B9. Motion on precribed trajectory
A moment of inertia corresponds to each axis of rotation
The rotation of the same rigid straight line with m1. m2, m3 is now to take place with the angular acceleration w about the axis A'A'. The masses m1. m2, m3 have from this axis the distances r1, r2,r3. Then, analogously, so that the rigid straight line with m1. m2, m3 as it rotates about the axis A'A' will attain at the end of unit time the angular velocity w, the force
must be applied at the distance of 1 cm from the axis A'A'. If you place the results relating to the axes A and A' side by side, you note that the difference between the two motions is expressed, respectively, by
The same mass points m move in both cases, but about different axes with the distances r and r. The difference between these sums manifests the difference of the grouping of the mass points of the same body with respect to different axes of rotation.
In the case of the rigid line in Fig. 107, the sum only involves three mass points, in the case of a body with arbitrary mass content, you must form the sum over all mass points and always for that axis about which the body then rotates. This expression is the moment of inertia of the body with respect to the instantaneous axis of rotation.
An example: If you want to shift the disk S (Fig. 108) as a whole, you are only interested in the number of kilograms it contains. However, if you want to keep two points of it fixed and rotate it about the line through these points, you are also interested in which two points are to be held fixed - in other words, you also want to know how its M kilograms are distributed around the axis. They are distributed differently about each of the axes AA' and BB'. If the radius of the disk is r cm, then, for example, in both cases the largest distance which a point can have from the axis is r cm. In the case of the axis AA', only two points (**) have the largest distance - we ignore the thickness of the disk! - , but in the case of the axis BB', all points on the rim have this distance. While in the first case, as many mass points have the distance zero from the axis. that is, remain at rest, as has one diameter (2r) of the disk, in the second case many fewer points remain at rest. In other words, as regards the distribution of the mass about the axis, the disk in the first cases is quite a different body from that in the second case. This fact is expressed by the moment of inertia Smr2 - we will omit the calculation - which is M·r2/4 in the first and M·r2/2 in the second case, that is, twice as large.
What is the meaning of these numbers? Answer: Set the disk into rotation by application of a force at 1 cm from its axis; it will have a certain angular acceleration. If you then imagine its mass replaced by a single mass point at the distance of 1 cm from the axis, you must - in order that the force will give this mass point the same angular acceleration, which it gives the disk - have at this point in the first case the mass M·r2/4, in the second case the mass M·r2/2 ; for example, if M = 5 kg and r = 5cm, you find in the first case 31,25 kg, in the second case 62.5 kg.
Parallel displacement of the axis of rotation
The moment of inertia of a body about every axis is related simply to that about a parallel axis through its centre of gravity.
For the sake of simplicity, imagine the body to be a thin disk (Fig. 109). It rotates about the axis through O, perpendicular to the plane of the drawing. Denote the distances of the mass points from this axis by r; its moment of inertia with respect to it is Tr = Smr2. If the body rotates about the parallel axis through its centre of gravity and you denote the distances of the mass points from the centre of gravity axis by r, then its moment of inertia about this axis is Tr = Smr2. The figure shows that r2 = r2 + a2 + 2ax, whence Smr2 = Smr2 + Sma2 + 2aSmx. However, Smx =0, because the distances x relates to the centre of gravity and the sum of the linear moments mx about each plane through the centre of gravity vanishes. Hence Tr = T r+ Ma2, that is, the moment of inertia about any axis A equals the moment of inertia about its parallel axis through the C.G. plus the moment of inertia of the total mass M, about the axis A. Hence the moment of inertia about the axis through the centre of gravity is the smallest.
The disk in Fig. 108 can rotate about an axis through the centre of gravity. If you imagine that the axis has been shifted to the dotted line, the moment of inertia increases by Mr2, so that in the first case Mr2/4 becomes 5Mr2/4, in the second case Mr2/2 becomes 3Mr2/2, that is, it becomes 5 times as large in the first case and 3 times as large in the second case.
