Lesson 21--Center of Mass and Moment of Inertia

Center of Mass and Moment of Inertia

Two concepts we’ve mentioned, but never explained thoroughly are center of mass and moment of inertia. Thus far, you have always treated center of mass as the center of an object and used moment of inertia as the rotational mass (without really knowing how to derive it).

CENTER OF MASS

The center of mass of an object is a position or point in space, about which all of the particles in the object are evenly distributed. It is an “average” position of all of the particles in an object.

Frequently in physics we use the center of mass of an object to represent the whole object, because it simplifies the analysis of the problem. Rather than following the twists and turns of the limbs of a student as it flies through the air, we can just examine the path of the center of mass to determine where it is going to land, how long it will be in the air, and so on. Since the whole body is moving with the center of mass the behavior of the whole object will ultimately follow whatever path the center of mass followed.

For the most part, assuming the center of mass is at the center of the object is fine, because most of our problems deal with uniform density and symmetrically shaped objects. However, most of the real world doesn’t function that way. The center of mass of a hammer, for example, is not at the center of the hammer but is located closer to the head.

Indeed, even more complicated systems exist. Imagine analyzing the center of something that bends and folds like a human being. In such a case the center of mass changes position as a person bends or stretches. Although this may seem strange, track and field athletes (high jumpers and pole vaulters) take advantage of the maneuverability of their center of mass. When high jumpers use the “flop” technique, they “fold” themselves around the bar that they are trying to clear. In doing so, they can clear the bar without having to raise their center of mass above the bar.

Well then, this begs the question: how do you compute the center of mass?

Systems of particles

Let’s start by examining the simplest case--the center of mass of point masses in one dimension where none of the masses are moving. It is probably easiest to see what to do by following an example. Take a look at the three masses on a number line below. We’ll define their positions as 0, 3, and 5 meters. To find the position of the center of mass you’ll take the product of each mass times its position, add them together, and divide by the total mass. The calculation would go as follows:





A two or three dimensional set of point masses would follow the same method of calculation as this, only using the y and z positions for the corresponding dimensions. Therefore, the equations for all three dimensions look like that shown below:



Solid objects

Most things are not point masses. In fact, the only time you’ll use point masses exclusively is in a physics or math class. However, we’ll shortly see that they can serve to help us find the center of mass of a solid object.

Suppose you have a sledge hammer whose head is 5 kg and handle is 0.5 kg. Both the head and handle have uniform densities (though different). The 0.92 m long handle is positioned in the center of the head that is 20 cm wide by 8 cm thick. So a rough sketch of the hammer might look like that below.



The trick to finding the center of mass of the sledge hammer is to replace the handle and head by point masses at the position of their center of mass. In this case the handle would be replaced by a point mass of 0.5 kg, 10 cm from the left side and 54 cm up. The head would be replaced by a 5 kg point mass, 10 cm from the left side, 4 cm up.

Since both point masses have the same horizontal position, we can conclude that the x position if the overall center of mass is 10 cm without performing any calculation.

All that remains is the y position. For this, we just perform the calculation as described by the point mass section.



So the center of mass of the sledge hammer is centered horizontally, and 8.2 cm from the bottom or 0.2 cm above the head in the handle.

In fact, when kinesiologists construct models of the human body, they use a technique similar to this to model the behavior of the body.

The calculus approach

Sometimes the shape of an object makes guessing the location of its center of mass a little tricky. Suppose you want to find the center of mass of a isosceles triangles of uniform density. It is fairly obvious that it is going to be along a line that bisects the base (as drawn). However, how high above the base is it?

Well, we could find the vertical position of every molecule in the triangle, multiply that by the mass of the molecule, add them together, and divide by the total mass to find the center of mass. Personally that sounds way too difficult and time consuming to me.

Instead of using that approach we can employ integration (since integration is a technique that adds an infinite number of infinitely small points together). The basic integration equation for finding the center of mass is:



In this equation r is the distance along the axis of consideration (y in our case), dm is a really small chuck of mass (essentially a point mass), and M is the total mass. The equation tells us to add up an infinite number of point mass along the axis we are considering, then divide by the total mass.

