moment of inertia experiment analysis

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19 Center of Mass; Moment of Inertia

19–1 Properties of the center of mass

In the previous chapter we found that if a great many forces are acting on a complicated mass of particles, whether the particles comprise a rigid or a nonrigid body, or a cloud of stars, or anything else, and we find the sum of all the forces (that is, of course, the external forces, because the internal forces balance out), then if we consider the body as a whole, and say it has a total mass $M$, there is a certain point “inside” the body, called the center of mass , such that the net resulting external force produces an acceleration of this point, just as though the whole mass were concentrated there. Let us now discuss the center of mass in a little more detail.

The location of the center of mass (abbreviated CM) is given by the equation \begin{equation} \label{Eq:I:19:1} \FLPR_{\text{CM}}=\frac{\sum m_i\FLPr_i}{\sum m_i}. \end{equation} This is, of course, a vector equation which is really three equations, one for each of the three directions. We shall consider only the $x$-direction, because if we can understand that one, we can understand the other two. What does $X_{\text{CM}} = \sum m_ix_i/\sum m_i$ mean? Suppose for a moment that the object is divided into little pieces, all of which have the same mass $m$; then the total mass is simply the number $N$ of pieces times the mass of one piece, say one gram, or any unit. Then this equation simply says that we add all the $x$’s, and then divide by the number of things that we have added: $X_{\text{CM}} =$ $m\sum x_i/mN =$ $\sum x_i/N$. In other words, $X_{\text{CM}}$ is the average of all the $x$’s, if the masses are equal. But suppose one of them were twice as heavy as the others. Then in the sum, that $x$ would come in twice. This is easy to understand, for we can think of this double mass as being split into two equal ones, just like the others; then in taking the average, of course, we have to count that $x$ twice because there are two masses there. Thus $X$ is the average position, in the $x$-direction, of all the masses, every mass being counted a number of times proportional to the mass, as though it were divided into “little grams.” From this it is easy to prove that $X$ must be somewhere between the largest and the smallest $x$, and, therefore lies inside the envelope including the entire body. It does not have to be in the material of the body, for the body could be a circle, like a hoop, and the center of mass is in the center of the hoop, not in the hoop itself.

Of course, if an object is symmetrical in some way, for instance, a rectangle, so that it has a plane of symmetry, the center of mass lies somewhere on the plane of symmetry. In the case of a rectangle there are two planes, and that locates it uniquely. But if it is just any symmetrical object, then the center of gravity lies somewhere on the axis of symmetry, because in those circumstances there are as many positive as negative $x$’s.

Another interesting proposition is the following very curious one. Suppose that we imagine an object to be made of two pieces, $A$ and $B$ (Fig.  19–1 ). Then the center of mass of the whole object can be calculated as follows. First, find the center of mass of piece $A$, and then of piece $B$. Also, find the total mass of each piece, $M_A$ and $M_B$. Then consider a new problem, in which a point mass $M_A$ is at the center of mass of object $A$, and another point mass $M_B$ is at the center of mass of object $B$. The center of mass of these two point masses is then the center of mass of the whole object. In other words, if the centers of mass of various parts of an object have been worked out, we do not have to start all over again to find the center of mass of the whole object; we just have to put the pieces together, treating each one as a point mass situated at the center of mass of that piece. Let us see why that is. Suppose that we wanted to calculate the center of mass of a complete object, some of whose particles are considered to be members of object $A$ and some members of object $B$. The total sum $\sum m_ix_i$ can then be split into two pieces—the sum $\sum_A m_ix_i$ for the $A$ object only, and the sum $\sum_B m_ix_i$ for object $B$ only. Now if we were computing the center of mass of object $A$ alone, we would have exactly the first of these sums, and we know that this by itself is $M_AX_A$, the total mass of all the particles in $A$ times the position of the center of mass of $A$, because that is the theorem of the center of mass, applied to object $A$. In the same manner, just by looking at object $B$, we get $M_BX_B$, and of course, adding the two yields $MX_{\text{CM}}$: \begin{align} MX_{\text{CM}} &=\sum_A m_ix_i+\sum_B m_ix_i\notag\\[.5ex] \label{Eq:I:19:2} &=M_AX_A+M_BX_B. \end{align} Now since $M$ is evidently the sum of $M_A$ and $M_B$, we see that Eq. ( 19.2 ) can be interpreted as a special example of the center of mass formula for two point objects, one of mass $M_A$ located at $X_A$ and the other of mass $M_B$ located at $X_B$.

The theorem concerning the motion of the center of mass is very interesting, and has played an important part in the development of our understanding of physics. Suppose we assume that Newton’s law is right for the small component parts of a much larger object. Then this theorem shows that Newton’s law is also correct for the larger object, even if we do not study the details of the object, but only the total force acting on it and its mass. In other words, Newton’s law has the peculiar property that if it is right on a certain small scale, then it will be right on a larger scale. If we do not consider a baseball as a tremendously complex thing, made of myriads of interacting particles, but study only the motion of the center of mass and the external forces on the ball, we find $\FLPF= m\FLPa$, where $\FLPF$ is the external force on the baseball, $m$ is its mass, and $\FLPa$ is the acceleration of its center of mass. So $\FLPF = m\FLPa$ is a law which reproduces itself on a larger scale. (There ought to be a good word, out of the Greek, perhaps, to describe a law which reproduces the same law on a larger scale.)

Of course, one might suspect that the first laws that would be discovered by human beings would be those that would reproduce themselves on a larger scale. Why? Because the actual scale of the fundamental gears and wheels of the universe are of atomic dimensions, which are so much finer than our observations that we are nowhere near that scale in our ordinary observations. So the first things that we would discover must be true for objects of no special size relative to an atomic scale. If the laws for small particles did not reproduce themselves on a larger scale, we would not discover those laws very easily. What about the reverse problem? Must the laws on a small scale be the same as those on a larger scale? Of course it is not necessarily so in nature, that at an atomic level the laws have to be the same as on a large scale. Suppose that the true laws of motion of atoms were given by some strange equation which does not have the property that when we go to a larger scale we reproduce the same law, but instead has the property that if we go to a larger scale, we can approximate it by a certain expression such that, if we extend that expression up and up, it keeps reproducing itself on a larger and larger scale. That is possible, and in fact that is the way it works. Newton’s laws are the “tail end” of the atomic laws, extrapolated to a very large size. The actual laws of motion of particles on a fine scale are very peculiar, but if we take large numbers of them and compound them, they approximate, but only approximate, Newton’s laws. Newton’s laws then permit us to go on to a higher and higher scale, and it still seems to be the same law. In fact, it becomes more and more accurate as the scale gets larger and larger. This self-reproducing factor of Newton’s laws is thus really not a fundamental feature of nature, but is an important historical feature. We would never discover the fundamental laws of the atomic particles at first observation because the first observations are much too crude. In fact, it turns out that the fundamental atomic laws, which we call quantum mechanics, are quite different from Newton’s laws, and are difficult to understand because all our direct experiences are with large-scale objects and the small-scale atoms behave like nothing we see on a large scale. So we cannot say, “An atom is just like a planet going around the sun,” or anything like that. It is like nothing we are familiar with because there is nothing like it . As we apply quantum mechanics to larger and larger things, the laws about the behavior of many atoms together do not reproduce themselves, but produce new laws , which are Newton’s laws, which then continue to reproduce themselves from, say, micro-microgram size, which still is billions and billions of atoms, on up to the size of the earth, and above.

