular velocity × RI².0182

But then if we wanted the total kinetic energy of the entire disk, we have to add up this kinetic energy for little tiny pieces of the disk to get the total.0197

Our total kinetic energy for it rotating is going to be the infinite sum for all those little pieces I of KI,0206

which is just going to be the sum over I of ½ MI ω² RI².0217

As I look at this, I have a couple constants here that I can pull out of the summation.0229

The ½ is a constant, it looks like the ω, the angular velocity is going to be the same for any points.0234

We can pull those out and say that this is going to be ω² / 2 to say our total K=ω² / 2 × the sum overall I of MI RI².0240

For my next step, what I am going to do is I'm going to define some constant capital I, constant for this problem, which we are going to call the moment of inertia.0258

The sum of all MR² we will call I.0268

I could rewrite this then as our total kinetic energy is going to be ½ I, some of them are squared, ω².0273

When I do that summation, our total kinetic energy is ½ the moment of inertia × the squared of the angular velocity.0289

If I know the moment of inertia, I do not have to worry about all these little pieces.0295

Notice how similar this is to our formula for translational kinetic energy ½ MV².0300

Instead of linear velocity, we now have angular velocity.0310

Instead of inertial mass, we are using rotational inertia or moment of inertia and we still have the ½ vector.0314

We have added one more variable that switches when we go to the rotational world instead of talking about mass and we talk about rotational inertia or moment of inertia.0321

How do we calculate this more generally?0332

Moment of inertia is the sum of all MR² for an object or if you got an infinite sum, we can take the integral of R² × the differential of mass.0335

The mass of one tiny piece.0345

As you go through this unit, it is helpful to memorize some moments of inertia of common objects.0350

Some of those include, well here is the general one, for anything you can use the sum overall MI R².0355

For a disk ½ MR², for a solid sphere 2/5 MR², for a rod about its center point 1/12 ML², L being its length.0362

For a hoop that is MR², for a hollow sphere its 2/3 mass × squared radius and for rod about its end it is 1/3 ML².0374

We will derive a couple of these but as you go through the course, it is usually helpful to memorize a couple of these to save you some time on some problems.0384

If you are asked to derive it, you already know the answers so you know you are doing things correctly.0392

These ones you will definitely want to know, the solid sphere, hollow sphere, hoops, and disk, 0397

they are all good ones to have in the back your mind.0402

Let us take a look at calculating MR² or moment of inertia with a couple of these.0406

Find the moment of inertia I of two 5 kg bowling balls joined by meter long rod of negligible mass when rotated about the center of the rods, rotated from right there.0412

Almost looks like a barbell.0422

Compare that to the moment of inertia of the object when rotated about one of the masses.0424

Before we even start, let us think about which one of these we think is going to be tougher to rotate.0429

It is going to be tougher to rotate this about its center point or this about its end?0434

Just by common sense and experience things I have seen in my life, I'm going to guess that this one is a little bit hard to rotate.0438

We are going to find that out.0445

Let us start here on the left, we will call this M1, this is M2, we will call this R1 and we will call R2.0447

Our moment of inertia is the sum of our MR² which is going to be M1 R1² + M2 R2².0459

The mass one is 5 so 5 kg in that distance if the whole thing is in meter, R1 must be half of meter so 0.5 m² + M2 5 × its length 0.5².0473

Moment of inertia is going to be 10 × 0.5² or 2.5 kg m².0489

Same setup, it rotates in about a different point on the right.0500

Moment of inertia is still the sum of all our MR² so we are going to have M1 R1² + M2 R2² where this is going to be R1.0504

It looks like R2 is going to be 0, moment of inertia is going to be 5 × 1m² + 5 × 0² or just 5 kg m².0517

Same object but rotated about a different point, we have twice the moment of inertia here on the right which makes sense.0534

As we said before we started here that we thought that it would be hard to give a rotational acceleration 2.0540

How about a slightly more complicated object?0549

Find the moment of inertia of the uniform rod about its end and about its center.0552

As we do this one, what we are going to do as far as a strategy is we are going to break the rod up into little tiny pieces 0558

with some mass VM at some distance R from our rotation point.0567

We will define the linear mass density as the total mass of our rod divided by its length.0575

