The Feynman Lectures on Physics Vol. I Ch. Center of Mass; Moment of Inertia
The moment of inertia, otherwise known as the angular mass or rotational inertia, of a rigid 9 See also; 10 References; 11 External links . This simple formula generalizes to define moment of inertia for an arbitrarily can be defined for an axis through its center of mass, with such a value that its moment of inertia is. The first is what comes out if you pivot about the center of mass with x You can use this method for things that vary linearly with distance (such as finding the torque, mass into the centre of mass throws out a lot of moment of inertia. Relation between inertia tensor and moment of inertia about an axis. Intuitively, this is because the mass is now carrying more momentum with it .. of simple shapes for which there exists a known equation for rotational inertia. is that the equations tell us the rotational inertia as found about the centroid of the.
However, in developing the ideas involved we need to assume a gravitational field and will speak of the center of gravity.
In developing the ideas for computing the location of the center of gravity we will view a body as an assemblage of individual particles. The earth exerts an attraction on each individual particle of a body and the weight of a body is the sum total of all the forces on all the particles making up the body.
Thus we will consider the problem of finding the location of the center of gravity for assemblages of particles in space. Consider a steel rod resting on a pivot as shown in Fig. If the pivot is directly below the center of gravity, the rod is balanced, and the sum of all the clockwise moments from particles to the right of the pivot is equal to the sum of all the counterclockwise moments from the particles to the left of the pivot.
The upward force F exerted by the pivot on the rod, as shown in Fig.
Center of Gravity and Newton's Second Law in Rotation
Now let us consider the situation in which the pivot is at the left end of the rod as shown in Fig. Suppose an upward force F is exerted directly below the center of gravity and equal in magnitude to the weight of the rod. Then the rod will be balanced, no force will be exerted on the pivot, and the sum of all of the clockwise moments from the particles of the rod will be equal to the counterclockwise moment Fd produced by force F where d is the distance from the pivot to the center of gravity, as shown in the figure.
Thus if we were able to compute the clockwise moments of the particles of the rod with respect to the pivot point and knew the weight of the rod, we would then be able to compute the distance d from the pivot to the center of gravity. Consider now a system of n point masses situated along a horizontal line, as shown in Fig. Suppose the weights of the masses are w1, w2, w3, Let M1 be the sum of the clockwise moments of the n masses with respect to the pivot point.
Then the counterclockwise moment M2 due to F is given by Fd where d is the distance from the pivot to the center of gravity. We lay an x-axis of a coordinate system along the rod with the origin at the pivot.Calculus 3 Lecture 14.4: Center of Mass (and Moments of Mass and Inertia) for Lamina in 2-D
We consider the rod to consist of an assemblage of particles m1, m2, The weight W of the rod is given by Since mass is proportional to weight, 3 can also be expressed in terms of mass.
In terms of mass it is where x is the distance of the infinitesimal element of mass dm from the pivot. We have considered a one-dimensional case where masses are distributed out in a line, as in a rod.
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.
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: 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.
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.
A right triangle and a right circular cone generated by rotating the triangle. This generates a cone. 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. 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. Spin it around this edge as axis to generate a sphere. 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. Note that it is not the three-dimensional distance, only the two-dimensional distance squared, even for a three-dimensional object.
As a simple example, consider a rod rotating about a perpendicular axis through one end Fig.
Mass, center of mass, and moment of inertia
But let us notice a few things about the moment of inertia. 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.