it ascend if projected vertically with this velocity at the surface of the moon, where g is 150 ?

Ans. 2426 centims. per second; 19620 centims.

Centrifugal Force.

33. A body moving in a curve must be regarded as continually falling away from a tangent. The accelera

tion with which it falls away is

v2

r

v denoting its velocity

The acceleration of a

and r the radius of curvature. body in any direction is always due to force urging it in that direction, this force being equal to the product of mass and acceleration. Hence the normal force on a body of m grammes moving in a curve of radius r centimetres, with velocity v centimetres per second, is dynes. This

mv2

The

force is directed towards the centre of curvature. equal and opposite force with which the body reacts is called centrifugal force.

If the body moves uniformly in a circle, the time. of revolution being T seconds, we have v =

hence

v2 2π\2

-=

r T

mr dynes.

60

n

T

2πη

Ꭲ ;

r, and the force acting on the body is

If n revolutions are made per minute, the value of T is

ηπ

, and the force is mr(3) dynes.

Examples.

1. A body of m grammes moves uniformly in a circle of radius 80 centims., the time of revolution being of a

second. Find the centrifugal force, and compare it with

The weight of the body (at a place where g is 981) is 981 m dynes. Hence the centrifugal force is about 52 times the weight of the body.

2. At a bend in a river, the velocity in a certain part of the surface is 170 centims. per second, and the radius of curvature of the lines of flow is 9100 centims. Find the slope of the surface in a section transverse to the lines of flow.

Ans. Here the centrifugal force for a gramme of the (170)2 water is = 3.176 dynes. If g be 981 the slope will

be

9100

3.176 1

=

; that is, the surface will slope upwards 981 309

from the concave side at a gradient of 1 in 309. The general rule applicable to questions of this kind is that the resultant of centrifugal force and gravity must be normal to the surface.

3. An open vessel of liquid is made to rotate rapidly round a vertical axis. Find the number of revolutions that must be made per minute in order to obtain a slope of 30° at a part of the surface distant 10 centims. from the axis, the value of g being 981.

Ans. We must have tan 30°

f

where f denotes the

g'

intensity of centrifugal force—that is, the centrifugal force

per unit mass. We have therefore

4. For the intensity of centrifugal force at the equator due to the earth's rotation, we have rearth's radius 6.38 × 108, T= 86164, being the number of seconds in a sidereal day.

=

If the earth were at rest, the value of g at the equator would be greater than at present by this amount. If the earth were revolving about 17 times as fast as at present, the value of g at the equator would be nil.

SUPPLEMENTAL SECTION.

On the help to be derived from Dimensions in investigating Physical Formulæ.

When one physical quantity is known to vary as some power of another physical quantity, it is often possible to find the exponent of this power by reasoning based on dimensions, and thus to anticipate the results--or some of the results-of a dynamical investigation.

Examples.

1. The time of vibration of a simple pendulum in a small arc depends on the length of the pendulum and the intensity of gravity. If we assume it to vary as the mth

power of the length, and as the nth power of g, and to be independent of everything else, the dimensions of a time must equal the mth power of a length, multiplied by the nth power of an acceleration, that is

T=L"(LT-2)" = L"L"T-2n

= L"+nT-2n

Since the dimensions of both members are to be identical, we have, by equating the exponents of T,

1=-2n, whence n =

and by equating the exponents of L,

m+ n = 0, whence m = };

that is, the time of vibration varies directly as the square root of the length, and inversely as the square root of g.

2. The velocity of sound in a gas depends only on the density D of the gas and its coefficient of elasticity E, and we shall assume it to vary as Dm E".

The dimensions of velocity are LT-1.

The dimensions of density, or

mass

volume'

are ML-3.

The dimensions of E, which will be explained in the

chapter on stress and strain, are

or ML-1T-2

The equation of dimensions is

force

or (MLT-2)L-2,

area

whence, by equating coefficients, we have the three equations

1

= 3m-n, - 1 = 2n, m + n = 0,

to determine the two unknowns m and n.

The second equation gives at once

n = 1.

The third then gives

m = 2,

and these values will be found to satisfy the first equation also.

The velocity, then, varies directly as the square root of E, and inversely as the square root of D.

3. The frequency of vibration f for a musical string (that is, the number of vibrations per unit time) depends on its length, its mass m, and the force with which it is stretched F.

The dimensions of ƒ are T-1.

4. The angular acceleration of a uniform disc round its axis depends on the applied couple G, the mass of the disc M, and its radius R.

Assume it to vary as G* M3 R2.

The dimensions of angular acceleration are T-2.