B11. Motion on precribed trajectory

Physical pendulum. Vibration point. Reversion pendulum

The formula t = p(l/g)^1/2 is valid for the mathematical pendulum. In reality, you have only physical ones with infinitely many mass points, each of which demands consideration of its distance from the axis (denoted by l for the mathematical pendulum). Every single point, considered on its own, is a mathematical pendulum.

The formula t = p(l/g)^1/2 shows that it oscillates more or less fast depending on whether its distance from the axis exceeds or is less than l. Hence the physical pendulum represents an infinity of mathematical pendulums with different periods. However, since they belong to the same rigid body, all of them are forced to execute their amplitudes during the same time: A given physical pendulum must have a definite period t.

Obviously, there exists a mathematical pendulum the period of which is the period t of that physical pendulum. It must have a length l, which together with t satisfies the equation
t = p(l/g)^1/2., that is, it must have the length l=g(t/p)t=p(l/g)^1/2, which is called the  reduced length of that physical pendulum.

We start this calculation from the formula R = (w Smr²)/a. You set R = Mg sin a , where M is the total mass of the physical pendulum; imagine M to be concentrated at the centre of gravity, where gravity acts, and assume that it lies at the distance s from the axis, that is, let a = s . Smr² is the moment of inertia q (assumed to be known). Denote it by Q, so that

 Mg sin a = w ·Q/s or g sin a = w (Q /s·M).

The corresponding equation for the mathematical pendulum of length is

g sin a= = w ·l.

Hence the angular acceleration of a physical pendulum with moment of inertia Q, mass M and distance s between its centre of gravity and its axis is at each instant equal to that of a mathematical pendulum (deflected by the same angle a from the position of rest) of length

l =Q /s.M.

That point of the physical pendulum, the distance of which is l and at which the mass of the pendulum could be concentrated without changing its period t, is called the vibration point, the line parallel to the axis vibration line. Each point of the pendulum lying on it has the period t, the same which it would also have, if it were isolated from the pendulum's other points at the distance l from the axis of rotation and vibrating suspended by a weightless, inextensible thread.

The reduced length l is larger than the distance s from the axis of the centre of gravity, for, if you place through the centre of gravity S an axis parallel to the axis of rotation and denote the moment of inertia about the axis of the cente of gravity by Q_s, then Q = Q_s + Ms² , that is, Q > M·s² and Q/M·s > s, that is, l> s, whence the vibration point lies below the centre of gravity.

A simple calculation shows that l =  s + k²/s, where s is the distance from the axis of the centre of gravity and k the radius of gyration . Since Q=Smr²+Ms²=Mk²+Ms²=
= M(k² + s²), that is l = [M(k² + s²)]/sM =  k²/s + s, whence also (l - s)s = k², that is, ss' = k², after setting l - s = s'.

The formula l =  s + k²/s for the distance between the axis of rotation and the line of vibration yields interesting and important information: You can make the line of vibration into the axis of rotation without changing the period of the pendulum; the line which was the axis of rotation then becomes the line of vibration (Fig. 122).

In fact, denoting the pendulum length belonging to the new vibration axis by l' and the corresponding distance of the centre of gravity by s', then l' =  s' + k²/s' and, since ss' = k², you obtain l' = s+ k²/s =l, a relationship suggests the following significance of the radius of gyration k:

If you choose in a vertical plane through the centre of gravity two axes such that the body, as it swings successively about the one and the other axis, has always the same period, then both axes must lie (equally distant from the centre of gravity or) at distances the product of which is k², because s + k²/s = s' + k²/s' yields s -s' =  k²(s - s')/ss'. Thus, if s -  s' does not vanish, you can divide both sides by this term and find, as above, ss' = k², that is, the distance of the two axes is the length of the (synchronous) mathematical pendulum which swings with the same period as the physical pendulum. This insight offers practical means for the measurement of this length.

