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188 CENTRES OF PERCUSSION AND OSCILLATION.

CENTRES OF PERCUSSION AND OSCILLATION.

THE Centres of Percussion and Oscillation being in the same point, their properties are the same, and their point is, that in a body revolving around a fixed axis, which, when stopped by any force, the whole motion, and tendency to motion, of the revolving body is stopped at the same time.

It is also that point of a revolving body which would strike any obstacle with the greatest effect, and from this property it has received the name of percussion.

As in bodies at rest, the whole weight may be considered as collected in the centre of gravity; so in bodies in motion, the whole force may be considered as concentrated in the centre of percussion therefore, the weight of a bar or rod, multiplied by the distance of the centre of gravity from the point of suspension, will be equal to the force of the rod, divided by the distance of the centre of percussion from the same point.

EXAMPLE. The length of a rod being 20 feet, and the weight of a foot in length equal 100 oz., having a ball attached at the under end weighing 1000 oz., at what point of the rod from the point of suspension will be the centre of percussion ?*

The weight of the rod is 20×100 2000 oz., which, multiplied by half its length, 2000×1020000, gives the momentum of the rod. The weight of the ball = 1000 oz., multiplied by the length of rod, 1000×20, gives the momentum of the ball. Now the weight of the rod multiplied by the square of the length, and divided by 2000×202 =266666, the force of the rod, and the weight of the ball multi3 plied by the square of the length of the rod, 1000×202400000, is the force of the 266666+-400000 ball: therefore, the centre of percussion =

3,

=

20000+20000

16.66 feet.

EXAMPLE.--Suppose a rod 12 feet long, and 2 lbs. each foot in length, with 2 balls of 3 lbs. each, one fixed 6 feet from the point of suspension, and the other at the end of the rod; what is the distance between the points of suspension and percussion?

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As the centre of percussion is the same with the centre of oscillation in the non-application to practical purposes, the following is the easiest and simplest mode of finding it in any beam, bar, &c.:

Suspend the body very freely by a fixed point, and make it vibrate in small arcs, counting the number of vibrations it makes in any time, as a minute, and let the number of vibrations made in a minute be called n; then shall the distance of the centre of oscillation from the point of suspension be

140850
n2

inches. For the

length of the pendulum vibrating seconds, or 60 times in a minute, being 39 inch

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CENTRES OF PERCUSSION AND OSCILLATION.

189

es, and the lengths of the pendulums being reciprocally as the square of the num 602×391 ber of vibrations made in the same time, therefore n2: 602 :: 39:

140850 n2

n2

=

being the length of the pendulum which vibrates n times in a minute, or

the distance of the centre of oscillation below the axis of motion.

There are many situations in which bodies are placed that prevent the application of the above rule, and for this reason the following data are given, which will be found useful when the bodies and the forms here given correspond:

1. If the body is a heavy, straight line of uniform density, and is suspended by one extremity, the distance of its centre of percussion is of its length.

2. In a slender rod of a cylindrical or prismatic shape, the breadth of which is very small compared with its length, the distance of its centre of percussion is nearly of its length from the axis of suspension.

If these rods were formed so that all the points of their transverse sections were equidistant from the axis of suspension, the distance of the centre of percussion would be exactly of their length.

3. In an Isosceles triangle, suspended by its apex, and vibrating in a plane perpendicular to itself, the distance of the centre of percussion is of its altitude. A Îine or rod, whose density varies as the distance from its extremity, or the point of suspension; also Fly-wheels, or wheels in general, have the same relation as the isosceles triangle, the centre of percussion being distant from the centre of suspen sion of its length.

4. In a very slender cone or pyramid, vibrating about its apex, the distance of its centre of percussion is nearly of its length.

The distance of either of these centres from the axis of motion is found thus: If the Axis of Motion be in the vertex of the figure, and the motion be flatwise; then, in a right line, it is of its length.

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EXAMPLE.-What must be the length of a rod without a weight, so that when hung by one end it shall vibrate seconds?

To vibrate seconds, the centre of oscillation must be 39.1013 inches from that of suspension; and as this must be of the rod,

Then 2:3:39.1013: 58.6519, Ans.

EXAMPLE. What is the centre of percussion of a rod 23 inches long? of 23 15.3 inches from the point of suspension or motion.

EXAMPLE. In a sphere of 10 inches diameter, the thread by which it is suspended being 20 inches, where is the centre of percussion or oscillation?

2X52 5(20+5)

50 125

+5+20=== +25=25.4, Ans..

These centres are in the same point only when the body is symmetrical with regard to the plane of motion, or when it is a solid of revolution, which is commonly the case.