Let us now return to the center of mass. The center of mass is sometimes called the center of gravity, for the reason that, in many cases, gravity may be considered uniform. Let us suppose that we have small enough dimensions that the gravitational force is not only proportional to the mass, but is everywhere parallel to some fixed line. Then consider an object in which there are gravitational forces on each of its constituent masses. Let $m_i$ be the mass of one part. Then the gravitational force on that part is $m_i$ times $g$. Now the question is, where can we apply a single force to balance the gravitational force on the whole thing, so that the entire object, if it is a rigid body, will not turn? The answer is that this force must go through the center of mass, and we show this in the following way. In order that the body will not turn, the torque produced by all the forces must add up to zero, because if there is a torque, there is a change of angular momentum, and thus a rotation. So we must calculate the total of all the torques on all the particles, and see how much torque there is about any given axis; it should be zero if this axis is at the center of mass. Now, measuring $x$ horizontally and $y$ vertically, we know that the torques are the forces in the $y$-direction, times the lever arm $x$ (that is to say, the force times the lever arm around which we want to measure the torque). Now the total torque is the sum \begin{equation} \label{Eq:I:19:3} \tau=\sum m_igx_i=g\sum m_ix_i, \end{equation} so if the total torque is to be zero, the sum $\sum m_ix_i$ must be zero. But $\sum m_ix_i = MX_{\text{CM}}$, the total mass times the distance of the center of mass from the axis. Thus the $x$-distance of the center of mass from the axis is zero.

Of course, we have checked the result only for the $x$-distance, but if we use the true center of mass the object will balance in any position, because if we turned it $90$ degrees, we would have $y$’s instead of $x$’s. In other words, when an object is supported at its center of mass, there is no torque on it because of a parallel gravitational field. In case the object is so large that the nonparallelism of the gravitational forces is significant, then the center where one must apply the balancing force is not simple to describe, and it departs slightly from the center of mass. That is why one must distinguish between the center of mass and the center of gravity. The fact that an object supported exactly at the center of mass will balance in all positions has another interesting consequence. If, instead of gravitation, we have a pseudo force due to acceleration, we may use exactly the same mathematical procedure to find the position to support it so that there are no torques produced by the inertial force of acceleration. Suppose that the object is held in some manner inside a box, and that the box, and everything contained in it, is accelerating. We know that, from the point of view of someone at rest relative to this accelerating box, there will be an effective force due to inertia. That is, to make the object go along with the box, we have to push on it to accelerate it, and this force is “balanced” by the “force of inertia,” which is a pseudo force equal to the mass times the acceleration of the box. To the man in the box, this is the same situation as if the object were in a uniform gravitational field whose “$g$” value is equal to the acceleration $a$. Thus the inertial force due to accelerating an object has no torque about the center of mass.

This fact has a very interesting consequence. In an inertial frame that is not accelerating, the torque is always equal to the rate of change of the angular momentum. However, about an axis through the center of mass of an object which is accelerating, it is still true that the torque is equal to the rate of change of the angular momentum. Even if the center of mass is accelerating, we may still choose one special axis, namely, one passing through the center of mass, such that it will still be true that the torque is equal to the rate of change of angular momentum around that axis. Thus the theorem that torque equals the rate of change of angular momentum is true in two general cases: (1) a fixed axis in inertial space, (2) an axis through the center of mass, even though the object may be accelerating.

19–2 Locating the center of mass

The mathematical techniques for the calculation of centers of mass are in the province of a mathematics course, and such problems provide good exercise in integral calculus. After one has learned calculus, however, and wants to know how to locate centers of mass, it is nice to know certain tricks which can be used to do so. One such trick makes use of what is called the theorem of Pappus. It works like this: if we take any closed area in a plane and generate a solid by moving it through space such that each point is always moved perpendicular to the plane of the area, the resulting solid has a total volume equal to the area of the cross section times the distance that the center of mass moved! Certainly this is true if we move the area in a straight line perpendicular to itself, but if we move it in a circle or in some other curve, then it generates a rather peculiar volume. For a curved path, the outside goes around farther, and the inside goes around less, and these effects balance out. So if we want to locate the center of mass of a plane sheet of uniform density, we can remember that the volume generated by spinning it about an axis is the distance that the center of mass goes around, times the area of the sheet.

For example, if we wish to find the center of mass of a right triangle of base $D$ and height $H$ (Fig.  19–2 ), we might solve the problem in the following way. Imagine an axis along $H$, and rotate the triangle about that axis through a full $360$ degrees. This generates a cone. The distance that the $x$-coordinate of the center of mass has moved is $2\pi x$. The area which is being moved is the area of the triangle, $\tfrac{1}{2}HD$. So the $x$-distance of the center of mass times the area of the triangle is the volume swept out, which is of course $\pi D^2H/3$. Thus $(2\pi x)(\tfrac{1}{2}HD) = \pi D^2H/3$, or $x = D/3$. In a similar manner, by rotating about the other axis, or by symmetry, we find $y = H/3$. In fact, the center of mass of any uniform triangular area is where the three medians, the lines from the vertices through the centers of the opposite sides, all meet. That point is $1/3$ of the way along each median. Clue: Slice the triangle up into a lot of little pieces, each parallel to a base. Note that the median line bisects every piece, and therefore the center of mass must lie on this line.

Now let us try a more complicated figure. Suppose that it is desired to find the position of the center of mass of a uniform semicircular disc—a disc sliced in half. Where is the center of mass? For a full disc, it is at the center, of course, but a half-disc is more difficult. Let $r$ be the radius and $x$ be the distance of the center of mass from the straight edge of the disc. Spin it around this edge as axis to generate a sphere. Then the center of mass has gone around $2\pi x$, the area is $\pi r^2/2$ (because it is only half a circle). The volume generated is, of course, $4\pi r^3/3$, from which we find that \begin{equation*} (2\pi x)(\tfrac{1}{2}\pi r^2) = 4\pi r^3/3, \end{equation*} or \begin{equation*} x=4r/3\pi. \end{equation*}

There is another theorem of Pappus which is a special case of the above one, and therefore equally true. Suppose that, instead of the solid semicircular disc, we have a semicircular piece of wire with uniform mass density along the wire, and we want to find its center of mass. In this case there is no mass in the interior, only on the wire. Then it turns out that the area which is swept by a plane curved line, when it moves as before, is the distance that the center of mass moves times the length of the line. (The line can be thought of as a very narrow area, and the previous theorem can be applied to it.)

19–3 Finding the moment of inertia

Now let us discuss the problem of finding the moments of inertia of various objects. The formula for the moment of inertia about the $z$-axis of an object is \begin{equation} I =\sum m_i(x_i^2+y_i^2)\notag \end{equation} or \begin{equation} \label{Eq:I:19:4} I =\int(x^2+y^2)\,dm=\int(x^2+y^2)\rho\,dV. \end{equation} That is, we must sum the masses, each one multiplied by the square of its distance $(x_i^2 + y_i^2)$ from the axis. Note that it is not the three-dimensional distance, only the two-dimensional distance squared, even for a three-dimensional object. For the most part, we shall restrict ourselves to two-dimensional objects, but the formula for rotation about the $z$-axis is just the same in three dimensions.

As a simple example, consider a rod rotating about a perpendicular axis through one end (Fig.  19–3 ). Now we must sum all the masses times the $x$-distances squared (the $y$’s being all zero in this case). What we mean by “the sum,” of course, is the integral of $x^2$ times the little elements of mass. If we divide the rod into small elements of length $dx$, the corresponding elements of mass are proportional to $dx$, and if $dx$ were the length of the whole rod the mass would be $M$. Therefore \begin{equation} dm = M\,dx/L\notag \end{equation} and so \begin{equation} \label{Eq:I:19:5} I=\int_0^Lx^2\,\frac{M\,dx}{L} = \frac{M}{L}\int_0^L x^2\,dx = \frac{ML^2}{3}. \end{equation} The dimensions of moment of inertia are always mass times length squared, so all we really had to work out was the factor $1/3$.

Now what is $I$ if the rotation axis is at the center of the rod? We could just do the integral over again, letting $x$ range from $-\tfrac{1}{2}L$ to $+\tfrac{1}{2}L$. But let us notice a few things about the moment of inertia. We can imagine the rod as two rods, each of mass $M/2$ and length $L/2$; the moments of inertia of the two small rods are equal, and are both given by the formula ( 19.5 ). Therefore the moment of inertia is \begin{equation} \label{Eq:I:19:6} I=\frac{2(M/2)(L/2)^2}{3}=\frac{ML^2}{12}. \end{equation} Thus it is much easier to turn a rod about its center, than to swing it around an end.

