to the point of cut-off; for up to that point the area of the rectangle MBba, is directly proportional to Ma, or A1, so that 14 is proportional to A1. The K. E. then increases at a constant rate up to cut-off; after cut-off the excess of effort over resistance becomes less, so that the rate of increase of K. E. is not so great, and this is shown by the curve beginning to curve downwards, e.g. between 2 and 3 the gain of K. E. is abba, which is less than the gain between 1 and 2, viz., abba. It must be particularly noticed, however, that between A and 4, that is so long as Effort > Resistance, the K. E. continuously increases, and with it of course the velocity, since K. E. varies as 2. A very common error is to imagine that because after C the effort falls off, the velocity must do likewise. The mistake lies in forgetting that the question whether the velocity will increase or not depends not on how much the effort exceeds the resistance, but on whether it exceed it at all. What is affected is, as we have seen above, the rate of increase.

At 4 the effort and resistance just balance, so that there is for the moment no increase of K. E., or the rate of increase has fallen to zero.

Next, we come to points past 4. And now the conditions have changed, for we have Effort < Resistance. Consider now the point 5. Then, for the whole motion from A,

The ordinate 5.5 will be less than 4.4, which represents MBCa4, by the area aab, so that for points to the right of 4 we obtain the heights of the ordinates by subtracting from 444, the lengths representing aab, arabe, etc. On the right of 4 then the curve begins to fall, and it will finally reach the axis AX at some point, which in our figure is marked 10, so 10 and 10 coincide.

P

We can easily see where 10 will be, for it is such that subtracting a4a10b10 from MBCa, leaves nothing, so to find 10 we make

a4a10b10 MBCα4,

10 being on the base line, shows that 10 is a point of zero K. E., so that 10 is the point at which the piston stops.

We now see what we have to do to find the cut-off when the stopping point is given. For given A10 we have, by trial, to find C, such that

MBCа4-a4a10b10.

And then the piston will stop at the required point. The curve Ac12 . . . 10 is a curve of K. E., but we have still to determine its scale.

Let now the scale of AB be 1 inch to m lbs., and that of A10, which represents the stroke, be 1 inch to n feet.

Then areas on the diagram represent energy on a scale I sq. inch to mn ft.-lbs. (page 70). The ordinate of the curve is, in linear inches, equal to the area on the diagram in sq. ins. (page 208), therefore the scale of the curve is

I inch=mn ft.-lbs.

The curve is on this scale a curve of Wv2/2g, but W/2g is a constant, so that the curve will on a proper scale be also a curve of v2. We can see what this scale is; for a line of 1 inch represents mn ft.-lbs., so that when the ordinate is I inch

If then we measure the ordinate on a scale of

the square root of this number of feet will give us the

value of v; we may say then the scale is 1 inch to 2gmn/W f.s. squared.

A further graphical construction can be used which will give a curve of velocity, but into this we have not space to inquire.

Turning Pairs—K. E. of Rotation.—When a body moves so that all the particles move in parallel straight lines, they all have necessarily the same velocity ข. In this case a small particle of weight w has K. E. ww2/2g, and the whole body a K. E. W72/2g, W being the total weight. All the cases we have so far considered have been of this type.

But now, when a body is rotating, no two particles have the same motion, thus (Fig. 146) a particle at B has a velocity Arg, A being the angular velocity about O, the centre of rotation; and one at C has a g velocity Arc. The velocities are respectively at right angles to OB and OC, so that even if rc=rB the directions of motion are different, although the amounts of the velocities are then equal.

0

Fig. 146.

Each heavy particle of which the body is composed has then its own K. E., and since we have seen that K. E. is independent of direction, it follows that the total K. E. of a rotating body is the arithmetic sum of the K. E.'s of all the particles composing it, irrespective of the direction of their motions.

The calculation of this quantity, then, requires that we divide the body up into indefinitely small particles, and sum up the K. E.'s of all these particles. This process involves the use of the Integral Calculus, and hence the student must, until he has mastered the use of this, accept the results given as facts.

Thin Ring.One case we can treat, viz., that of a thin ring or cylinder, rotating about its axis. Let r be

the mean radius.

Then, if the thickness be very small compared to r, all the particles are practically at the same distance r from the If then

centre.

and each particle will have the same K. E., so that

W being the weight of the ring.

This result can be used without serious error for most fly-wheels, W being the weight of rim, its mean radius, and an allowance of two or three per cent added for the K. E. of boss and arms.

Radius of Gyration.—Consider now the case of an actual fly-wheel with a definite thickness of rim. R and R' be the inner and outer radii.

Take now a radius

intermediate be

Let

tween R and R', and consider the particles which lie on a very thin ring of radius r. Then this ring has a K. E. due to the velocity Ar. All the particles outside this ring have velocities greater than Ar, and so their total K. E. will be greater than if they all lay on the ring; other hand, those which lie inside have velocities less than Ar, and their total K. E. is less than if they lay on the ring.

Comparing now the actual K. E. of the wheel, with what it would be if all the particles were concentrated on the ring of radius r, it appears there is an excess due to the particles outside the ring, and loss due to those

inside. It is not difficult then to see that by selecting properly, we can make the gain and loss balance, and if this be done, then the actual K. E. of the wheel is the same as if the whole mass were concentrated on a circle of radius r.

The value of r which satisfies the above condition is called the Radius of Gyration, and when by use of the calculus the value of the radius of gyration is found, then calling it r, the K. E. of the rotating body is WA22/2g, where A is the angular velocity.

Although we cannot calculate the value of the radius of gyration without the calculus, yet we can show the method of proceeding, and this we will now do.

We will take the fly-wheel already considered. Divide it up into a very large number of thin rings. Let their radii be in order

r1 being very nearly R, and rn nearly R'. Let the weights of the rings be

.. Total K. E. = =42 { wyry2+wary2 + wyry2+...+ww3n2}.

A2 2g

It is for the summation of the quantity in brackets, when n is indefinitely large, that we require to use the calculus. We shall meet a similar expression further on in the book, and it is known as the Moment of Inertia.

But if r be the radius of gyration,