CENTRAL FORCES.

ALL bodies moving around a centre or fixed point have a tendency to fly off in a straight line: this is called the Centrifugal Force; it is opposed to the Centripetal Force, or that power which maintains the body in its curvilineal path.

The centrifugal force of a body, moving with different velocities in the same circle, is proportional to the square of the velocity. Thus, the centrifugal force of a body making 10 revolutions in a minute is four times as great as the centrifugal force of the same body making 5 revolutions in a minute.

To find the Centrifugal Force of any Body.

RULE 1. Divide the velocity in feet per second by 4.01, also the square of the quotient by the diameter of the circle; the quotient is the centrifugal force, assuming the weight of the body as 1. Then this, multiplied by the weight of the body, is the centrifugal force.

EXAMPLE.-What is the centrifugal force of the rim of a flywheel 10 feet in diameter, running with a velocity of 30 feet in a second?

30-4.01X7.48÷105.59 times the weight of the rim, Ans. NOTE.-When great accuracy is required, find the centre of gyration of the body and take twice the distance of it from the axis for the diameter.

RULE 2.-Multiply the square of the number of revolutions in a minute by the diameter of the circle in feet, and divide the product by the constant number 5870; the quotient is the centrifugal force when the weight of the body is 1. Then, as in the previous rule, this quotient, multiplied by the weight of the body, is the centrifugal force.

EXAMPLE.-What is the centrifugal force of a grindstone, weighing 1200 lbs., 42 inches in diameter, and turning with a velocity of 400 revolutions in a minute?

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The central forces are as the radii of the circles directly, and the squares of the times inversely; also, the squares of the times are as the cubes of the distances. Hence, let v represent velocity of body in feet per second,

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Dr. Brewster has furnished the following:

1. The centrifugal forces of two unequal bodies, having the same velocity, and at the same distance from the central body, are to one another as the respective quantities of matter in the two bodies.

2. The centrifugal forces of two equal bodies, which perform their revolutions in the same time, but are different distances from their axis, are to one another as their respective distances from their axis.

3. The centrifugal forces of two bodies, which perform their revolutions in the same time, and whose quantities of matter are inversely as their distances from the centre, are equal to one another.

4. The centrifugal forces of two equal bodies, moving at equal distances from the central body, but with different velocities, are to one another as the squares of their velocities.

5. The centrifugal forces of two equal bodies, moving with equal velocities at different distances from the centre, are inversely as their distances from the centre.

6. The centrifugal forces of two unequal bodies, moving with equal velocities at different distances from the centre, are to one another as their quantities of matter multiplied by their respective distances from the centre.

7. The centrifugal forces of two unequal bodies, having unequal velocities, and at different distances from their axis, are, in the compound ratio of their quantities of matter, the squares of their velocities, and their distances from the centre.

The weight of the rim of a fly-wheel for a 20 horse engine is 6000 lbs., the diam eter 16 feet, and the revolutions 45; what is its centrifugal force?

33120 lbs., Ans. SUMMARY.-Let b represent any particle of a body B, and d its distance from the axis of motion, S.

G, O, R, the centres of Gravity, Oscillation, and Gyration

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FLY-WHEELS.

To find the Weight of Fly-wheels.

RULE.-Multiply the horses' power of the engine by 2240, and divide the product by the square of the velocity of the circumference of the wheel in feet per second; the quotient will be the weight in 100 lbs.

EXAMPLE. The power of an engine is 35 horses, the diameter of the wheel 14 feet, and the revolutions 40; what should be the weight of the wheel?

35x2240-40×14×3.1416÷60

2 78400
858.5

X100 9130 lbs.

The weight of the wheel in engines that are subjected to irregu lar motion, as in the cotton-press, rolling-mill, &c., must be greater than in others where so sudden a check is not experienced, and 3000 would be a better multiplier in such cases.

GOVERNORS.

THE Governor acts upon the principle of central forces.

When the balls diverge, the ring or the vertical shaft raises, and that in proportion to the increase of the velocity squared; or, the square roots of the distances of the ring from the top, corresponding to two velocities, will be as these velocities.

EXAMPLE. If a governor make 6 revolutions in a second when the ring is 16 inches from the top, what will be the distance of the ring when the speed is increased to 10 revolutions in the same time?

As 10 6 16 inches: 2.4 inches, which, squared, is 5.76 inches, the distance of the ring from the top.

A governor performs in one minute half as many revolutions as a pendulum vibrates, the length of which is the perpendicular distance between the plane in which the balls move and the centre of suspension.

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