Of course, we could go on to compute the moments of inertia of various other bodies of interest. However, while such computations provide a certain amount of important exercise in the calculus, they are not basically of interest to us as such. There is, however, an interesting theorem which is very useful. Suppose we have an object, and we want to find its moment of inertia around some axis. That means we want the inertia needed to carry it by rotation about that axis. Now if we support the object on pivots at the center of mass, so that the object does not turn as it rotates about the axis (because there is no torque on it from inertial effects, and therefore it will not turn when we start moving it), then the forces needed to swing it around are the same as though all the mass were concentrated at the center of mass, and the moment of inertia would be simply $I_1 = MR_{\text{CM}}^2$, where $R_{\text{CM}}$ is the distance from the axis to the center of mass. But of course that is not the right formula for the moment of inertia of an object which is really being rotated as it revolves, because not only is the center of it moving in a circle, which would contribute an amount $I_1$ to the moment of inertia, but also we must turn it about its center of mass. So it is not unreasonable that we must add to $I_1$ the moment of inertia $I_c$ about the center of mass. So it is a good guess that the total moment of inertia about any axis will be \begin{equation} \label{Eq:I:19:7} I=I_c+MR_{\text{CM}}^2. \end{equation}

This theorem is called the parallel-axis theorem , and may be easily proved. The moment of inertia about any axis is the mass times the sum of the $x_i$’s and the $y_i$’s, each squared: $I=\sum (x_i^2 + y_i^2)m_i$. We shall concentrate on the $x$’s, but of course the $y$’s work the same way. Now $x$ is the distance of a particular point mass from the origin, but let us consider how it would look if we measured $x'$ from the CM, instead of $x$ from the origin. To get ready for this analysis, we write \begin{equation*} x_i=x_i'+X_{\text{CM}}. \end{equation*} Then we just square this to find \begin{equation*} x_i^2=x_i'^2+2X_{\text{CM}}x_i'+X_{\text{CM}}^2. \end{equation*} So, when this is multiplied by $m_i$ and summed over all $i$, what happens? Taking the constants outside the summation sign, we get \begin{equation*} I_x=\sum m_ix_i'^2+2X_{\text{CM}}\sum m_ix_i'+ X_{\text{CM}}^2\sum m_i. \end{equation*} The third sum is easy; it is just $MX_{\text{CM}}^2$. In the second sum there are two pieces, one of them is $\sum m_ix_i'$, which is the total mass times the $x'$-coordinate of the center of mass. But this contributes nothing, because $x'$ is measured from the center of mass, and in these axes the average position of all the particles, weighted by the masses, is zero. The first sum, of course, is the $x$ part of $I_c$. Thus we arrive at Eq. ( 19.7 ), just as we guessed.

Let us check ( 19.7 ) for one example. Let us just see whether it works for the rod. For an axis through one end, the moment of inertia should be $ML^2/3$, for we calculated that. The center of mass of a rod, of course, is in the center of the rod, at a distance $L/2$. Therefore we should find that $ML^2/3 = ML^2/12 + M(L/2)^2$. Since one-quarter plus one-twelfth is one-third, we have made no fundamental error.

Incidentally, we did not really need to use an integral to find the moment of inertia ( 19.5 ). If we simply assume that it is $ML^2$ times $\gamma$, an unknown coefficient, and then use the argument about the two halves to get $\tfrac{1}{4}\gamma$ for ( 19.6 ), then from our argument about transferring the axes we could prove that $\gamma = \tfrac{1}{4}\gamma + \tfrac{1}{4}$, so $\gamma$ must be $1/3$. There is always another way to do it!

In applying the parallel-axis theorem, it is of course important to remember that the axis for $I_c$ must be parallel to the axis about which the moment of inertia is wanted.

One further property of the moment of inertia is worth mentioning because it is often helpful in finding the moment of inertia of certain kinds of objects. This property is that if one has a plane figure and a set of coordinate axes with origin in the plane and $z$-axis perpendicular to the plane, then the moment of inertia of this figure about the $z$-axis is equal to the sum of the moments of inertia about the $x$- and $y$-axes. This is easily proved by noting that \begin{equation*} I_x=\sum m_i(y_i^2+z_i^2)=\sum m_iy_i^2 \end{equation*} (since $z_i = 0$). Similarly, \begin{equation*} I_y = \sum m_i(x_i^2 +z_i^2)=\sum m_ix_i^2, \end{equation*} but \begin{align*} I_z=\sum m_i(x_i^2+y_i^2)&=\sum m_ix_i^2 + \sum m_iy_i^2\\[1ex] &=I_x+I_y. \end{align*}

As an example, the moment of inertia of a uniform rectangular plate of mass $M$, width $w$, and length $L$, about an axis perpendicular to the plate and through its center is simply \begin{equation*} I = M(w^2 + L^2)/12, \end{equation*} because its moment of inertia about an axis in its plane and parallel to its length is $Mw^2/12$, i.e., just as for a rod of length $w$, and the moment of inertia about the other axis in its plane is $ML^2/12$, just as for a rod of length $L$.

To summarize, the moment of inertia of an object about a given axis, which we shall call the $z$-axis, has the following properties:

• The moment of inertia is \begin{equation*} I_z = \sum_i m_i(x_i^2 + y_i^2) = \int(x^2 + y^2)\,dm. \end{equation*}
• If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces.
• The moment of inertia about any given axis is equal to the moment of inertia about a parallel axis through the CM plus the total mass times the square of the distance from the axis to the CM.
• If the object is a plane figure, the moment of inertia about an axis perpendicular to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and intersecting at the perpendicular axis.

The moments of inertia of a number of elementary shapes having uniform mass densities are given in Table  19–1 , and the moments of inertia of some other objects, which may be deduced from Table  19–1 , using the above properties, are given in Table  19–2 .

19–4 Rotational kinetic energy

Now let us go on to discuss dynamics further. In the analogy between linear motion and angular motion that we discussed in Chapter  18 , we used the work theorem, but we did not talk about kinetic energy. What is the kinetic energy of a rigid body, rotating about a certain axis with an angular velocity $\omega$? We can immediately guess the correct answer by using our analogies. The moment of inertia corresponds to the mass, angular velocity corresponds to velocity, and so the kinetic energy ought to be $\tfrac{1}{2}I\omega^2$, and indeed it is, as will now be demonstrated. Suppose the object is rotating about some axis so that each point has a velocity whose magnitude is $\omega r_i$, where $r_i$ is the radius from the particular point to the axis. Then if $m_i$ is the mass of that point, the total kinetic energy of the whole thing is just the sum of the kinetic energies of all of the little pieces: \begin{equation*} T=\tfrac{1}{2}\sum m_iv_i^2= \tfrac{1}{2}\sum m_i(r_i\omega)^2. \end{equation*} Now $\omega^2$ is a constant, the same for all points. Thus \begin{equation} \label{Eq:I:19:8} T=\tfrac{1}{2}\omega^2\sum m_ir_i^2=\tfrac{1}{2}I\omega^2. \end{equation}

At the end of Chapter  18 we pointed out that there are some interesting phenomena associated with an object which is not rigid, but which changes from one rigid condition with a definite moment of inertia, to another rigid condition. Namely, in our example of the turntable, we had a certain moment of inertia $I_1$ with our arms stretched out, and a certain angular velocity $\omega_1$. When we pulled our arms in, we had a different moment of inertia, $I_2$, and a different angular velocity, $\omega_2$, but again we were “rigid.” The angular momentum remained constant, since there was no torque about the vertical axis of the turntable. This means that $I_1\omega_1 = I_2\omega_2$. Now what about the energy? That is an interesting question. With our arms pulled in, we turn faster, but our moment of inertia is less, and it looks as though the energies might be equal. But they are not, because what does balance is $I\omega$, not $I\omega^2$. So if we compare the kinetic energy before and after, the kinetic energy before is $\tfrac{1}{2}I_1\omega_1^2 = \tfrac{1}{2}L\omega_1$, where $L=$ $I_1\omega_1=$ $I_2\omega_2$ is the angular momentum. Afterward, by the same argument, we have $T = \tfrac{1}{2}L\omega_2$ and since $\omega_2 > \omega_1$ the kinetic energy of rotation is greater than it was before. So we had a certain energy when our arms were out, and when we pulled them in, we were turning faster and had more kinetic energy. What happened to the theorem of the conservation of energy? Somebody must have done some work. We did work! When did we do any work? When we move a weight horizontally, we do not do any work. If we hold a thing out and pull it in, we do not do any work. But that is when we are not rotating! When we are rotating, there is centrifugal force on the weights. They are trying to fly out, so when we are going around we have to pull the weights in against the centrifugal force. So, the work we do against the centrifugal force ought to agree with the difference in rotational energy, and of course it does. That is where the extra kinetic energy comes from.