A pendulum that has been specially designed for this purpose (Fig. 123) has two prismatic suspension fixtures with knife edges: One edge at the top as respective rotation axis, one edge at the bottom through the vibration point. It hangs with the edge of the prism on a corresponding base. You only have to turn it over in order to make the axis of rotation into the line of vibration, and change he distance between the knife edges until you obtain equal periods (reversion pendulum, Henry Kater 1777-1835, J.G.F. v. Bohnenberger). This type of pendulum is used for the determination of the pendulum with the period of one second. Such a pendulum does not demand knowledge of its moment of inertia nor of its mass distribution; only the relative distance between  the edges is required.

Foucault's pendulum

Like the mass points of all bodies which rotate about an axis, also those of the physical pendulum tendas a result of their inertia to maintain unchanged the position of their circular orbits in space, that is, the position of their plane of oscillation, unless an external force acts. Foucault saw in this conservation of the plane of oscillation a foundation for a proof of Earth's rotation.

Imagine a pendulum suspended over one of Earth's Poles and, for example, swinging in the direction of the Zero Meridian, then an observer looking in that direction will see the pendulum coming directly at and away from  him. After a quarter of a day, that is, after each point on Earth has completed a quarter of its circular orbit about the axis, the observer's position has been changed by 90^o. However, since only gravity acts on it, the pendulum swings unchanged in its original plane.

The observer, seeing it now in front of himself swinging from right to left and left to right, has therefore the impression that the plane of oscillation of the pendulum has turned. He would see this plane completing during one day a full turn about the position of rest of the pendulum, which is at the pole identical with the axis of Earth's rotation.

The phenomenon  is explained by the fact that Earth and with it the observer complete in 24 hours a full turn about its axis and the position of rest of the pendulum. The apparent rotation of the plane of oscillation at any point on Earth's surface is proportional to the sine of the geographical latitude, that is, it is the smaller the closer the observer is to the Equator; at Berlin, it is only 285^o 36', at the Equator 0.

In 1851, Foucault employed in Paris a sphere made out of copper, weighing 2 kg on a 17 m long steel wire with a period of 16.40 ".

Initially, the body of the pendulum must not have a velocity with respect to Earth. Observing definite precautions, you ensure therefore that is starts to swing from its extreme position without disturbance. The pendulum has also Earth's angular velocity, that is, it has a motion component at right angle to the plane of oscillation. Its orbit, projected on to the horizontal plane, is therefore not a straight line, but a extremely flat oval which expands as the amplitudes shorten. (Fig. 123a)

Pendulum clock. Metronome. Seconds pendulum

Since a pendulum uses during each oscillation the same time, it is employed for the measurement of time. For example, if a given pendulum uses 1 second for each oscillation and you count the number of oscillations during a given time interval, the answer tells the length of the time interval.

The pendulum clock ( Huygens 1656) also utilizes the equality of the oscillations of a pendulum; in fact, in order to stop for a split second the motion of a system of wheels after equally large time intervals. This stopping is the soul of the clock (balance wheel) (Fig. 124).

Clocks have the task to turn a pointer (we are here concerned only with one, say, the minutes pointer) in front of the clock's face after equally large time intervals by equally large angles. The length of the arc described by the point of the clock's hand indicates the time that has passed. The pointer sits on the axle of a wheel R behind the clocks face; it can only then turn at equal time intervals by equal angles when this wheel does so. The wheel is moved by a weight (omitted in Fig. 124) which hangs from a string wrapped around this axle.

The weight falls and thereby turns the wheel and with it the hand. If the wheel and the pointer were on the axle without another mechanism, the weight would drop to the deepest possible point and while doing so make the pointer turn in front of the clock's face, at first slowly, and then faster and faster - and atime measurement would not be possible.

This accelerated motion of the wheel is stopped shortly after it started, while the velocity is still very small, and then the wheel is again released, etc. The equality of the time intervals between halts in pendulum clocks is caused by a swinging pendulum:

You make the wheel into a cogged wheel and supply the pendulum with an anchor nm, the stopper, which is linked tightly to the pendulum by the frame abo and partakes in the oscillations. It grabs successively with the claws n, m in between the teeths of the cogged wheel and thus stops thereby its rotation and the fall of the weight.

