W2 (P — R1) } W:(P — R2). P — RI . P — R2 . (8o)
WIW2 ' Wr W2
If RI =R2, which is the case when the resistance, as well as the effort, arises from the mutual actions of the two bodies, the above becomes,
E : El: E2
::WrW2:W2:WI ' (81) that is to say, the energy is exerted on the bodies in shares inversely proportional to their weights; and they receive accelerations inversely proportional to their weights, according to the principle of dynamics, already quoted in a note to § 110, that the mutual actions of a system of bodies do not affect the motion of their common centre of gravity.
For example, if the weight of a gun be 16o times that of its ball H? of the energy exerted by the powder in exploding will be employed' in propelling the ball, and i h in producing the recoil of the gun, provided the gun up to the instant of the ball's quitting the muzzle meets with no resistance to its recoil except the friction of the ball.
§ 124. Centre of Percussion.—It is obviously desirable that the deviations or changes of motion of oscillating pieces in machinery should, as far as possible, be effected by forces applied at their centres of percussion.
If the deviation be a translation—that is, an equal change of motion of all the particles of the body—the centre of percussion is obviously the centre of gravity itself ; and, according to the second law of motion, if dv be the deviation of velocity to be produced in the interval dt, and W the weight of the body, then
pgdt
is the unbalanced effort required.
If the deviation be a rotation about an axis traversing the centre of gravity, there is no centre of percussion; for such a deviation can only be produced by a couple of forces, and not by any single force. Let da be the deviation of angular velocity to be produced in the interval dt, and I the moment of the inertia of the body about an axis through its centre of gravity; then ild(a2)=Iada is the variation of the body's actual energy. Let M be the moment of the unbalanced couple required to produce the deviation; then by equation 57, § 104, the energy exerted by this couple in the interval dt is Madt, which, being equated to the variation of energy, gives
da R2W da M = I = g dt'
R is called the radius of gyration of the body with regard to an axis through its centre of gravity.
Now (fig. 133) let the required deviation be a rotation of the body BB about an axis 0, not traversing the centre of gravity G, da
(77)
(82)
(83)
being, as before, the deviation of angular velocity to be produced in the interval dl. A rotation with the angular velocity a about an axis 0 may be considered as compounded of a rotation with the same angular velocity about an axis drawn through G parallel to 0 and a translation with the velocity a. OG, OG being the perpendicular distance between the two axes. Hence the required deviation may be regarded as compounded of a deviation of translation dv=OG. da, to produce which there would be required, according to equation (82), a force applied at G perpendicular to the plane OG
P fig . OG . da (84)
and a deviation da of rotation about an
produce which there would be required a couple of the moment M given by equation (83). According to the principles of statics, the resultant of the force P, applied at G perpendicular to the plane OG, and the couple M is a force equal and parallel to P, but applied at a distance GC from G, in the prolongation of the perpendicular OG, whose value is
GC = M/P = R'/OG. (85) Thus is determined the position of the centre of percussion C, corresponding to the axis of rotation 0. It is obvious from this equation that, for an axis of rotation parallel to 0 traversing C, the centre of percussion is at the point where the perpendicular OG meets O.
§ 125.* To find the moment of inertia of a body about an axis through its centre of gravity experimentally.—Suspend the body from any
conveniently selected axis 0 (fig. 48) and hang near it a small plumb bob. Adjust the length of the plumbline until it and the body oscillate together in unison. The length of the plumbline, measured from its point of suspension to the centre of the bob, is for all practical purposes equal to the length OC, C being therefore the centre of percussion corresponding to the selected axis O. From equation (85)
R2 = CG X OG = (OC —OG) OG.
The position of G can be found experimentally; hence OG is known, and the quantity R2 can be calculated, from which and the ascertained weight W of the body the moment of inertia about an axis through G, namely, W/gXR2, can be computed.
§ 126.* To find the force competent to produce the instantaneous
acceleration of any link of a mechanism.—In many practical problems
it is necessary to know the magnitude and position of the forces
acting to produce the accelerations of the several links of a mechan
ism. For a given link, this force is the resultant of all the accelerating
forces distributed through the substance of the material of the link
required to produce the requisite acceleration of each particle. and
the determination of this force depends upon the principles of the
two preceding sections. The investigation of the distribution of
the forces through the material and the stress consequently pro
duced belongs to the subject of the STRENGTH OF MATERIALS (q. v.).
Let BK (fig. 134) be any link moving in any manner in a plane, and
let G be its centre of gravity.
Then its motion may be an
alysed into (I) a translation of
its centre of gravity; and (2) a
rotation about an axis through
its centre of gravity perpen
dicular to its plane of motion.