There is still another interesting feature which we can treat only descriptively, as a matter of general interest. This feature is a little more advanced, but is worth pointing out because it is quite curious and produces many interesting effects.

Consider that turntable experiment again. Consider the body and the arms separately, from the point of view of the man who is rotating. After the weights are pulled in, the whole object is spinning faster, but observe, the central part of the body is not changed , yet it is turning faster after the event than before. So, if we were to draw a circle around the inner body, and consider only objects inside the circle, their angular momentum would change ; they are going faster. Therefore there must be a torque exerted on the body while we pull in our arms. No torque can be exerted by the centrifugal force, because that is radial. So that means that among the forces that are developed in a rotating system, centrifugal force is not the entire story, there is another force . This other force is called Coriolis force , and it has the very strange property that when we move something in a rotating system, it seems to be pushed sidewise. Like the centrifugal force, it is an apparent force. But if we live in a system that is rotating, and move something radially, we find that we must also push it sidewise to move it radially. This sidewise push which we have to exert is what turned our body around.

Now let us develop a formula to show how this Coriolis force really works. Suppose Moe is sitting on a carousel that appears to him to be stationary. But from the point of view of Joe, who is standing on the ground and who knows the right laws of mechanics, the carousel is going around. Suppose that we have drawn a radial line on the carousel, and that Moe is moving some mass radially along this line. We would like to demonstrate that a sidewise force is required to do that. We can do this by paying attention to the angular momentum of the mass. It is always going around with the same angular velocity $\omega$, so that the angular momentum is \begin{equation*} L=mv_{\text{tang}}r=m\omega r\cdot r=m\omega r^2. \end{equation*} So when the mass is close to the center, it has relatively little angular momentum, but if we move it to a new position farther out, if we increase $r$, $m$ has more angular momentum, so a torque must be exerted in order to move it along the radius. (To walk along the radius in a carousel, one has to lean over and push sidewise. Try it sometime.) The torque that is required is the rate of change of $L$ with time as $m$ moves along the radius. If $m$ moves only along the radius, omega stays constant, so that the torque is \begin{equation*} \tau=F_cr=\ddt{L}{t}=\ddt{(m\omega r^2)}{t}=2m\omega r\,\ddt{r}{t}, \end{equation*} where $F_c$ is the Coriolis force. What we really want to know is what sidewise force has to be exerted by Moe in order to move $m$ out at speed $v_r = dr/dt$. This is $F_c =$ $\tau/r =$ $2m\omega v_r$.

Now that we have a formula for the Coriolis force, let us look at the situation a little more carefully, to see whether we can understand the origin of this force from a more elementary point of view. We note that the Coriolis force is the same at every radius, and is evidently present even at the origin! But it is especially easy to understand it at the origin, just by looking at what happens from the inertial system of Joe, who is standing on the ground. Figure  19–4 shows three successive views of $m$ just as it passes the origin at $t = 0$. Because of the rotation of the carousel, we see that $m$ does not move in a straight line, but in a curved path tangent to a diameter of the carousel where $r= 0$. In order for $m$ to go in a curve, there must be a force to accelerate it in absolute space. This is the Coriolis force.

This is not the only case in which the Coriolis force occurs. We can also show that if an object is moving with constant speed around the circumference of a circle, there is also a Coriolis force. Why? Moe sees a velocity $v_M$ around the circle. On the other hand, Joe sees $m$ going around the circle with the velocity $v_J = v_M + \omega r$, because $m$ is also carried by the carousel. Therefore we know what the force really is, namely, the total centripetal force due to the velocity $v_J$, or $mv_J^2/r$; that is the actual force. Now from Moe’s point of view, this centripetal force has three pieces. We may write it all out as follows: \begin{equation*} F_r=-\frac{mv_J^2}{r}=-\frac{mv_M^2}{r}- 2mv_M\omega-m\omega^2r. \end{equation*} Now, $F_r$ is the force that Moe would see. Let us try to understand it. Would Moe appreciate the first term? “Yes,” he would say, “even if I were not turning, there would be a centripetal force if I were to run around a circle with velocity $v_M$.” This is simply the centripetal force that Moe would expect, having nothing to do with rotation. In addition, Moe is quite aware that there is another centripetal force that would act even on objects which are standing still on his carousel. This is the third term. But there is another term in addition to these, namely the second term, which is again $2m\omega v$. The Coriolis force $F_c$ was tangential when the velocity was radial, and now it is radial when the velocity is tangential. In fact, one expression has a minus sign relative to the other. The force is always in the same direction, relative to the velocity, no matter in which direction the velocity is. The force is at right angles to the velocity, and of magnitude $2m\omega v$.

10.5 Calculating Moments of Inertia

Learning objectives.

By the end of this section, you will be able to:

• Calculate the moment of inertia for uniformly shaped, rigid bodies
• Apply the parallel axis theorem to find the moment of inertia about any axis parallel to one already known
• Calculate the moment of inertia for compound objects

In the preceding section, we defined the moment of inertia but did not show how to calculate it. In this section, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. This section is very useful for seeing how to apply a general equation to complex objects (a skill that is critical for more advanced physics and engineering courses).

Moment of Inertia

We defined the moment of inertia I of an object to be I = ∑ i m i r i 2 I = ∑ i m i r i 2 for all the point masses that make up the object. Because r is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. To see this, let’s take a simple example of two masses at the end of a massless (negligibly small mass) rod ( Figure 10.23 ) and calculate the moment of inertia about two different axes. In this case, the summation over the masses is simple because the two masses at the end of the barbell can be approximated as point masses, and the sum therefore has only two terms.

In the case with the axis in the center of the barbell, each of the two masses m is a distance R away from the axis, giving a moment of inertia of

In the case with the axis at the end of the barbell—passing through one of the masses—the moment of inertia is

From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center.

In this example, we had two point masses and the sum was simple to calculate. However, to deal with objects that are not point-like, we need to think carefully about each of the terms in the equation. The equation asks us to sum over each ‘piece of mass’ a certain distance from the axis of rotation. But what exactly does each ‘piece of mass’ mean? Recall that in our derivation of this equation, each piece of mass had the same magnitude of velocity, which means the whole piece had to have a single distance r to the axis of rotation. However, this is not possible unless we take an infinitesimally small piece of mass dm , as shown in Figure 10.24 .

The need to use an infinitesimally small piece of mass dm suggests that we can write the moment of inertia by evaluating an integral over infinitesimal masses rather than doing a discrete sum over finite masses:

This, in fact, is the form we need to generalize the equation for complex shapes. It is best to work out specific examples in detail to get a feel for how to calculate the moment of inertia for specific shapes. This is the focus of most of the rest of this section.

