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for the result.

This result possesses the remarkable property, that the kinetic energy of the motion expressed by it is less than that of any other motion which fulfils the prescribed condition as to velocity. For, if έ, n,,,, etc., denote the impulses required to produce any other motion, &,,,,,, etc., and T, the corresponding kinetic energy, we have, by (9),

2T, = §‚¿‚ + n‚Å‚ + §‚8, + etc.

But by (11),

=

...

§4+n,$ +3,0 + etc. = ↓,, since, by (15), we have 7 = 0, 0, etc. Hence 2T, = §4, + §, (4, − 4) + n, (¿, − $) + 5, (0, − 0) + Now let also this second case (4, 4,,...) of motion fulfil the prescribed velocity-condition = A. We shall have % (2 - 4) + (B - B) + % (0-0) + .

...

1=

− (§, − §)(↓, − ¿) + (n, − n)($, − $) + (3, − 3)(0, − Ô) +

A, etc.............(16)

=

since
0, 0, 0,.... Hence if denote the kinetic
energy of the differential motion (4,− 4, 4, − 4,.......) we have

27 = 2T+2T.....

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........(17); but is essentially positive and therefore T, the kinetic energy of any motion fulfilling the prescribed velocity-condition, but differing from the actual motion, is greater than T the kinetic energy of the actual motion; and the amount, T, of the difference is given by the equation

2T = n, (4, − $) + ¿, (ė, ė) + etc... (18),

or in words,

Kinetic

317. The solution of the problem is this:-The motion

a

minimum actually taken by the system is the motion which has less

in this case.

kinetic energy than any other fulfilling the prescribed velocityconditions. And the excess of the energy of any other such motion, above that of the actual motion, is equal to the energy of the motion which must be compounded with either to produce the other.

energy

in this case.

In dealing with cases it may often happen that the use of the Kinetic co-ordinate system required for the application of the solution minimum (16) is not convenient; but in all cases, even in such as in examples (2) and (3) below, which involve an infinite number of degrees of freedom, the minimum property now proved affords an easy solution.

a smooth

of infinite

The mass on a free rigid

rest.

Example (1). Let a smooth plane, constrained to keep moving Impact of with a given normal velocity, q, come in contact with a free rigid plane inelastic rigid body at rest: to find the motion produced. velocity-condition here is, that the motion shall consist of any body at motion whatever giving to the point of the body which is struck a stated velocity, q, perpendicular to the impinging plane, compounded with any motion whatever giving to the same point any velocity parallel to this plane. To express this condition, let u, v, w be rectangular component linear velocities of the centre of gravity, and let w, p, σ be component angular velocities round axes through the centre of gravity parallel to the line of reference. Thus, if x, y, z denote the co-ordinates of the point struck relatively to these axes through the centre of gravity, and if l, m, n be the direction cosines of the normal to the impinging plane, the prescribed velocity-condition becomes

-

(u + pz − σy) l + (v + σx − wz) m + (w + wy− px) n = — — q....................... (a), the negative sign being placed before q on the understanding that the motion of the impinging plane is obliquely, if not directly, towards the centre of gravity, when l, m, n are each positive. If, now, we suppose the rectangular axes through the centre of gravity to be principal axes of the body, and denote by Mƒ3, Mg3, Mh the moments of inertia round them, we have

T= } M (u* + v2 + w3 +f2w3 +g3p2 + h3σ3). .................................. (b). This must be made a minimum subject to the equation of condition (a). Hence, by the ordinary method of indeterminate multipliers,

...

X

Mu+l=0, Mv + λm = 0, Mw + λn = 0 Mƒ3w + λ (ny—mz) = 0, My3p+λ(lz−nx) = 0, Mh3o+λ(mx−ly) = 1 = 0}(c). These six equations give each of them explicitly the value of one of the six unknown quantities u, v, w, π, p, o, in terms of A and data. Using the values thus found in (a), we have an equation to determine ; and thus the solution is completed. The first three of equations (c) show that A, which has entered as all

Generation of motion

by impulse in an inextensible cord or chain.

indeterminate multiplier, is to be interpreted as the measure of the amount of the impulse.

Example (2). A stated velocity in a stated direction is communicated impulsively to each end of a flexible inextensible cord forming any curvilineal arc: it is required to find the initial motion of the whole cord.