But how are you going to integrate this sucker? Integrating along a distance is the standard practice, how do you integrate with respect to mass? The answer is, you don’t. You need to develop an expression for dm in terms of the position along the y axis.

Since the triangle has uniform density that means the point mass, dm, has a density of:





The area of the point mass dm, is the area of the line in the drawing (essentially a rectangle). Looking closely at the piece dm, we can see that its area will be:



Since we are calling the distance along the axis, y. the integration part of the problem becomes:



We can’t integrate this just yet because x changes as y changes. However, using similar triangles we can find an expression to substitute for x.



But we can get a cleaner expression by substituting for M, by looking at density once again.



Substituting for M and canceling, we find that the position of the center of mass is



And you thought there was no application for calculus.

MOMENTUM AND THE CENTER OF MASS REFERENCE FRAME

A reference frame is a viewpoint or perspective where the viewer sees themselves as stationary and views other objects accordingly. In most cases, the reference frame we use is the laboratory frame where the floor is stationary and the objects moving across the floor are considered to be moving.

In truth, all motion is relative so you can choose any perspective as stationary and adjust the behavior of all the objects under consideration accordingly. To illustrate, imagine you are driving your car 60 miles per hour through a school zone. Relative to you the front bumper of the car seems stationary--I mean it's neither getting closer nor further away from you, so it seems like its velocity is zero. But to the little girl in the crosswalk, you’d better believe it seems like it is moving. However, from your perspective the girl seems like she is coming toward you.

Now you might argue that you can tell by observing your surroundings who is moving and who is stationary. To which, I reply, “wanna bet?” In the previous scene, your car was actually driving on a high tech treadmill, with great graphics surrounding it. You had to keep your car speedometer reading 60 miles per hour just to keep from going backwards. The little girl is stationary relative to the treadmill, but she is coming at you at 60 miles per hour. Than blamo!! She is killed by a stationary object. Huh?

From a perspective and result point of view these two situations are identical. The main point is to illustrate that different points of view see things moving at different velocities. However, within any reference frame (point of view) the laws of physics must hold.

How does this apply to momentum and the center of mass reference frame? Well, let’s first look at the motion of the center of mass from the laboratory reference frame. Let’s say two particles are moving along a line such that they are both moving to the right and the one farthest right has a mass of 10 kg and a speed of 4 m/s, while the other has a mass of 6 kg and a speed of 6 m/s.

The center of mass is going to exist somewhere between these two objects. Why? If it didn’t, it wouldn’t be centered, duh. It’s velocity can be computed in a similar fashion to the technique used to find the position of the center of mass. That is,



So, from the laboratory perspective we can compute the velocity of the center of mass as being:



Thus, it is closing in on the mass on the right at a rate of 0.75 m/s, while the mass on the left is catching up to it at a rate of 1.25 m/s.

But, things change when we go to the center of mass reference frame. There, the center of mass is stationary. From its perspective both objects are moving toward it. The mass to the right of the center of mass has a velocity of 0.75 m/s to the left, while the mass to the left of the center of mass has a velocity of 1.25 m/s to the right.

What then, it the momentum of the center of mass?



Zero, in fact, is always the total momentum of a closed system from the center of mass point of view. This quirk, allows for easier computations in certain situations.

Moment of Inertia

Although we’ve discussed moment of inertia, we haven’t attempted to analyze its origin, or calculate it from a ground up standpoint. You ought to realize that it is the rotational analog of mass and the way it differs from mass is in that the distance from the axis of rotation affects the size of the moment of inertia.

Point masses

To calculate moment of inertia using the point mass method you follow similar strategies as you did in finding center of mass by the point mass method. Although there are slight differences in the equations, the basic problem is plug and chug.

When turning something, the difficulty you encounter as you turn the object depends on two things: the mass of the object and the distance of the mass from the axis of rotation. The significant thing about the distance’s contribution to the moment of inertia is that moment of inertia varies with the square of the distance. Consequently, as the mass gets farther away from the axis of rotation, it becomes much harder to turn the object.

The point mass equation looks like this:



To calculate the moment of inertia of the particles at the right. You would calculate the moment of inertia for each particle individually.



Notice how much greater the moments of inertia were for the 5 kg particles than the 3 kg particles.