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in Figure 10.25 . We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z -axis is the axis of rotation and the x -axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x -axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density λ λ of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

If we take the differential of each side of this equation, we find

since λ λ is constant. We chose to orient the rod along the x -axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x -axis and has a length dx ; in fact, d L = d x d L = d x in this situation. We can therefore write d m = λ ( d x ) d m = λ ( d x ) , giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x , as shown in the figure. Putting this all together, we obtain

The last step is to be careful about our limits of integration. The rod extends from x = − L / 2 x = − L / 2 to x = L / 2 x = L / 2 , since the axis is in the middle of the rod at x = 0 x = 0 . This gives us

Next, we calculate the moment of inertia for the same uniform thin rod but with a different axis choice so we can compare the results. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. This happens because more mass is distributed farther from the axis of rotation.

A uniform thin rod with axis at the end

Now consider the same uniform thin rod of mass M and length L , but this time we move the axis of rotation to the end of the rod. We wish to find the moment of inertia about this new axis ( Figure 10.26 ). The quantity dm is again defined to be a small element of mass making up the rod. Just as before, we obtain

However, this time we have different limits of integration. The rod extends from x = 0 x = 0 to x = L x = L , since the axis is at the end of the rod at x = 0 x = 0 . Therefore we find

Note the rotational inertia of the rod about its endpoint is larger than the rotational inertia about its center (consistent with the barbell example) by a factor of four.

The Parallel-Axis Theorem

The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass. Such an axis is called a parallel axis . There is a theorem for this, called the parallel-axis theorem , which we state here but do not derive in this text.

Parallel-Axis Theorem

Let m be the mass of an object and let d be the distance from an axis through the object’s center of mass to a new axis. Then we have

Let’s apply this to the rod examples solved above:

This result agrees with our more lengthy calculation from above. This is a useful equation that we apply in some of the examples and problems.

Check Your Understanding 10.5

What is the moment of inertia of a cylinder of radius R and mass m about an axis through a point on the surface, as shown below?

A uniform thin disk about an axis through the center

Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center ( Figure 10.27 ).

Since the disk is thin, we can take the mass as distributed entirely in the xy -plane. We again start with the relationship for the surface mass density , which is the mass per unit surface area. Since it is uniform, the surface mass density σ σ is constant:

Now we use a simplification for the area. The area can be thought of as made up of a series of thin rings, where each ring is a mass increment dm of radius r equidistant from the axis, as shown in part (b) of the figure. The infinitesimal area of each ring dA is therefore given by the length of each ring ( 2 π r 2 π r ) times the infinitesimal width of each ring dr :

The full area of the disk is then made up from adding all the thin rings with a radius range from 0 to R . This radius range then becomes our limits of integration for dr , that is, we integrate from r = 0 r = 0 to r = R r = R . Putting this all together, we have

Note that this agrees with the value given in Figure 10.20 .

Calculating the moment of inertia for compound objects

Now consider a compound object such as that in Figure 10.28 , which depicts a thin disk at the end of a thin rod. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object. However, if we go back to the initial definition of moment of inertia as a summation, we can reason that a compound object’s moment of inertia can be found from the sum of each part of the object:

It is important to note that the moments of inertia of the objects in Equation 10.21 are about a common axis . In the case of this object, that would be a rod of length L rotating about its end, and a thin disk of radius R rotating about an axis shifted off of the center by a distance L + R L + R , where R is the radius of the disk. Let’s define the mass of the rod to be m r m r and the mass of the disk to be m d . m d .

The moment of inertia of the rod is simply 1 3 m r L 2 1 3 m r L 2 , but we have to use the parallel-axis theorem to find the moment of inertia of the disk about the axis shown. The moment of inertia of the disk about its center is 1 2 m d R 2 1 2 m d R 2 and we apply the parallel-axis theorem I parallel-axis = I center of mass + m d 2 I parallel-axis = I center of mass + m d 2 to find

Adding the moment of inertia of the rod plus the moment of inertia of the disk with a shifted axis of rotation, we find the moment of inertia for the compound object to be

Applying moment of inertia calculations to solve problems

Now let’s examine some practical applications of moment of inertia calculations.

Example 10.11

Person on a merry-go-round.

Our goal is to find I total = ∑ i I i I total = ∑ i I i .

Significance

Example 10.12, rod and solid sphere.

• I total = ∑ i I i = I Rod + I Sphere I total = ∑ i I i = I Rod + I Sphere ; I Sphere = I center of mass + m Sphere ( L + R ) 2 = 2 5 m Sphere R 2 + m Sphere ( L + R ) 2 I Sphere = I center of mass + m Sphere ( L + R ) 2 = 2 5 m Sphere R 2 + m Sphere ( L + R ) 2 ; I total = I Rod + I Sphere = 1 3 m Rod L 2 + 2 5 m Sphere R 2 + m Sphere ( L + R ) 2 ; I total = I Rod + I Sphere = 1 3 m Rod L 2 + 2 5 m Sphere R 2 + m Sphere ( L + R ) 2 ; I total = 1 3 ( 2.0 kg ) ( 0.5 m ) 2 + 2 5 ( 1.0 kg ) ( 0.2 m ) 2 + ( 1.0 kg ) ( 0.5 m + 0.2 m ) 2 ; I total = 1 3 ( 2.0 kg ) ( 0.5 m ) 2 + 2 5 ( 1.0 kg ) ( 0.2 m ) 2 + ( 1.0 kg ) ( 0.5 m + 0.2 m ) 2 ; I total = ( 0.167 + 0.016 + 0.490 ) kg · m 2 = 0.673 kg · m 2 . I total = ( 0.167 + 0.016 + 0.490 ) kg · m 2 = 0.673 kg · m 2 .
• I Sphere = 2 5 m Sphere R 2 + m Sphere R 2 I Sphere = 2 5 m Sphere R 2 + m Sphere R 2 ; I total = I Rod + I Sphere = 1 3 m Rod L 2 + 2 5 m Sphere R 2 + m Sphere R 2 I total = I Rod + I Sphere = 1 3 m Rod L 2 + 2 5 m Sphere R 2 + m Sphere R 2 ; I total = 1 3 ( 2.0 kg ) ( 0.5 m ) 2 + 2 5 ( 1.0 kg ) ( 0.2 m ) 2 + ( 1.0 kg ) ( 0.2 m ) 2 I total = 1 3 ( 2.0 kg ) ( 0.5 m ) 2 + 2 5 ( 1.0 kg ) ( 0.2 m ) 2 + ( 1.0 kg ) ( 0.2 m ) 2 ; I total = ( 0.167 + 0.016 + 0.04 ) kg · m 2 = 0.223 kg · m 2 . I total = ( 0.167 + 0.016 + 0.04 ) kg · m 2 = 0.223 kg · m 2 .

Example 10.13

Angular velocity of a pendulum.

At the top of the swing: U = m g h cm = m g L 2 ( cos θ ) U = m g h cm = m g L 2 ( cos θ ) . At the bottom of the swing, U = m g L 2 U = m g L 2 with respect to the lowest point the rod swings.

At the top of the swing, the rotational kinetic energy is K = 0 K = 0 . At the bottom of the swing, K = 1 2 I ω 2 K = 1 2 I ω 2 . Therefore:

Solving for ω ω , we have

Inserting numerical values, we have

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18.7: Moments of Inertia via Composite Parts and Parallel Axis Theorem

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• Jacob Moore & Contributors
• Pennsylvania State University Mont Alto via Mechanics Map

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As an alternative to integration, both area and mass moments of inertia can be calculated via the method of composite parts, similar to what we did with centroids. In this method we will break down a complex shape into simple parts, look up the moments of inertia for these parts in a table, adjust the moments of inertia for position, and finally add the adjusted values together to find the overall moment of inertia. This method is known as the method of composite parts .