During each oscillation, soon after the pendulum has passed the vertical position, one claw releases a tooth of the wheel, so that it can turn; in doing so, the tooth slides along an inclined plane of the claw and by its pressure gives the pendulum the required push, in order to make up for the loss of energy by friction and air resistance.

Immediately afterwards, the other claw catches another tooth and stops the rotation and the fall of the weight so long until the pendulum on its way back restarts the process. Hence the wheel can only turn while the pendulum travels from the vertical to its return point. Therefore it turns only by one tooth; during that interval, it is accelerated, but the time is so short, that it turns almost uniformly, that is, during equal times by equal angles. The equality of the pendulum oscillations thus converts by the fall of the weight the accelerated motion of the system of wheels into almost uniform motion.

You hear in pendulum clocks the arrival of the pendulum at a return point, the clicking of the stopping mechanism, whence you can also sense the equality of time intervals with you ears. In the case of a clock (except during certain physical measurements), no notice is taken of the sound, but in the case of a metronome (Johann Nepomuk Maelzel 1772-1838 1815) it is its main purpose. It beats the time and is listened to by musicians, in order to fix the rate of performance of a piece of music. It is a pendulum which gives loud ticks. Its body can be shifted along its bar. By this shift you change the distance of its body's centre of gravity from the axle, that is the length of the reduced pendulum, and hence the time intervals between the ticks - the metronome's rate.

The length of the pendulum with the period of one second (seconds pendulum) is known exactly, whence the formula t = p (l/g)^1/2  yields a very exact measurement of g. If the seconds pendulum is l₁ cm long, since t = 1, you find g =p ²·l₁ cm. The acceleration of the body of the pendulum by gravity is seen to be independent of its mass; this only means: All masses fall equally fast. The identity of g for all masses can be proved experimentally, as is shown by the formula g =p ²·l₁, since the length of the seconds pendulum - the reduced length of the pendulum - is the same for all bodies. By observations of pendulums you can therefore prove the identity of all masses more readily than by fall experiments.

Horizontal pendulum

A pendulum in special suspension reacts readily to earthquakes and similar motions of Earth. It is used therefore in the seismometer (Greek: seismos = earthquake) where it is to indicate the period, direction and amplitude of Earth's motion. However, for that purpose, you must arrange for it to have as long a duration of oscillation as possible.

If the axis about which the pendulum swings is horizontal (as has been assumed hitherto ), then gravity acts on the pendulum entirely by a rotation and its period is t=p(l/g)^1/2 (l is the reduced pendulum length). In order to increase the duration of the oscillation arbitrarily, you must, since lack of space limits an enlargement of l, decrease the effect of gravity, that is, let only a component of g act on the pendulum.

You can do so by inclining the axis of rotation with respect to the horizontal direction by a certain angle (Fig. 125). The arm QQ' - linked to the frame G at o so that it can turn - carries in bearings at D and D' (indicated by arrows) a triangular frame, at the point A of which is the body of the pendulum.

If you incline the axis of rotation by the angle i with respect to the horizontal direction, then gravity only acts with the component g·cos i on the pendulum and the period becomes t_i=p(l/g cos i)^1/2. The larger is i, the larger will be t_i. If you make i = 90º, t_i becomes infinitely large, that is, the pendulum with a perpendicular axis is in indifferent equilibrium. The weight of the pendulum must therefore swing only in an almost horizontal plane, for which reason the name horizontal pendulum, which Johann Karl Friedrich Zöllner 1834-1882 gave it, is not quite appropriate.

Fig. 126 shows the horizontal pendulum in the form in which it is used. Its motion is similar to that of a door which swings at a not quite vertical door post about the line through the hinges. If the hinges are exactly vertical on top of each other, the door rests in every position ( i = 90º), otherwise it rests only in a definite position or it swings about this position if the friction is not too large.

The position of rest is characterized by the fact that the centre of gravity lies in that plane which can be placed through the axis of rotation and direction of gravity. If you incline the door post and thereby the axis of rotation with respect to the position of rest perpendicular plane just a little bit, the pendulum finds another position of rest which differs from the former by a considerable angle.