The calculus method

Most objects are not point masses. Solid objects have a nearly infinite number of “point” masses that make them up. You could find the mass and position of each atom and follow the previous technique to find the moment of inertia. As with center of mass calculations, this would be idiotic because you’d spend the rest of your life on that calculation. Calculus provides us with a technique that will shorten the calculation considerably.

Whenever you find moment of inertia you are adding the products of all the masses times distance squared. Calculus would look at really puny masses--point masses. We’ll call these dm. So, the calculus equation for moment of inertia is:



Once again, dm is not something you are going to integrate. Rather you are going to find a distance dependent expression for dm and substitute before solving for I.

When computing moment of inertia calculus style, you need to select dm carefully. Be sure that all the mass in dm is nearly equidistant to the axis of rotation (± dr--the thickness of the piece).

Let’s do a couple of examples, because this takes a little getting used to.

Example #1: Find the moment of inertia of a stick (length L and mass M) of uniform density, rotating about an axis through its center of mass.

The illustration shows a picture of this stick. The chunk of mass, dm is exaggerated in size for clarity. Notice that the center of the chunk is a distance x away from the axis and is dx wide. Because the drawing is exaggerated, dx looks pretty wide. However, it is actually infinitely thin.

Naturally, we’ll start with the equation:



The limits exist as shown because half the stick is on either side of the axis.



Before we can integrate we need to substitute for dm. We can arrive at an expression for dm by using density.



Since r is x according to our illustration, then the integral becomes:



Finally, using the density relationship,



, we can substitute and express I in terms of mass and length.



This process is much less painful than the atom by atom method. Even so, it is typically more work than most people care to go through. Fortunately, the most common geometries have already been calculated. Unfortunately for you, I think it is useful and mind expanding for you to derive these moments of inertia. So you have something to look forward to in the problem set part of the packet.



PARALLEL AXIS THEOREM

A handy way of figuring out moment of inertia for something that doesn’t rotate through it’s center of mass is by employing the parallel axis theorem.

Consider an object of mass M revolving around an axis, such as a sphere swinging from a massless string. Its center of mass is some distance from the axis of rotation, let’s say 3R. One might think that the total moment of inertia would just be the mass of the sphere times the distance from the center of mass to the axis of rotation squared. So the moment of inertia would be 9MR2. However, if you checked this out experimentally, you’d find that this value is too low. So you need to re-examine your thinking.



Upon careful consideration, you would realize that the sphere has dimension and is not a dimensionless point. After repeated observations you could observe that the sphere makes one rotation every revolution. That is, it is not only turning around the axis of rotation, it is also turning about its center of mass. Thus, you decide to add the moment of inertia of the sphere about its center of mass to your previous calculation, getting I = 9.4MR2. Quickly, you rush to repeat the experiment and, joy upon joys, success! Everything works perfectly and you win the Noble Prize for high school physics (a little known prize, actually, that has no monetary value or esteemed recognition--not unlike valedictorian or high honors on the Golden State Exam).

This technique works for any object of any shape. The main problems are that you need to know the moment of inertia about the center of mass of the object and the location of the center of mass. If these are given, the problem should be fairly routine.

An equation for moment of inertia using the parallel axis theorem is:



Since you are going to have to calculate the moment of inertia of a rod rotating about one end using calculus, I’ll show you how easy it is to do it using the parallel axis theorem. If you look at the chart you can see that the moment of inertia for a thin stick about its center of mass is 1/12 ML2. The distance from the axis of rotation to the center of mass is L/2. Substituting these into the parallel axis equation, you get:





Now, go back to the table and you’ll see that this is exactly the answer we were trying to find. Wasn’t that easy. Ha ha!

You have now seen three ways to compute moment of inertia. For this problem set you will get some practice working with these techniques, plus a little review of the semester to this point.



CENTER OF MASS PROBLEMS:

1.Is your center of mass always in the same place? Why?
Need a hint?


Check your solution?

2. What is a point mass?


Need a hint?


Check your solution?

3.Can an object's center of mass lie outside of its physical boundaries? Explain.


Need a hint?


Check your solution?

4.What can you do to simplify a solid body problem?


Need a hint?


Check your solution?

5. How is the center of mass useful in simplifying physics problems?


Need a hint?