A key part to this process that was not present in centroid calculations is the adjustment for position. As discussed on the previous pages, the area and mass moments of inertia are dependent upon the chosen axis of rotation. Moments of inertia for the parts of the body can only be added when they are taken about the same axis . However, the moments of inertia in the table are generally listed relative to that shape's centroid. Because each part has its own individual centroid coordinate, we cannot simply add these numbers. We will use something called the Parallel Axis Theorem to adjust the moments of inertia so that they are all taken about some standard axis or point. Once the moments of inertia are adjusted with the Parallel Axis Theorem, then we can add them together using the method of composite parts.

The Parallel Axis Theorem

When we calculated the area and mass moments of inertia via integration, one of the first things we had to do was to select a point or axis we were going to take the moment of inertia about. We then measured all distances from that point or axis, where the distances were the moment arms in our moment integrals. Because the centroid of a shape is the geometric center of an area or volume, the average distance from the centroid to any one point in a body is at a minimum. If we pick a different point or axis to take the moment of inertia about, then on average all the distances in our moment integral will be a little bit bigger. Specifically, the further we move from the centroid, the larger the average distances become.

Though this complicates our analysis, the nice thing is that the change in the moment of inertia is predictable. It will always be at a minimum when we take the moment of inertia about the centroid, or an axis going through the centroid. This minimum, which we will call \(I_C\), is the value we will look up in our moment of inertia table. From this minimum, or unadjusted value, we can find the moment of inertia value about any point \(I_P\) by adding an an adjustment factor equal to the area times distance squared for area moments of inertia, or mass times distance squared for mass moments of inertia.

\[ I_{xxP} = I_{xxC} + A * r^2 \]

\[ I_{xxP} = I_{xxC} + m * r^2 \]

This adjustment process with the equations above is the parallel axis theorem . The area or mass terms simply represent the area or mass of the part you are looking at, while the distance (\(r\)) represents the distance that we are moving the axis about which we are taking the moment of inertia. This may be a vertical distance, a horizontal distance, or a diagonal depending on the axis the moment of inertia is taken about.

Say we are trying to find the moments of inertia of the rectangle above about point \(P\). We would start by looking up \(I_{xx}\), \(I_{yy}\), and \(J_{zz}\) about the centroid of the rectangle (\(C\)) in the moment of inertia table. Then we would add on an area-times-distance-squared term to each to find the adjusted moments of inertia about \(P\). The distance we are moving the \(x\) axis for \(I_{xx}\) is the vertical distance \(r_x\), the distance we are moving the \(y\)-axis for \(I_{yy}\) is the horizontal distance \(r_y\), and the distance we would move the \(z\)-axis (which is pointing out of the page) for \(J_{zz}\) is the diagonal distance \(r_z\).

Center of mass adjustments follow a similar logic, using mass times distance squared, where the distance represents how far you are moving the axis of rotation in three-dimensional space.

Using the Method of Composite Parts to Find the Moment of Inertia

To find the moment of inertia of a body using the method of composite parts, you need to start by breaking your area or volume down into simple shapes. Make sure each individual shape is available in the moment of inertia table, and you can treat holes or cutouts as negative area or mass.

Next you are going to create a table to keep track of values. Devote a row to each part that your numbered earlier, and include a final "total" row that will be used for some values. Most of the work of the method of composite parts is filling in this table. The columns will vary slightly with what you are looking for, but you will generally need the following.

• The area or mass for each piece (area for area moments of inertia or mass for mass moments of inertia). Remember that cutouts should be listed as negative areas or masses.
• The centroid or center of mass locations (\(x\), \(y\) and possibly \(z\) coordinates). Most of the time, we will be finding the moment of inertia about centroid of the composite shape, and if that is not explicitly given to you, you will need to find that before going further. For more details on this, see the page Centroids and Centers of Mass via Method of Composite Parts .
• The moment of inertia values about each shape's centroid. To find these values you will plug numbers for height, radius, mass, etc. into formulas on the moment of inertia table. Do not use these formulas blindly though, as you may need to mentally rotate the body, and thus switch equations, if the orientation of the shape in the table does not match the orientation of the shape in your diagram.
• The adjustment distances (\(r\)) for each shape. For this value you will want to determine how far the \(x\)-axis, \(y\)-axis, or \(z\)-axis moves to go from the centroid of the piece to the overall centroid or point you are taking the moment of inertia about. To calculate these values, generally you will be finding the horizontal, vertical, or diagonal distances between piece centroids and the overall centroids that you have listed earlier in the table. See the parallel axis theorem section of this page earlier for more details.
• Finally, you will have a column of the adjusted moments of inertia. Take the original moment of inertia about the centroid, then simply add your area times \(r^2\) term or mass times \(r^2\) term for this adjusted value.

The overall moment of inertia of your composite body is simply the sum of all of the adjusted moments of inertia for the pieces, which will be the sum of the values in the last column (or columns, if you are finding the moments of inertia about more than one axis).

Video \(\PageIndex{1}\): Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/zulGTSWF6xs .

Example \(\PageIndex{1}\)

Use the parallel axis theorem to find the mass moment of inertia of this slender rod with mass \(m\) and length \(L\) about the \(z\)-axis at its endpoint.

Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/4oN0kgDO3Yw .

Example \(\PageIndex{2}\)

A beam is made by connecting two 2" x 4" beams in a T-pattern with the cross section as shown below. Determine the location of the centroid of this combined cross section and then find the rectangular area moment of inertia about the \(x\)-axis through the centroid point.

Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/lgXlp2lRaiA .

Example \(\PageIndex{3}\)

A dumbbell consists of two spheres of diameter 0.2 meter, each with a mass of 40 kg, attached to the ends of a 0.6-meter-long slender rod of mass 20 kg. Determine the mass moment of inertia of the dumbbell about the \(y\)-axis shown in the diagram.

Video \(\PageIndex{4}\): Worked solution to example problem \(\PageIndex{3}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/ufewJ7CmvIs .

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Moment of Inertia

Moment of inertia, also known as rotational inertia or angular mass, is a physical quantity that resists a rigid body’s rotational motion . It is analogous to mass in translational motion . It determines the torque required to rotate an object by a given angular acceleration . Moment of inertia does not restrict itself to a rigid body only. It also applies to a system of particles rotating about a common axis.

How to Calculate Moment of Inertia

For a point mass (single body), the moment of inertia formula is given by the product of mass and the square of the object’s perpendicular distance from the axis of rotation.

I is the moment of inertia

m is the mass

r is the perpendicular distance from the axis of rotation

System of Particles

Suppose n particles of masses m 1 , m 2 , …, m n rotate about a common axis at perpendicular distances r 1 , r 2 , …, r n from the axis. Then the moment of inertia of the system is

The moment of inertia of a system of particles is the sum of the moments of inertia of the individual particles taken about a common axis.

Units and Dimensions

The SI unit of moment of inertia is kgˑm 2 , and the cgs unit is gˑcm 2 . The dimension is [M L 2 T 0 ].

Integral Moment of Inertia

For a continuous mass distribution, the integral form of moment of inertia is given by

Where dm is the instantaneous mass.

The above equation can also be written in terms of density ρ and instantaneous volume dV as follows.

Moment of Inertia Table

Several everyday objects, such as rotating disks, cylinders, and spheres, have well-defined moment of inertia formulas. A chart consisting of the different formulas is listed in the table below. Here M represents mass, R represents radius, and L represents the length.

