Hence equating the two values of the K. E.,

The results of this calculation for some particular cases are as follows:

Solid cylinder of radius a rotating round its axis 2=a2/2

A rod, length 1, rotating round an axis through

its end, perpendicular to its length

2=12/3

Parallelogram of height h rotating round its base r2=h2/3

Do. about a centre line parallel to its base

A circular plate, radius a, diameter h, rotating about a diameter

A triangle, height h, about its base

Do., about an axis through its CG, parallel to

its base

Do., about an axis through its vertex, parallel

to its base

r2=a2/4=h2/16

Most of these results are not required for the present, but they are collected for future reference.

We are now in a position to calculate accurately the K. E. of the rim of a fly-wheel, and so see how near our approximate result would be.

We proceed thus

Let R1 and R2 be the inner and outer radii, then

K. E. of wheel=K. E. of cylinder radius R1

- K. E. of cylinder radius R2.

Let W1 and W2 be the weights of these cylinders,

then

1

2

which decreases as R, approaches R1.

2g

This method of finding the K. E. of a body, by considering it as the difference of two other bodies, or in other cases as their sum, can be often used, but we must be careful to take the K. E.'s all about the one given axis.

Motion of an Unconstrained Body. The principle of work enables us to solve questions relating to the motion of an unconstrained body, i.e. a body whose motion is not defined by its connection to other bodies, e.g. a shot after leaving the gun.

This case is of interest, and we will briefly examine it. The shot lies in the gun, and the powder burning behind it causes a gaseous pressure which forces it along the bore; during this period the motion of the shot is defined by the nature of the pairing between it and the bore of the gun. It now issues from the mouth of the gun with a certain velocity and K. E., and, during the remainder of its motion before reaching the earth or target, its motion is free from any constraint other

than that exerted by gravity. We have then the reverse of our usual problem, for generally we know the path of the body and require to determine the forces acting, while here we know the force and not the path. Let

W=weight of shot,

V1 = velocity on leaving the muzzle,

and consider the motion between the moment of leaving the muzzle and that when the shot has attained a height h. Then

v being the velocity the shot has at the moment the height his reached.

This example shows well the importance of clear definition of the period of time chosen (page 137).

For

suppose we take the period to be from the moment of igniting the powder. Then we have

Energy exerted = whole energy of powder.

Then, assuming for simplicity that the bore of the gun offers no resistance to the shot, we have

and this is correct, for this equation is the form taken by the principle of work for the period during which the shot is being pushed along the bore, since we then have

Energy exerted energy of powder,

Work done=0,

If the bore offer a resistance R and its length be 7,

then the work done will be increased in the two preceding equations by the term R7, and the K. E. W7j2|2g will be accordingly diminished.

Potential Energy.-Returning to the consideration of the motion when unconstrained, the principle of work is sometimes stated in a new form.

The only resistance is gravity (neglecting the friction of the air), and this is a reversible resistance (page 41). The work done against gravity during the rise h, viz. Wh, will be restored again when the shot falls. This fact is sometimes expressed by saying that the shot has Potential Energy Wh, due to being at a height h. The equation

then becomes

0=Wh+

W&2 Way2

2g 2g

0= Final Pot. E+ Final K. E. - Initial K. E., or, remembering that the initial potential energy is zero, and that the equation refers to any point during the flight, since h may have any value, we have at any point of the flight

Pot. E. + K. E. Initial Pot. E. + Initial K. E.,

=

so that the sum of the potential and kinetic energies is constant for all points of the flight.

The principle of work may then be written for this

case as

Potential Energy + Kinetic Energy = constant.

EXAMPLES.

1. A slider weighing 100 lbs. rests on a table, it is moved as in Fig. 143 by a weight of 20 lbs., and when it has moved 2 ft. its velocity is observed to be 2 f.s. Find the coefficient of friction. Ans. .163.

2. In question 1, page 90, the rider goes 60 yds. from rest before getting the speed up. Find the mean moment he exerts. Ans. 34.3 lbs.-ft.

3. A train weighs 60 tons and the engine 25 tons. It is ascending an incline of 1 in 100 at 30 miles per hour when the

draw bar breaks. Taking the resistance at 16 lbs. per ton, find how far the carriages will run before stopping; also what speed would the engine finally attain if it continued to exert the same H. P. as before the breakage, and its resistance were unaltered? Ans.mile; 102 miles per hour.

4. If in the preceding the brakes did not act, what would be the speed of the train when running back and passing the point at which the breakage occurred?

Ans. 12 miles per hour.

5. In question 3, the guard's van weighs 18 tons, and he applies his brake, skidding the wheels, directly the breakage occurs. Find in what distance the train is brought up. cient of friction between wheel and rail .18.

CoeffiAns.mile.

6. The piston and pump rods of a "Bull" engine weigh 18 lbs. per sq. in. of piston. The initial steam pressure is 50 lbs. absolute, and the cut-off takes place when the piston has travelled one foot of the up-stroke. Assuming hyperbolic expansion, find the least length of cylinder, that the piston may not strike the cover. Back pressure 2 lbs. Ans. 7 ft. 7 ins. 7. In the preceding find the position of maximum piston velocity, and the corresponding K. E. of the moving parts per sq. in. of piston.

Ans. 2 ft. 6 ins. from commencement; 45 ft. -lbs.

8. A body weighing 112 lbs. is fastened to a rope passing over an axle 2 ins. diameter, on which is a fly-wheel 2 ft. diameter. Find the weight of the fly-wheel rim, so that the body after falling 40 ft. may have a velocity of only 4 f.s.

Ans. 124 lbs. 9. Find the K. E. of a disc running at a speed V f.s. along a plane. Radius r ft. Ans. The disc is moving as a whole at velocity V f.s., and also rotating with an angular velocity V/r. The K. E. is the sum of that due to each motion separately, which can be proved analytically, or may be seen as follows: Suppose the disc to have a loose axle through its centre, then by stopping this axle we can take out the K. E. Wo/2g; but the disc will then still rotate about the axle with its original angular velocity; hence the result.