If we do that then, the differential of mass, the amount of mass that in that little tiny bit, 0582

DM is going to be the linear mass density × dx as we integrate from 0 to L.0588

If we do this about its end, our moment of inertia is going to be using our formula R² dm which is going to be the integral from 0 to L.0597

Our R is just our x coordinate that is x² and dm we said was λ dx.0612

This implies then that the moment of inertia is going to be equal to, λ is a constant in this case so we can pull λ out of the integral sign, 0622

integral from 0 to L of x² dx, integral of x² is x³/3 evaluated from 0 to L which is going to be λ L³.0630

But we also said that λ was M/L so our moment of inertia I is going to be, we will replace λ with M/L, we still have an L³/3.0647

I can make a ratio of 1 that becomes L² so we end up with ML² / 3.0661

There is the moment of inertia of that rod rotated about its end.0671

Let us do it about its center.0676

We have a different starting point, instead of rotating about this end, we are going to rotate it about the middle.0682

Same basic calculation but set up just a little bit differently.0688

Moment of inertia is R² dm which is now going to be the integral from -L /2 to L /2 calling our center point here 0 of x² λ dx, 0691

which implies then that our moment of inertia is going to be, we can pull our λ out again and 0708

we are going to have x³/3 evaluated from-L /2 to L/2, which is going to be λ /3 × we will have L /2³/3 - -L /2³/3.0713

Our moment of inertia is going to be equal to λ/3 all of this is going to be equal to λ³/8 - - λ³/8, 2 λ³/8, or λ³/4.0733

Once again, we can take a look at our λ which we defined as M/L.0752

If λ is M/L that will be M/L, for λ we have still got a L³ and we got a 12 down here L that becomes L².0757

And I end up with 1/12 ML², much smaller moment of inertia to rotate about that center point.0769

You can verify those as when I said you probably have to memorize from our previous formula screen.0779

All right looking at another object, let us take a look at a solid cylinder.0786

Find the moment of inertia of a uniform solid cylinder about an axis through its center.0791

This is kind of our soda can that is spinning through its point on the center.0795

We are going to assume it has a uniform density because it is a uniform solid cylinder.0799

Its volume mass density is going to be its total mass divided by its volume, which would be its mass divided by, its volume will be the area × its length which is π R² L.0804

Our strategy is going to be as we integrate it, we are going to take little tiny pieces of the can and think of them as very thin slices.0824

As we integrate from the tiny once all the way out, we are going to get our total cylinder.0834

We need to figure out the differential of mass that is in one tiny piece of the can of that size.0839

To take that, imagine we take this hollow can, what you are going to do is cut it and spread it out to find its mass.0845

To do that then the differential of mass, the mass included there is going to be the area of the rectangle of material we make × its thickness.0853

Its area is going to be the circumference 2π R × its length L × the mass density and then the thickness 0864

is going to be our little dr is we integrate from 0 all the way out to R.0877

We are going to make these infinitesimally thin.0882

Our differential of mass inside our little hollow, it is in that piece of little hollow cylinder is 2π RL × our volume mass density × dr the thickness.0885

As if we have cut that and spread it out to make a tiny thin rectangle of material.0898

Once we got that set up, the actual integration piece is pretty straightforward.0903

Our moments of inertia is the integral of R² dm which is going to be the integral from R =0 to R, 0909

the radius of our entire cylinder of R² and our dm we just defined as 2π RL ρ dr.0921

Our moment of inertia then, we can pull out our constants.0937

2π is a constant, our volume mass density ρ is a constant because it is uniform, L is a constant.0941

That will leave us with the integral from 0 to R of, we have here R³ dr which is 2π ρl.0950

The integral of R³ is R⁴/4 evaluated from 0 to R which implies then that our moment of inertia is going to be 2π ρ L.0960

We will have R⁴ / 4 -0⁴/4 which is 0.0973

But we also know that our ρ is M/π R² L.0981

We will substitute that in so this is going to be equal to, we will not imply, we will say that is equal to, we have 2π, our ρ is M/ir² L.0990

We also still have our L R⁴/4 so this implies then that our moment of inertia is going to be, let us see what we can cancel out of here.1010

We have got an L, we have got an L, we got a π and π, R² and R⁴ that becomes R in the second, 2 becomes a 2.1022

