Page:EB1911 - Volume 17.djvu/997

of the velocity of the moving point at the instant t. Obviously OV→ is parallel to the tangent to the path at P, and its magnitude is ds/dt, where s is the arc. If we project OV→ on the co-ordinate axes (rectangular or oblique) in the usual manner, the projections u, v, w are called the component velocities parallel to the axes. If x, y, z be the co-ordinates of P it is easily proved that

u =	dx	,   v =	dy	,   w =	dz	.
	dt		dt		dt

(1)

The momentum of a particle is the vector obtained by multiplying the velocity by the mass m. The impulse of a force in any infinitely small interval of time δt is the product of the force into δt; it is to be regarded as a vector. The total impulse in any finite interval of time is the integral of the impulses corresponding to the infinitesimal elements δt into which the interval may be subdivided; the summation of which the integral is the limit is of course to be understood in the vectorial sense.

Newton’s Second Law asserts that change of momentum is equal to the impulse; this is a statement as to equality of vectors and so implies identity of direction as well as of magnitude. If X, Y, Z are the components of force, then considering the changes in an infinitely short time δt we have, by projection on the co-ordinate axes, δ(mu) = Xδt, and so on, or

m	du	= X,   m	dv	= Y,   m	dw	= Z.
	dt		dt		dt

(2)

For example, the path of a particle projected anyhow under gravity will obviously be confined to the vertical plane through the initial direction of motion. Taking this as the plane xy, with the axis of x drawn horizontally, and that of y vertically upwards, we have X = 0, Y = −mg; so that

d 2x	= 0,	d 2y	= −g.
dt 2		dt 2

(3)

The solution is

x = At + B,   y = −1/2gt ² + Ct + D.

(4)

If the initial values of x, y, ẋ, ẏ are given, we have four conditions to determine the four arbitrary constants A, B, C, D. Thus if the particle start at time t = 0 from the origin, with the component velocities u₀, v₀, we have

x = u₀t,   y = v₀t − 1/2gt ².

(5)

Eliminating t we have the equation of the path, viz.

y =	v0	x −	gx2	.
	u0		2u2

(6)

This is a parabola with vertical axis, of latus-rectum 2u₀²/g. The range on a horizontal plane through O is got by putting y = 0, viz. it is 2u₀v₀/g. we denote the resultant velocity at any instant by ṡ we have

ṡ² = ẋ² + ẏ² = ṡ₀² − 2gy.

(7)

Another important example is that of a particle subject to an acceleration which is directed always towards a fixed point O and is proportional to the distance from O. The motion will evidently be in one plane, which we take as the plane z = 0. If μ be the acceleration at unit distance, the component accelerations parallel to axes of x and y through O as origin will be −μx, −μy, whence

d 2x	= −μx,	d 2y	= − μy.
dt 2		dt 2

(8)

The solution is

x = A cos nt + B sin nt,   y = C cos nt + D sin nt,

(9)

where n = √μ. If P be the initial position of the particle, we may conveniently take OP as axis of x, and draw Oy parallel to the direction of motion at P. If OP = a, and ṡ₀ be the velocity at P, we have, initially, x = a, y = 0, ẋ = 0, ẏ = ṡ₀ whence

x = a cos nt,   y = b sin nt,

(10)

if b = ṡ₀/n. The path is therefore an ellipse of which a, b are conjugate semi-diameters, and is described in the period 2π/√μ; moreover, the velocity at any point P is equal to √μ·OD, where OD is the semi-diameter conjugate to OP. This type of motion is called elliptic harmonic. If the co-ordinate axes are the principal axes of the ellipse, the angle nt in (10) is identical with the “excentric angle.” The motion of the bob of a “spherical pendulum,” i.e. a simple pendulum whose oscillations are not confined to one vertical plane, is of this character, provided the extreme inclination of the string to the vertical be small. The acceleration is towards the vertical through the point of suspension, and is equal to gr/l, approximately, if r denote distance from this vertical. Hence the path is approximately an ellipse, and the period is 2π √(l/g).

Fig. 65.	Fig. 66.

The above problem is identical with that of the oscillation of a particle in a smooth spherical bowl, in the neighbourhood of the lowest point. If the bowl has any other shape, the axes Ox, Oy may be taken tangential to the lines of curvature at the lowest point O; the equations of small motion then are

d 2x	= −g	x	,	d 2y	= −g	y	,
dt 2		ρ1		dt 2		ρ2

(11)

where ρ₁, ρ₂, are the principal radii of curvature at O. The motion is therefore the resultant of two simple vibrations in perpendicular directions, of periods 2π √(ρ₁/g), 2π √(ρ₂/g). The circumstances are realized in “Blackburn’s pendulum,” which consists of a weight P hanging from a point C of a string ACB whose ends A, B are fixed. If E be the point in which the line of the string meets AB, we have ρ₁ = CP, ρ₂ = EP. Many contrivances for actually drawing the resulting curves have been devised.

It is sometimes convenient to resolve the accelerations in directions having a more intrinsic relation to the path. Thus, in a plane path, let P, Q be two consecutive positions, corresponding to the times t, t + δt; and let the normals at P, Q meet in C, making an angle δψ. Let v (= ṡ) be the velocity at P, v + δv that at Q. In the time δt the velocity parallel to the tangent at P changes from v to v + δv, ultimately, and the tangential acceleration at P is therefore dv/dt or s̈. Again, the velocity parallel to the normal at P changes from 0 to vδψ, ultimately, so that the normal acceleration is vdψ/dt. Since

dv

=

dv

ds

= v

dv

,   v

dψ

= v

dψ

ds

=

v 2

,

dt

ds

dt

ds

dt

ds

dt

ρ

(12)

where ρ is the radius of curvature of the path at P, the tangential and normal accelerations are also expressed by v dv/ds and v ²/ρ, respectively. Take, for example, the case of a particle moving on a smooth curve in a vertical plane, under the action of gravity and the pressure R of the curve. If the axes of x and y be drawn horizontal and vertical (upwards), and if ψ be the inclination of the tangent to the horizontal, we have

mv	dv	= − mg sin ψ = − mg	dy	,	mv 2	= − mg cos ψ + R.
	ds		ds		ρ

(13)

The former equation gives

v ² = C − 2gy,

(14)

and the latter then determines R.

In the case of the pendulum the tension of the string takes the place of the pressure of the curve. If l be the length of the string, ψ its inclination to the downward vertical, we have δs = lδψ, so that v = ldψ/dt. The tangential resolution then gives

l	d 2ψ	= − g sin ψ.
	dt 2

(15)

If we multiply by 2dψ/dt and integrate, we obtain

(	dψ	)2 =	2g	cos ψ + const.,
	dt		l

(16)

which is seen to be equivalent to (14). If the pendulum oscillate between the limits ψ = ±α, we have

(	δψ	)2 =	2g	(cos ψ − cos α) =	4g	(sin2 1/2α − sin2 1/2ψ);
	dt		l		l

(17)

and, putting sin 1/2ψ = sin 1/2α. sin φ, we find for the period (τ) of a complete oscillation

τ = 4 $\int _{0}^{{\frac {1}{2}}\pi }$	dt	dφ = 4√	l	· $\int _{0}^{{\frac {1}{2}}\pi }$	dφ
	dφ		g		√(1 − sin2 1/2α · sin2 φ)

= 4√	l	· F1 (sin 1/2α),
	g

(18)