The horizontal pendulum therefore enlarges greatly very small changes in the inclination of its axis and equally changes in the direction of gravitational action. For example, if the axis of rotation of a horizontal pendulum deviates only by 2' from the vertical and inclines the axis only by 0.01" in the plane, perpendicular to the position of rest, the pendulum changes to a new position of rest which forms already with the former an angle of 17".

This horizontal pendulum corresponds in its sensitivity to a vertical pendulum of enormous length. - More recently (the author writes in 1935!) Zöllner's suspension has been used instead of pins: A light metal tube (approximately 25 cm long) is held almost horizontally by two thin tight wires which are fixed to a frame above and below of the tube. The line linking the two fixing points lying above another is the axis of rotation. - The horizontal pendulum was described for the first time by Hengler 1806-1958 (1833) as astronomical pendulum balance.

Dependence of weight on the rotation of Earth about its axis. Flattening of Earth

The seconds pendulum does not have the same length everywhere on Earth. For example:  

Hence the acceleration g and therefore also the weight of a mass increases towards the Equator and Poles. This is explained as follows: Not the entire force of attraction of Earth on a mass manifests itself by pressure of the mass on its base, one part of it reduces the centrifugal force due to Earth's rotation. The centrifugal acceleration with r the radius of Earth's sphere is

at the Equator (Latitude 0)   f_o = 4p²r/T²
at Latitude f   f_j^- = 4p²r/T² cos j = f_o^-·cos j,

where T is the time of a full revolution of Earth about its axis. At the equator (Fig. 127), the entire acceleration f_o acts along the line along which C attracts a mass, in contrast, at Latitude j, only the component

f_fcosj = f_o·cos² f of f_j.

Assuming Earth to be at rest, a perfect sphere and its attraction at each point of its surface equally large, denote its attraction at each point on its surface by G. During its rotation, f_o and f_j ^·cos j will reduce G, since they act in the opposite direction. Hence, denoting by g_o and g_j the accelerations by gravity during rotations, you have: 

The centrifugal force due to Earth's rotation therefore reduces G most at the Equator, the larger is j - towards the poles - it does so less and at the poles not at all. This is the reason why g is smallest at the Equator and increases towards the Poles, and with it the weight of a mass m. Substituting in f_o = 4p²r/T² for r the radius at the Equator in metres and for T the duration of a revolution of Earth in seconds, you obtain f_o = 34 mm·sec^-², that is g_o = G - 34 mm·sec^-², whence at the Equator the gravitational acceleration is reduced due to the centrifugal force by 34 mm·sec^-². Pendulum observations at the Equator have yielded g_o=9.7806 m·sec^-².

Since f_o/G is approximately 1/289, the reduction of f_o at the Equator is 1/289·G, whence the attracting force of Earth acting on a mass at the Equator is decreased by 1/289 of its true weight. - If this reduction were 289 times as large, the mass would have no weight. However, for the centrifugal acceleration to be 289 times as large as it really is, the square of the rate of rotation of Earth would have to be 289 times as large, that is, the angular velocity would have to be 17 times as large as it is, that is, a day would only last 1 hour and 25 minutes in our time.

If you employ f_o/G = 1/289 to determine how the length of the seconds pendulum changes with the geographical latitude, you obtain a result which differs from the experimental one. Pendulum experiments have yielded lengths which yield f_o/G = 1/192 (not 1/289). You can demonstrate this difference as follows: If only the centrifugal force were the cause of the change in the weight, then g would have to be smaller by 34 mm·sec^-2 at the Equator than at the Poles; however, these experiments have shown that it is smaller by 52 mm·sec^-2. The cause for this deviation is: Earth is not a perfect sphere, it is flattened. - Measurements of degrees show that the polar diameter is shorter by 1/289 compared with that at the Equator. - Th increase in the equatorial diameter compared with the polar one is also explained by the fact that the centrifugal force at the Equator is larger than at points between the Equator and the Pole and by the action of the centrifugal force on Earth's mass. While it still was plastic, the sphere was changed into a rotational ellipsoid.

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