Among all the axes passing through a given point there is one for which the moment of inertia is a maximum, and another one, perpendicular to it, for which it is a minimum. These axes and a third axis, perpendicular to them, are called the principal axes of inertia of the body with respect to the given point; the corresponding moments of inertia are the principal moments of inertia. If you plot (Louis Poinsot 1777-1859) the reciprocal square roots (T)-1/2 of the moments of inertia T of a body with respect to all axes passing through a given point as segments to both sides of the corresponding axis, the end points of these segments lie on an ellipsoid, called the ellipsoid of inertia with respect to that point as centre (Fig. 110). The ellipsoid of inertia, the centre of which is the centre of gravity, is called the central ellipsoid of inertia, its axes and moments are the principal axes of inertia and the principal moments of inertia of the centre of gravity. Corresponding to the theorem of Jakob Steiner 1796-1863, according to which the moment of inertia about an axis through the centre of gravity is the smallest of all moments of inertia with respect to all axes parallel to it, the reciprocal value (T)-1/2 has its maximum, and so has the ellipsoid about it. The ellipsoids become smaller the further away the reference point lies from the centre of gravity (Fig. 110). - It can happen that the ellipsoid, corresponding to some point, is an ellipsoid of rotation; then all lines through its centre, which are perpendicular to the axis of rotation of the ellipsoid, are equal and can therefore be considered to be principal axes. If the ellipsoid of inertia happens to be a sphere, every line through its centre is a principal axis of inertia. - A principal axis of inertia of the centre of gravity is also for every other point of the body on it a principal axis of inertia.
You can image the moment of inertia Smr2 of a body with the total mass Sm = M to have been replaced by the moment of inertia of a ring which - with the mass m and radius r - has the same moment of inertia with respect to the same axis. Then the product mr2 is constant and equals Smr2. But m differs depending on the choice of r (and r depending on the choice of m). The mass which you obtain for a given r is called reduced mass. That value of r for which the reduced mass equals M is called the radius of gyration and denoted by k, whence always Smr2 = M·k2 - the same consideration lets you imagine that the total mass has been joined at one point at the distance k from the axis.
Equation of motion of a rotating body
Let the place of the rigid line (Fig. 107) be taken by an arbitrary body which can rotate about a given axis . If it has with respect to this axis the moment of inertia Smr2 and it is to acquire in unit time at uniform acceleration the angular acceleration w by application of a force R at the distance a from the axis of rotation, you must satisfy the condition:
This equation does for rotation what the equation P = M·v does for displacement; their similarity explains the significance of the moment of inertia: If the force which gave the body the angular acceleration w (about the axis with the moment of inertia Smr2) is applied at unit distance from the axis, then you must have R = w ·Smr2. The analogy with the expression P = v·M becomes now clearer: Each, R1,P and w , v are now forces and accelerations, respectively. A force equals mass times an acceleration, and the formula P = M·v tells just that. This lets us also view Smr2 as a symbol for a mass of magnitude Smr2. Denote it by m, whence R1 = w·m, that is, the force has given the mass m the path velocity w during unit time.
However, this path velocity w has one mass point only, if it lies 1 cm from the axis. Hence we must imagine the mass m concentrated and the force to be attacking there; we can then interpret Smr2 as a mass which is concentrated at a point at distance 1 cm from the axis and rigidly attached to it. The symbol Smr2 is used for problems of rotation in the same way as the symbol Sm for problems of displacement. - The unit of the moment of inertia about a given axis has the point mass 1 g at the distance 1 cm from the axis and the dimensional formula: [l2·m].
Rotary impulse. Impulsive moment
The same considerations, which led us earlier
to the concept of impulse, take us here to rotary impulse and impulsive
moment. We understood by the angular
acceleration w
the angular velocity which was reached at the
end of 1 second from a state of rest. If you replace w by the ratio of the in an arbitrary time interval t attained
angular velocity O to the length of this time interval,
that is, by O/t and let 
to be
of the order of magnitude of an instant, then the rotatory body
receives an instantaneous rotary impulse.
The impulsive moment of a mass point m about a point of reference is called the product of the impulse m·v and the lever arm r of the impulsive vector (perpendicular from the point of reference to the line of the impulsive vector). You call the sum of the impulsive moments of all its individual mass points about this point of reference the impulsive momentum of a mass system (body) about a point of reference Applied to the rotation of a rigid body about an axis through the point of reference, the impulsive moment is called a rotary impulse.The vector of the rotary impulse is the rotary impulse which you must give the body about the axis in the direction of the rotation and its strength is such that it brings it instantaneously from rest to its present angular velocity (or which the body would exercise, if you were to stop it abruptly). The rotary impulse is the motor force, inherent in the body.