Check your solution?

6.What is density and how is it useful in calculating the center of mass of an object?


Need a hint?


Check your solution?

7. Find the center of mass of the objects pictured below.




Need a hint?


Check your solution?

8.An 8 kg mass has coordinates (-2, 3, 4)m, a 5 kg mass has coordinates (4, 2, -6)m, and a 10 kg mass has coordinates ( 3, -4, 4)m. What is the center of mass of the particles described in this situation?


Need a hint?


Check your solution?

9.A projectile is fired at a 37° angle above the horizontal at 25 m/s. When it reaches the top of its trajectory, it explodes into two equal pieces--in such a way that only their horizontal velocities change. One of the two pieces lands 90 meters from where the projectile was launched. Where does the other piece land? Hint: find where it would have landed had the projectile not exploded.


Need a hint?


Check your solution?

10.Show, using calculus, that the center of mass of a stick of length, L, and uniform density, r, is at L/2. Where r is the linear density M/L.

11. What is the center of mass of the object shown below, given that the object has uniform density?




Need a hint?


Check your solution?

12. What is the total momentum for the center of mass reference frame?


Need a hint?


Check your solution?

13. Can the center of mass ever be traveling faster than all of the objects it considers?


Need a hint?


Check your solution?

14. Three objects are rolling along a straight line. A 5 kg mass is going 3 m/s to the right, a 4 kg mass is going 2 m/s to the left, and an 8 kg mass is traveling 8 m/s to the right. How fast is each going relative to the center of mass?


Need a hint?


Check your solution?

MOMENT OF INERTIA PROBLEMS:

1.1. If the moment of inertia of a point mass m rotating at a distance of R from the axis of rotation is I, what is the moment of inertia when:

a. the mass rotates at 2R
b. the mass rotates at 5R
c. the mass rotates at 0.5R
d. the mass rotates at 0.1R
e. another mass of 2M rotates at R
f. another mass of 0.5 M rotates at R
g. another mass of 3M rotates at 2R
h. another mass of 0.2M rotates at 0.4R


Need a hint?


Check your solution?

Problems 2-4 refer to the drawing below.



2. Which of the point masses has the greatest moment of inertia?


Need a hint?


Check your solution?

3.How many times greater is the largest moment of inertia than the smallest moment of inertia?


Need a hint?


Check your solution?

4.What percent of the total moment of inertia is due to mass M?


Need a hint?


Check your solution?

5. Using calculus find the moment of inertia of the following objects.

a solid sphere rotating about its center of mass
a stick rotating about one end
a solid cylinder rotating about its center of mass


Need a hint?


Check your solution?

6.How would problems 2, 3, and 4 change if each mass was a sphere of radius L instead of a point mass?


Need a hint?


Check your solution?

7. If each object in the illustration for #2-4 is a solid disk of radius L, what are the new moments of inertia for each object. What is the new moment of inertia for each object.


Need a hint?


Check your solution?

8.A light string is wrapped around a pulley of mass M and radius R. Suspended from the end of the string is a mass of 3M.



a. What is the moment of inertia of the pulley (think disk or cylinder).

b. Show that the acceleration of the suspended mass will be "6g/7" when released.




Need a hint?


Check your solution?

9.What is the angular velocity of the disk in #8 after t seconds have passed? Express your answer in terms of g, R, and t. Assume the initial angular velocity is zero.


Need a hint?


Check your solution?

10.Hopefully you remember the period of a physical pendulum is



. If you don’t, the period of a physical pendulum is



--I thought that I’d remind you. If the physical pendulum is a stick rotating about one end, show that it’s period could be expressed as




Need a hint?


Check your solution?

11. What conditions are necessary for harmonic motion to occur?


Need a hint?


Check your solution?

12. How many penguins can fit into an average igloo? Yes, you really have to answer this question--it is an estimation problem.


Need a hint?


Check your solution?

13. Given the following graph for the position of an object of mass m undergoing harmonic motion, find the equation for the net force on the the object.




Need a hint?


Check your solution?

14. What similarities exist between gravitational force and electrostatic force (Coulomb's Law)? What differences exist?


Need a hint?


Check your solution?




Questions? Send comments to Like I care
Main Home Page
AP Physics Home Page