I end up with MR² /2 moment of inertia for our solid cylinder.1034

You get the idea of the procedure you go to define moments of inertia of these continuous or more complicated objects.1046

Let us take a look at the parallel axis theorem, this is a really cool, 1054

helpful formula that will help you with moment of inertia when you are not talking about the center point.1059

If you know the moment of inertia, I of any objects through an axis that intersects the center of mass of the object, we will call that axis L.1064

You can find a moment of inertia around any axis that is parallel to that current axis of rotation we will call L prime.1066

If this is our object of some sort, we know its moment of inertia through L.1083

We want to know its moment of inertia through L prime at some distance D, away in parallel to that initial one.1088

Assuming the initial one goes to the center of mass, the way to find the moment of inertia about L prime is just going to be the moment of inertia 1094

about that center of mass + mass × the distance between those two axis².1101

Let us take a look at how we can use that to solve a problem.1116

Find the moment of inertia of a rod of mass M and length L about one end of the rod using the parallel axis theorem.1121

We have already done this sort of problem before but let us find about the end since we already know what it is, about its center.1128

About its center, we know that the moment of inertia about the center of mass is 1/12 ML² around about its center.1135

And this distance must be L/2 because we are going to move from here to here.1145

Our distance D is L/2 so the moment of inertia of the rod about its end is the moment of inertia about its center of mass + mass × the square of its distance, 1152

which is, we have moment of inertia about the center of mass 1/12 ML² + mass and our D is L /2² which will be ML² / 12 + ML² / 4, 1165

which implies that the moment of inertia about the end is going to ML² / 12 + this will be 3 ML² /12 to give ourselves a common denominator1183

which is 4 ML² / 12 or ML² /3.1196

Another way you can find a moment of inertia about an object once you know its moment of inertia about the center of mass, 1203

as long as that new axis is parallel to the one where you know already, it is nice, straightforward, easy calculation.1209

Alright, to calculate the moment of inertia of a hollow sphere with a mass of 10 kg and a radius of 0.2 m.1218

Here we are going to assume that you have memorized your moment of inertia for common objects.1226

So this would be 2/3 M R² for a hollow sphere which is 2/3 × mass 10 kg and our radius 0.2 m² or 0.27 kg m².1232

How about for a long, thin rod?1255

Find the moment of inertia for a long, thin rod with the mass of 2 kg and a length of 1 m rotating about the center of its length.1258

Let us take a look and assume it is uniform so that is 1/12 ML² about its center which is 1/12 × 2 kg × 1 m² or about 0.17 kg m².1268

What is its moment of inertia when rotating about its end?1288

That is just going to be 1/3 ML² or 1/2 × at 2 kg × L² 1 which is just going to be 0.67 kg m².1291

An object with uniform mass density is rotated about an axle which may be in position A, B, C, or D.1312

Rank the objects moment of inertia from smallest to largest based on axle position.1319

From smallest to largest, we are going where it is easiest to accelerate it rotationally towards its toughest to accelerate rotationally.1326

As we know it is going to be easiest when you got it at the center point here, C we have the smallest moment of inertia.1334

As you move away from that center point, it gets tougher and tougher C, B, D and finally if you are rotating it about A.1340

I would go C, B, D, A, for the ranking of the moment of inertia from smallest to largest.1351

Alright let us do one more, a uniform rod of length L has a moment of inertia I 0 when rotated about its midpoint.1358

A sphere of mass M is added to each of the rod, what is the new moment of inertia of the rod ball system?1365

Over here, moment of inertia is I 0, here we need to figure out its new moment of inertia.1372

The moment of inertia on the right is going to be the moment of inertia of the rod + we have to add up our M².1380

The moment of inertia will be I 0 + we have M × its distance from our center point that is going to be L /2.1391

We got L /2² + same thing on the right hand side ML /2².1403

I is going to be I initial + ML² /4 + ML² /4. 1411

Our moment of inertia is going to be I initial + ML² /2.1420

Alright, hopefully that gets you a good start on moment of inertia or rotational inertia.1430

We will be using it quite extensively in the next few lessons.1434

Thanks for joining us at www.educator.com and make it a great day everyone.1437