The impulsive force gives to the freely movable mass point m in the very short instant t the impulse mv, that is, to the about a fixed axis rotatable mass point m at the distance r from the axis of rotation the impulsive moment mv·r=mr·w·r=mr2w. If you now proceed from a single mass point to a system of such points - a body of mass M - you obtain the impulse Mv and the impulsive moment (rotary impulse) M'w, where M' is the moment of inertia of the body with respect to the axis of rotation. The impulse M manifests itself on the freely movable mass completely as a velocity, the rotary impulse M'w on the about a fixed axis rotatable mass only then, if the axis is a central one, otherwise only partly as rotation of the mass, partly as pressure of the mass on the axis.
Conservation of angular momentum
If no external force act on a rotating body, it rotates with constant angular velocity, that is, wSmr2 remains constant (during a shift of the mass m, the product v·m remains constant). However, what will happen if its mass points shift with respect to each other during the rotation? (This question could not arise in the case of a rigid body, as the mass depended then only on the number of mass points and not on their relative positions). When mass points in a rotating body shift with respect to each other, their distances from the axis change, that is, then its moment of inertia changes. What is the consequence for the rotating body?
The simplest case is shown in Fig. 100. The mass point m travels about a centre; it changes during the motion its distance r from it and simultaneously its velocity; these two changes are interconnected in that during each time unit its radius vector covers a constant area. You say: The area velocity of the circulating mass point remains constant (Theorem of conservation of angular momentum). The closer the mass point is to the centre, the faster it travels, the further it is away from it, the more slowly it travels - its angular velocity changes in such a manner that it is at each instant inversely proportional to the square (r2) of its distance from the centre, whence, if w and w1 denote the angular velocities and r and the corresponding distances of two points from the axis, then: w /w 1 = r2/r12.
Hence also the angular velocity is at each instant inversely proportional to its moment of inertia (the circulating mass consists only of a single mass point!). Using the same notation and denoting the point's mass by m, one has w · r2 = w1·r12. Thus, the product of the instantaneous angular velocity and the instantaneous moment of inertia is constant. - This theorem (area theorem) is the simplest case of the theorem of conservation of angular momentum, which essentially is for the rotation of a body what the theorem of inertia is for the translation of a body.
What holds here for a single mass point which travels with changing distance around a centre, can also be proved for a similar system of mass points. Its angular velocity and its moment of inertia vary inversely proportional to one another, so that their product remains constant. This is shown by the Isotomeograph (Fig. 112): A symmetric beam in horizontal bifilar suspension, along which, symmetrically with respect to its suspension, two equal supplementary masses may be moved. Both of these can be shifted simultaneously inwards from the ends of the beam to its centre or outwards from the centre. The beam, although relative to its environment at rest, has angular velocity, since its environment takes part in Earth's rotation about its axis. (If it were to hang over the North Pole, when its suspension would be in the direction of Earth's axis, its angular velocity would be the same as that of Earth; at Latitude j it only has the corresponding component.)
To start with, let the beam with
the supplementary masses at its ends hang at rest. Let w denote
its angular velocity and J its moment of inertia so that
its impulsive momentum is Jw. Shift suddenly
the masses towards the the centre of the beam, when the moment of
inertia becomes suddenly J'. The angular velocity then
becomes w'= (J/J')w, since the product
of the angular velocity and moment of inertia must remain
constant, that is, J
. Thus, the beam takes suddenly from its environment
an angular velocity different from that of its environment,
whence it turns at the first instant clearly noticeably relative to its environment. The rotation is immediately influenced by the
reactive moment of the bifilar suspension and causes
horizontal rotary oscillations. J.G.Hagen 1847-1930 has used this
apparatus with success (1910, 1919) to prove the existence of
Earth's rotation.
In an arrangement, which is similar to that of the Isotomeograph, you can sense on your body the action of conservation of angular momentum. Place yourself at the centre of a disk, rotating uniformly about a vertical axis, hold your arms stretched out away from your body with weights in your hands; as you now let the weights come closer to your body by bending your arms, you will sense an acceleration of the rotation.
