22A: Center of Mass, Moment of Inertia - Physics LibreTexts
Using a string through a tube, a mass is moved in a horizontal circle with of force, a circular orbit can be produced by a force acting toward the center. That point mass relationship becomes the basis for all other moments of inertia since. The center of mass of a boat must be low enough for the boat to be stable. . center of mass for the rod above will have different coordinates, but it will always be. Moving the masses closer to the center reduces the moment of inertia, which increases the angular acceleration. For a cylinder rotating about its center-of- mass, where the rotation axis The pulley gets us the third equation we need.
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. 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.
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.
The Moment of Inertia of Two Balls — Collection of Solved Problems
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. In fact, it becomes more and more accurate as the scale gets larger and larger. We would never discover the fundamental laws of the atomic particles at first observation because the first observations are much too crude. It is like nothing we are familiar with because there is nothing like it. 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.
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. 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.
10.5: Calculating Moments of Inertia
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. 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.Physical Pendulum Problems - Moment of Inertia - Physics
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: 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: 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.
PhysicsLAB: Discrete Masses: Center of Mass and Moment of Inertia
Calculation of the moment of inertia I for a uniform thin rod about an axis through the center of the rod. 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.
- 22A: Center of Mass, Moment of Inertia
- List of moments of inertia
- Rotational inertia
We therefore need to find a way to relate mass to spatial variables. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure.
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. 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. Calculating the moment of inertia for a thin disk about an axis through its center. 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. This cannot be easily integrated to find the moment of inertia because it is not a uniformly shaped object. Compound object consisting of a disk at the end of a rod.
The merry-go-round can be approximated as a uniform solid disk with a mass of kg and a radius of 2. Find the moment of inertia of this system. Calculating the moment of inertia for a child on a merry-go-round. Strategy This problem involves the calculation of a moment of inertia. We are given the mass and distance to the axis of rotation of the child as well as the mass and radius of the merry-go-round.
Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass.
Torque Formula (Moment of Inertia and Angular Acceleration)
Rod and Solid Sphere Find the moment of inertia of the rod and solid sphere combination about the two axes as shown below. The rod has length 0. The radius of the sphere is Strategy Since we have a compound object in both cases, we can use the parallel-axis theorem to find the moment of inertia about each axis.