Let x, y, z be the co-ordinates of any point P in it, and ¿, ý, ż the components of the required initial velocity. Let also s be the length from one end to the point P.

If the cord were extensible, the rate per unit of time of the stretching per unit of length which it would experience at P, in virtue of the motion x, y, z, would be

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Hence, as the cord is inextensible, by hypothesis,

dx dx dy dy, dz dz
ds ds ds ds ds ds

+

+

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= 0.....

(a).

Subject to this, the kinematical condition of the system, and

x = u

x = u'

y=v when s=0, y=v' when s = l,
2=
z=w']

=w

=

7 denoting the length of the cord, and (u, v, w), (u', v', w'), the components of the given velocities at its two ends: it is required to find x, y, at every point, so as to make

(b)

a minimum, μ denoting the mass of the string per unit of length, at the point P, which need not be uniform from point to point; and of course

μ (x2 + y2+) ds...

.......

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ds=(dx2 + dy2 + dzo)3..

..(c). Multiplying (a) by λ, an indeterminate multiplier, and proceeding as usual according to the method of variations, we have

dx ddx

(dx

dy dồý, dz dôż

S' {μ (¿âï + ÿôÿ + ¿ôï) + λ

+
+
ds ds ds ds ds ds

ds = 0,

in which we may regard x, y, z as known functions of s, and this it is convenient we should make independent variable. Inte

of motion

grating "by parts" the portion of the first member which contains Generation A, and attending to the terminal conditions, we find, according to by impulse the regular process, for the equations containing the solution.

in an inextensible cord or chain.

d

dz

μα

put = d (Add), my = (d), p = d (ada).

dx ds ds

ds

ds

ds

These three equations with (a) suffice to determine the four unknown quantities, x, y, ż, and A. Using (d) to eliminate ✯, ý, ž from (a), we have

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Taking now 8 for independent variable, and performing the differentiation here indicated, with attention to the following relations:

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and the expression (§ 9) for p, the radius of curvature, we find

18x d (2)

μάδε

dλ λ

.(d).

ds ds μα

(e),

a linear differential equation of the second order to determine
λ, when
and P are given functions of s.

μ

=

0.......

of

The interpretation of (d) is very obvious. It shows that A is the impulsive tension at the point P of the string; and that the velocity which this point acquires instantaneously is the resultant 1 dλ λ tangential, and towards the centre of curvature. μ d8 ρμ The differential equation (e) therefore shows the law of transmission of the instantaneous tension along the string, and proves that it depends solely on the mass of the cord per unit of length in each part, and the curvature from point to point, but not at all on the plane of curvature, of the initial form. Thus, for instance, it will be the same along a helix as along a circle of the same curvature.

Generation of motion by impulse in an inextensible cord or chain.

With reference to the fulfilling of the six terminal equations, a difficulty occurs inasmuch as ¿, ý, ż are expressed by (d) immediately, without the introduction of fresh arbitrary constants, in terms of A, which, as the solution of a differential equation of the second degree, involves only two arbitrary constants. The explanation is, that at any point of the cord, at any instant, any velocity in any direction perpendicular to the tangent may be generated without at all altering the condition of the cord even at points infinitely near it. This, which seems clear enough without proof, may be demonstrated analytically by transforming the kinematical equation (a) thus. Let f be the component tangential velocity, q the component velocity towards the centre of curvature, and p the component velocity perpendicular to the osculating plane. Using the elementary formulas for the direction cosines of these lines (§ 9), and remembering that 8 is now independent variable, we have

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Substituting these in (a) and reducing, we find

df 9
ds

P

=

, y = etc.

(ƒ),

a form of the kinematical equation of a flexible line which will be of much use to us later.

We see, therefore, that if the tangential components of the impressed terminal velocities have any prescribed values, we may give besides, to the ends, any velocities whatever perpendicular to the tangents, without altering the motion acquired by any part of the cord. From this it is clear also, that the directions of the terminal impulses are necessarily tangential; or, in other words, that an impulse inclined to the tangent at either end, would generate an infinite transverse velocity.

To express, then, the terminal conditions, let F and F" be the tangential velocities produced at the ends, which we suppose We have, for any point, P, as seen above from (d),

known.

(g),

1 αλ

f
μ 18

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