would do against opposing forces before it would come to rest.

measure.

31. Work, like force, is often expressed in gravitationGravitation units of work, such as the footpound and kilogramme-metre, vary with locality, being proportional to the value of g.

One gramme-centimetre is equal to g ergs.

One kilogramme-metre is equal to 100,000 g ergs.

One foot-poundal is 453·59 × (30·4797)2 = 421390 ergs. One foot-pound is 13,825 g ergs, which, if g be taken as 981, is 1.356 × 107 ergs.

32. The C.G.S. unit rate of working is 1 erg per second. Watt's "horse-power" is defined as 550 foot-pounds per second. This is 7.46 × 109 ergs per second. The "force de cheval" is defined as 75 kilogrammetres per second. This is 7.36 × 109 ergs per second. X We here assume g = 981.

A new unit of rate of working has been lately introduced for convenience in certain electrical calculations. It is called the Watt, and is defined as 107 ergs per second. A thousand watts make a kilowatt. The following tabular statement will be useful for reference.

1 Watt = 107 ergs per second = 00134 horse-power =737 foot-pounds per second = 1019 kilogrammetres per second.

=

1 Kilowatt = 1.34 horse-power.

1 Horse-power = 550 foot-pounds per second = 76.0 kilogrammetres per second = 746 watts = 1.01385 force de cheval.

1 Force de cheval 75 kilogrammetres per second

=

542.48 foot-pounds per second

=

=9863 horse-power.

=

736 watts

Examples.

1. If a spring balance is graduated so as to show the masses of bodies in pounds or grammes when used at the equator, what will be its error when used at the poles, neglecting effects of temperature?

Ans. Its indications will be too high by about the total weight.

1

of

196

2. A cannon-ball, of 10,000 grammes, is discharged with a velocity of 45,000 centims. per second. Find its kinetic energy.

Ans. × 10000 × (45000)2 = 1.0125 × 1013 ergs.

3. In last question find the mean force exerted upon the ball by the powder, the length of the barrel being 200 centims.

Ans. 5.0625 × 1010 dynes.

4. Given that 42 million ergs are equivalent to 1 gramme-degree of heat, and that a gramme of lead at 10° C. requires 15.6 gramme-degrees of heat to melt it; find the velocity with which a leaden bullet must strike a target that it may just be melted by the collision, supposing all the mechanical energy of the motion to be converted into heat and to be taken up by the bullet.

=

We have 15.6 × J, where J = 42 × 106. Hence v2 = 1310 v = 36.2 thousand centims. per

second.

millions;

5. With what velocity must a stone be thrown vertically upwards at a place where g is 981 that it may rise to a height of 3000 centims.? and to what height would it ascend if projected vertically with this velocity at the surface of the moon, where 9 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 v denoting its velocity and

the radius of curvature. The acceleration of a 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
r

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

γ

2

2πη

;

T

r, and the force acting on the body is

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

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.

Ans. The centrifugal force is m x (笑)

× 80 = 50532 m dynes.

× 80= m × 64π2

The weight of the body (at a place where g is 981) is 981 m dynes. Hence the centrifugal force is about 521 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 = 3.176 dynes. If g be 981 the slope will

water is

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°

where ƒ denotes the

intensity of centrifugal force-that is, the centrifugal force per unit mass. We have therefore

C

4. For the intensity of centrifugal force at the equator due to the earth's rotation, we have r= earth'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 at the equator would be nil.

9

SUPPLEMENTAL SECTION.

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

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