Let a be the acceleration of
the centre of gravity and let A
be the angular acceleration
about the axis through the
centre of gravity; then the
force required to produce the
translation of the centre of
gravity is F =Wa/g, and the
couple required to produce the
angular acceleration about the
centre of gravity is M = IA/g,
GB = kg:gb. Og is then the acceleration of the centre of gravity and the force F can therefore be immediately calculated. To find the angular acceleration A, draw kt, bt respectively parallel to and at right angles to the link KB. Then tb represents the angular acceleration of the point B relatively to the point K and hence tb/KB is the value of A, the angular acceleration of the link. Its moment of inertia about G can be found experimentally by the method explained in § 125, and then the value of the couple M can be computed. The value of x is found immediately from the quotient M/F. Hence the magnitude F and the position of F relatively to the centre of gravity of the link, necessary to give rise to the couple M, are known, and this force is therefore the resultant force required.
§ 127.* Alternative construction for finding the position of F relatively to the centre of gravity of the link.—Let B and K be any two
points in the link which for greater generality are taken in fig. 135, so that the centre of gravity G is not in the line joining them. First find the value of R experimentally. Then produce the given directions of acceleration of B and K to meet in 0; draw a circle through the three points B, K and 0; produce the line joining 0 and G to cut the circle in Y; and take a point Z on the line OY so that
G X GZ = R2. 'Then Z is a point in the line of action of the force F. This useful theorem is due to G. T. Bennett, of Emmanuel College, Cambridge. A proof
of it and three corollaries are given in appendix 4 of the second edition of Dalby's Balancing of Engines (London, 1906). It is to be noticed that only the directions of the accelerations of two points are required to find the point Z.
For an example of the application of the principles of the two preceding sections to a practical problem see Valve and Valve Gear Mechanisms, by W. E. Dalby (London, 1906), where the inertia stresses brought upon the several links of a Joy valve gear, belonging to an express passenger engine of the Lancashire & Yorkshire railway, are investigated for an enginespeed of 68 m. an hour.
§ 128.* The Connecting Rod Problem.—A particular problem of practical importance is the determination of the force producing the motion of the connecting rod of a steamengine mechanism of the usual type. The methods of the two preceding sections may be used when the acceleration of two points in the rod are known. In this problem it is usually assumed that the crank pin K (fig. 136)
moves with uniform velocity, so that if a is its angular velocity and r its radius, the acceleration is a'r in a direction along the crank arm from the crank pin to the centre of the shaft. Thus the accelera tion of one point K is known completely. The acceleration of a second point, usually taken at the centre of the crosshead pin, can be found by the principles of § 82, but several special geometrical constructions have been devised for this purpose, notably the construction of Klein,' discovered also independently by Kirsch.' But probably the most convenient is the construction due to G. T. Bennett' which is as follows: Let OK be the crank and KB the connecting rod. On the connecting rod take a point L such that KL X KB =KO'. Then, the crank standing at any angle with the line of stroke, draw LP at right angles to the connecting rod, PN at right angles to the line of stroke OB and NA at right angles to the connecting rod; then AO is the acceleration of the point B to the scale on which KO represents the acceleration of the point K. The proof of this construction is given in The
Balancing of Engines.
The finding of F may be continued thus: join AK, then AK is the acceleration image of the rod, OKA being the acceleration diagram. Through G, the centre of gravity of the rod, draw Gg parallel to the line of stroke, thus dividing the image at g in the proportion that the connecting rod is divided by G. Hence Og represents the acceleration of the centre of gravity and, the weight of the connecting
1 J. F. Klein, " New Constructions of the Force of Inertia of Connecting Rods and Couplers and Constructions of the Pressures on their Pins," Journ. Franklin Inst., vol. 132 (Sept. and Oct., 1891).
2 Prof. Kirsch, " Ober die graphische Bestimmung der Kolbenbeschleunigung," Zeitsch. Verein deutsche Ingen. (18 0), p. 1320.
Dalby, The Balancing of Engines (London, 1906), app. 1.
rod being ascertained, F can be immediately calculated. To find a point in its line of action, take a point Q on the rod such that KG X GQ = R2, R having been determined experimentally by the method of § 125; join G with 0 and through Q draw a line parallel to BO to cut GO in Z. Z is a point in the line of action of the resultant force F; hence through Z draw a line parallel to 0g. The force F acts in this line, and thus the problem is completely solved. The above construction for Z is a corollary of the general theorem given in § 127.
§ 129. Impact. Impact or collision is a pressure of short duration exerted between two bodies.
The effects of impact are sometimes an alteration of the distribution of actual energy between the two bodies, and always a loss of a portion of that energy, depending on the imperfection of the elasticity of the bodies, in permanently altering their figures, and producing heat. The determination of the distribution of the actual energy after collision and of the loss of energy is effected by means of the following principles:
I. The motion of the common centre of gravity of the two bodies is unchanged by the collision.
II. The loss of energy consists of a certain proportion of that part of the actual energy of the bodies which is due to their motion relatively to their common centre of gravity.
Unless there is some special reason for using impact in machines, it ought to be avoided, on account not only of the waste of energy which it causes, but from the damage which it occasions to the frame and mechanism. (W. J. M. R.; W. E. D.)
End of Article: W2 (P — R1) 

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