Moment of Inertia and Rotational Kinematics

The moment of inertia of a rotating object about a fixed axis is useful in calculating a few key quantities in rotational motion. Newton’s second law for rotation gives a relationship between torque, moment of inertia, and angular acceleration . According to this law,

τ is the applied torque

α is the angular acceleration

The relationship between moment of inertia and rotational kinetic energy is given by

And with angular momentum , the equation is

K is the rotational kinetic energy

L is the angular momentum

ω is the angular velocity

Factors Affecting Moment of Inertia

The moment of inertia depends upon

• The shape of the object
• Mass distribution or density
• Location of the axis of rotation
• Moment of Inertia – Hyperphysics.phy-astr.gsu.edu
• Rotational Inertia – Khanacademy.org
• Moment of Inertia – Isaacphysics.org
• What Is Moment of Inertia in Physics? – Thoughtco.com
• Center of Mass; Moment of Inertia – Feynmanlectures.caltech.edu
• Calculating Moment of Inertia – Phys.libretexts.org

Article was last reviewed on Wednesday, August 2, 2023

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moment of inertia

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• Khan Academy - More on moment of inertia
• University of Central Florida Pressbooks - Calculating Moments of Inertia
• Academia - Moment of inertia
• BCcampus Open Publishing - Moment of Inertia and Rotational Kinetic Energy
• Physics LibreTexts - Moments of Inertia
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moment of inertia , in physics , quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed. The moment of inertia ( I ), however, is always specified with respect to that axis and is defined as the sum of the products obtained by multiplying the mass of each particle of matter in a given body by the square of its distance from the axis. In calculating angular momentum for a rigid body, the moment of inertia is analogous to mass in linear momentum . For linear momentum, the momentum p is equal to the mass m times the velocity v ; whereas for angular momentum, the angular momentum L is equal to the moment of inertia I times the angular velocity ω.

The figure shows two steel balls that are welded to a rod AB that is attached to a bar OQ at C . Neglecting the mass of AB and assuming that all particles of the mass m of each ball are concentrated at a distance r from OQ , the moment of inertia is given by I = 2 mr 2 .

The unit of moment of inertia is a composite unit of measure. In the International System (SI), m is expressed in kilograms and r in metres, with I (moment of inertia) having the dimension kilogram-metre square. In the U.S. customary system, m is in slugs (1 slug = 32.2 pounds) and r in feet, with I expressed in terms of slug-foot square.

The moment of inertia of any body having a shape that can be described by a mathematical formula is commonly calculated by the integral calculus . The moment of inertia of the disk in the figure about OQ could be approximated by cutting it into a number of thin concentric rings, finding their masses, multiplying the masses by the squares of their distances from OQ , and adding up these products. Using the integral calculus, the summation process is carried out automatically; the answer is I = ( mR 2 )/2. (See mechanics ; torque .)

For a body with a mathematically indescribable shape, the moment of inertia can be obtained by experiment. One of the experimental procedures employs the relation between the period (time) of oscillation of a torsion pendulum and the moment of inertia of the suspended mass. If the disk in the figure were suspended by a wire OC fixed at O , it would oscillate about OC if twisted and released. The time for one complete oscillation would depend on the stiffness of the wire and the moment of inertia of the disk; the larger the inertia, the longer the time.

• Moment of Inertia

Moment of inertia also known as the angular mass or rotational inertia can be defined w.r.t. rotation axis, as a quantity that decides the amount of torque required for a desired angular acceleration or a property of a body due to which it resists angular acceleration. The formula for the moment of inertia is the “sum of the product of mass” of each particle with the “square of its distance from the axis of the rotation”. The formula of Moment of Inertia is expressed as I = Σ m i r i 2 .

What is Moment of Inertia

The moment of inertia of an object is a determined measurement for a rigid body rotating around a fixed axis. The axis might be internal or external, and it can be fixed or not. However, the moment of inertia (I) is always described in relation to that axis.

The moment of Inertia depends on the distribution of the mass around its axis of rotation. MOI varies depending upon the position of the axis that is chosen. That is, depending on the location and direction of the axis of rotation, the same item might have various moment of inertia values.

Angular mass or rotational inertia are other names for the moment of inertia.

The SI unit of moment of inertia is kg m 2 .

Suggested Videos

Moment of inertia example.

Imagine you are on a bus right now. You find a seat and sit down. The bus starts moving forward. After a few minutes, you arrive at a bus stop and the bus stops. What did you experience at this point? Yes. When the bus stopped, your upper body moved forward whereas your lower body did not move.

Why is that? It is because of Inertia .  Your lower body is in contact with the bus but your upper body is not in contact with the bus directly. Therefore, when the bus stopped, your lower body stopped with the bus but your upper body kept moving forward, that is, it resisted change in its state.

Similarly, when you board a moving train, you experience a force that pushes you backwards. That is because before boarding the train you were at rest. As soon as you board the moving train, your lower body comes in contact with the train but your upper body is still at rest. Therefore, it gets pushed backwards, that is, it resists change in its state.

Understand the Theorem of Parallel and Perpendicular Axis here in detail.

Browse more Topics Under System Of Particles And Rotational Dynamics

• Introduction to Rotational Dynamics
• Vector Product of Two Vectors
• Centre of Mass
• Motion of Centre of Mass
• Theorems of Parallel and Perpendicular Axis
• Rolling Motion
• Angular Velocity and Angular Acceleration
• Linear Momentum of System of Particles
• Torque and Angular Momentum
• Equilibrium of a Rigid Body
• Angular Momentum in Case of Rotation About a Fixed Axis
• Dynamics of Rotational Motion About a Fixed Axis
• Kinematics of Rotation Motion about a Fixed Axis

What is Inertia?

What is Inertia? It is the property of a body by virtue of which it resists change in its state of rest or motion. But what causes inertia in a body? Let’s find out.

Inertia in a body is due to its mass. More the mass of a body more is the inertia. For instance, it is easier to throw a small stone farther than a heavier one. Because the heavier one has more mass, it resists change more, that is, it has more inertia.

Moment of Inertia Definition

So we have studied that inertia is basically mass. In rotational motion, a body rotates about a fixed axis. Each particle in the body moves in a circle with linear velocity, that is, each particle moves with an angular acceleration. Moment of inertia is the property of the body due to which it resists angular acceleration, which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

Moment of Inertia Formula

Moment of Inertia (I) = Σ m i r i 2

m = Sum of the product of the mass.

r = Distance from the axis of the rotation.

And the Integral form of MOI is as follows:

I = ∫ d I = ∫ 0 M  r 2  dm

dm = The mass of an infinitesimally small component of the body

r = (perpendicular) distance between the point mass and the axis of rotation

Moment of Inertia of a System of Particles

For a system of point particles revolving about a fixed axis, the moment of inertia is:

where r i is the perpendicular distance from the axis to the i th particle which has mass m i .

A system of point particles is shown in the following figure. Each particle has a mass of 0.3 kg and they all lie in the same plane. What is the moment of inertia of the system about the given axis?

The mass of all the particles is equal. Hence, m = 0.3 kg

I = Σ m i r i 2 = m Σ r i 2 = 0.3 [0.6 2 + 0.4 2 + 0.2 2 ]  …..(Converting the distance of the particles to metre)

I = 0.168 kg m 2

Moments of Inertia for Different Objects

• For Cylinder

• For Rectangular objects

• For Ring/Disc

The moment of inertia, as we all know, is affected by the axis of rotation. After selecting two distinct axes, you will notice that the object resists the rotational change differently. As a result, the following theorems can be used to calculate the moment of inertia along any given axis:

Parallel Axis Theorem

• Perpendicular Axis Theorem

The Parallel Axis Theorem states,

The moment of inertia of a body about an axis parallel to the body passing through its centre is equal to the sum of moment of inertia of the body about the axis passing through the centre and product of the mass of the body times the square of the distance between the two axes.

Its formula is,

I = I c + Mh 2

I = moment of inertia of the body

I c = moment of inertia about the center

M = mass of the body

h 2 = square of the distance between 2 axes

Let us see how the Parallel Axis Theorem helps us to determine the moment of inertia of a rod whose axis is parallel to the axis of the rod and it passes through the center of the rod.

Moment of inertia of rod is given as:

I = (1/3) ML 2

The distance between the end of the rod and its centre is:

Therefore, the parallel axis theorem of the rod is:

I c = (1/3) ML 2 – M(L/2) 2

I c = (1/3) ML 2 – ML 2 /4

I c = (1/12) ML 2

Kinetic Energy in Rotational Motion

What is the analogue of mass in rotational motion? To answer this question, we have to derive the equation of kinetic energy in rotational motion.