The equation: Instantaneous angular velocity times instantaneous moment of inertia = const for a rotating body, not subject to external forces, is clearly analogous to the theorem of conservation of impulse (mv) of a mass m, which solely following its inertia translates into its instantaneous velocity.
Pendulum. Mathematical pendulum
So far, we have excluded the effect of gravity on the rotating body by letting it rotate about a vertical axis at rest. The rotation of a body about a resting, inclined axis also allows to a certain extent elementary treatment. If the axis is inclined (Fig. 113), one component of the gravitational force acts perpendicularly to the axis and can rotate the body. The component is the greater, the closer the angle between the direction of the axis and that of gravity is to a right angle. (horizontal pendulum)
Now place the axis horizontally, so that gravity acts perpendicular to it. It cannot cause rotation, as it intersects the axis, its point of attack, the centre of gravity S of the body, lies on the axis (Fig.114 (a)) or perpendicularly below it (Fig.114 (b)) or above it (Fig.114 (c)). In the case (a), the equilibrium is indifferent, in (c) unstable, in (b) the centre of gravity lies perpendicularly below the axis, that is, at the deepest point it can occupy, If then the body is at rest, only gravity acts on it, that is, it stays at rest. Gravity alone can not initiate rotation.
However, if the body is turned sidewards by another force (shown by a dotted line) and then released, it becomes subject to the rotating action of gravity, because now (dotted line) its line of action no longer intersects the axis. Therefore the centre of gravity tends to return to its deepest position, arrives there with a certain velocity and continues beyond due to its inertia, loses gradually its velocity, comes to a halt and again falls towards it position of rest, passes it , etc. - it oscillates. All other points of the pendulum do the same on parallel circular arcs.
A body which is rotatable about a horizontal axis under the action of gravity (Fig. 114 b) is called a physical or a compound pendulum in contrast to the mathematical or simple pendulum, that is, a mathematical concept by which the law of pendulum oscillations can be formulated. Imagine the mass of the physical pendulum concentrated at its centre of gravity S and through it a rigid, weightless straight line SC, which represents its perpendicular distance from the axis and links it to the axis; this is the mathematical pendulum (Fig. 115) with all characteristic details of a pendulum.
The rigid line SC allots to the mass point S its distance from the axis and forces it into a motion along a circle about the horizontal axis, that is, along a vertically oriented circular arc with radius SC. You can also conceive the simple pendulum as shown in Fig. 115 and Fig. 116 where the latter shows a body rolling to and fro in a circular grove of radius SC.
Assume that the body has been lifted to S1 and then released. It falls back towards its position of rest, arrives there with a certain velocity and rises to a position beyond. How far will it rise? Obviously, this will depend on the velocity at which it arrives at S, It passes on the path from S1 to S the straight elements of the curve, out of which you can imagine every curved line to consist; that is, at each instant, it moves along an inclined plane the inclination of which with respect to the base ST1 is given by the tangent at its position. The velocity at which a point arrives at the base of an inclined plane depends only on the height of the plane and equals the velocity at which it arrives from its height at the horizontal plane in free fall. The pendulum passes along a row of inclined planes which continue each other and have the total height . The velocity at which it arrives at S is therefore equal to the velocity which it would have attained during a free fall through the distance S1T1. (During the transition from one inclined plane to another the direction changes, but this change is caused continuously by the centripetal force and therefore does not affect the velocity of the pendulum). This velocity is sufficient to lift it by the same distance S2T2 = S1T1. If there is no friction, the pendulum would never come to rest. The only points, at which it could stay at rest, lie vertically above the axis - which it cannot reach, because it cannot rise higher on the other side than the height from which it started - and vertically below the axis - which it always reaches at a non-zero velocity, so that it must pass through it. However, a frictionless pendulum is only a mathematical concept.
Since the pendulum swings equally far to both sides, that is, the height it climbs to equals that from which it falls, the period during which it falls equals that during which it climbs, that is, it takes the same time to travel from S1 to S as from S to S2 or from S2 to S or from S to S1. The path between the points of return is called the amplitude , the time taken to travel from S1 to S2 the period of the vibration. Since it is always the same, the pendulum is used to measure time; you count the number of times the pendulum travels between its return points during the time you want to measure. - In order to express the time to be measured in terms of seconds, you must first of all know the period of your pendulum. Its computation is the next task.
