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 casethe 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: 
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 massespoint 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 (± drthe 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 9MR^{2}. However, if you checked this out experimentally, you’d find
that this value is too low. So you need to reexamine 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 = 9.4MR^{2}. 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 recognitionnot 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
ML^{2}. 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?
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5. 
How is the center of mass useful in simplifying physics problems?
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6.  What is density and how is it useful in calculating the center of mass of an object?
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7. 
Find the center of mass of the objects pictured below.
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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?
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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 piecesin 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.
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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?
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12. 
What is the total momentum for the center of mass reference frame?
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13. 
Can the center of mass ever be traveling faster than all of the objects it
considers?
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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?
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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 24 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?
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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?
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7. 
If each object in the illustration for #24 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.
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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. 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.
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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?
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12. 
How many penguins can fit into an average igloo? Yes, you really have to answer this questionit is an estimation problem.
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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.
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14. 
What similarities exist between gravitational force and electrostatic force (Coulomb's Law)? What differences exist?
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