Consider a particle of mass m  at a distance from the axis with linear velocity = v i = r i ω.  Therefore, the kinetic energy of this particle is,

k i = 1/2 m i v i 2 = 1/2  m i (r i  ) 2 ω 2

where  m i is the mass of the particle. The total kinetic energy K of the body is thus the sum of the kinetic energies of individual particles. ∴ K = Σ k i  = 1/2 ( Σ  m i (r i  ) 2 ω 2 )   where  n  is the number of particles in the body. We know that angular acceleration  ω is the same for all particles. Therefore, taking  ω out of the sum, we get,

K = 1/2 ω 2  (Σ  m i (r i  ) 2  )  ………………….(I)

Let Σ  m i (r i  ) 2   be I which is a new parameter characterising the rigid body known as the Moment of Inertia .   Therefore, with this definition,

K = 1/2I ω 2  …………….. (II)

The parameter I is independent of the magnitude of the angular velocity. It is the characteristic of the rigid body and the axis about which it rotates. We already know that linear velocity in linear motion is analogous to angular acceleration in rotational motion.

On comparing equation II with the formula of the kinetic energy of the whole rotating body in linear motion, it is evident that mass in linear motion is analogous to the moment of inertia in rotational motion. Hence, the question is answered. In simple words, the moment of inertia is the measure of the way in which different parts of the body are distributed at different distances from the axis.

• Radius of Gyration

As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we define a new parameter known as the radius of gyration . It is related to the moment of inertia and the total mass of the body. Notice that we can write I = M k 2  where  k has the dimension of length.

Therefore, the radius of gyration is the distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis. Therefore, the moment of inertia depends not only on the mass, shape, and size of the body but also on the distribution of mass in the body about the axis of rotation.

Learn more about Torque and Angular Momentum here .

Here are a few Solved Questions for You

Question 1. The moment of inertia depends on:

a) mass of the body                                                     b) shape and size of the body

c) distribution of mass about the axis of rotation       d) all of the above

Answer:  d) all of the above. All the factors together determine the moment of inertia of a body.

Question 2:  Find the moment of inertia of a disc of mass 5 kg and radius 30 cm about the following axes.

• Axis passing through the center and perpendicular to the plane of the disc.
• Axis touching the edge and perpendicular to the plane of the disc.

Given: Mass (M) = 5 kg, Radius (R) = 30 cm = 0.3 m

Question 3:  Determine the moment of inertia about the geometric centre of the given construction, which consists of a single thin rod connecting two identical solid spheres as shown in the figure.

The above image consists of 3 structures: 1 thin rod and 2 solid spheres.

The mass of the rod (M) = 3 kg, total length of the rod (l) = 80 cm = 0.8 m

The moment of inertia of the rod about its center of mass is,

I rod = (1/12) x Ml 2 = (1/12) x 3 x (0.8) 2 = 0.16 kg m 2

The mass of the sphere (M) = 5 kg, radius of the sphere (R) = 10 cm = 0.1 m

The moment of inertia of the sphere about its center of mass is,

I c = (2/5) MR 2

But the moment of inertia of the sphere about the geometric center of the structure is,

I sph = I c + Md 2

where, d = 40 cm + 10 cm = 50 cm = 0.5 m

I sph = (2/5) MR 2 + Md 2

I sph = (2/5) x 5 x (0.1) 2 + 5 x (0.5) 2 = 1.27 kg m 2

As there is one thin rod and 2 similar solid spheres,

Total Moment of Inertia (I total )= I rod + (2 x I sph )

I total = (0.16) x (2 x 1.27) = 2.7 kg m 2

Hence, the moment of inertia of the given construction figure is 2.7 kg m 2 .

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JSmol Viewer

Dynamic characteristic analysis of underwater suspended docking station for resident uuvs.

1. Introduction

• This study proposes an SMSD model using a hybrid dynamic overlapping grid technique through DFBI, featuring a guiding funnel, a suspended body, and a catenary. Moreover, a UUV dock within the SMSD model is developed with contact coupling. These models successfully simulate the SMSD’s equilibrium stability and dynamic contact response under ocean flow conditions.
• We analyze the effects of mass, catenary stiffness, and flow velocity on SMSD stability. In addition, an investigation is conducted regarding how volume, moment of inertia, mass, and catenary stiffness impact UUV docking. This improves understanding of flexible mooring methods and contact interactions, guiding the optimization of SMSD design.
• Through sea trials and numerical simulation results, we verify the stability and contact reliability of the suspended moored structure of the SMSD. The well-considered design enhances the SMSD’s adaptability to environmental conditions and stability during UUV docking.

2. Numerical Methodology

2.1. catenary equations, 2.2. contact coupling and forces, 2.3. dfbi motion, 3. numerical model, 3.1. calculation model, 3.2. grid division, 3.3. numerical model validation, 4. results and discussion, 4.1. attitude of smsd, 4.1.1. effects of the smsd mass, 4.1.2. effects of the catenary stiffness, 4.2. motion response during uuv docking process, 4.3. effects of various design parameters on uuv docking, 4.3.1. effects of the smsd volume, 4.3.2. effects of the moment of inertia of the smsd, 4.3.3. effects of the smsd mass, 4.3.4. effects of the catenary stiffness, 4.3.5. discussion.

• Docking time is primarily influenced by volume and mass.
• Contact force is mainly affected by volume and the moment of inertia.
• The number of contacts is sensitive to volume and mass; it first decreases, then stabilizes as the moment of inertia and catenary stiffness increase.
• The angle between the UUV’s central axis and the guiding funnel’s central axis on the horizontal plane is negatively correlated with volume and the moment of inertia and positively correlated with mass; the influence of catenary stiffness is negligible.
• The variation of the SMSD yaw angle is mainly influenced by volume and the moment of inertia (negatively correlated with both), and volume is satisfied within a specific range of conditions. High volume causes large contact forces, making the SMSD yaw angle unstable.
• The variation of the UUV yaw angle is mostly affected by volume and mass, with a negative correlation with mass.

5. Sea Experiment

6. conclusions.

• The pitch of the SMSD should be kept within 10°. Increasing the net buoyancy force and incorporating stabilizer fins can improve SMSD stability and reduce the pitch angle of the SMSD.
• Low stiffness in the mooring chain facilitates the SMSD in quickly reaching equilibrium, resulting in a stable state.
• The SMSD’s mooring connection allows for flexibility, absorbing the contact impacts and resulting in low UUV angular velocity, high UUV velocity, and reduced contact force, lowering damage risk.
• The key parameters affecting UUV docking, in order of impact, are volume, moment of inertia, and mass. Catenary stiffness is negligible.
• Optimal design parameters involve a careful trade-off. Moderate volume balances docking efficiency and UUV navigation. Increased mass enhances both docking efficiency and SMSD stability. A suitable moment of inertia maintains SMSD flexibility, aiding docking accuracy and safety. High catenary stiffness increases contact force.

Author Contributions

Institutional review board statement, data availability statement, conflicts of interest.

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Guo, J.; Meng, L.; Feng, M.; Liu, J.; Peng, Z.; Feng, W.; Cui, J.-H. Dynamic Characteristic Analysis of Underwater Suspended Docking Station for Resident UUVs. J. Mar. Sci. Eng. 2024 , 12 , 1493. https://doi.org/10.3390/jmse12091493

Guo J, Meng L, Feng M, Liu J, Peng Z, Feng W, Cui J-H. Dynamic Characteristic Analysis of Underwater Suspended Docking Station for Resident UUVs. Journal of Marine Science and Engineering . 2024; 12(9):1493. https://doi.org/10.3390/jmse12091493

Guo, Jingqian, Lingshuai Meng, Mengmeng Feng, Jun Liu, Zheng Peng, Wei Feng, and Jun-Hong Cui. 2024. "Dynamic Characteristic Analysis of Underwater Suspended Docking Station for Resident UUVs" Journal of Marine Science and Engineering 12, no. 9: 1493. https://doi.org/10.3390/jmse12091493

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