49

Rule for finding the accelerating force of a body.

Divide the force by the mass (remembering that mass is equal to weight divided by 3219) or the velocity by the time, either quotient will give the accelerating force.

EXAMPLE. A force of 25 lbs. acts on a body whose weight is 84 lbs. Find the accelerating force...

84

The mass=

= 2.6 nearly;

32-19

25

= 9.62 nearly.

2-6.

The velocity at the end of 10 seconds = 962 × 10 = 96*2.

Time of a Body falling down an Inclined Plane.

Let ABC be an inclined plane, BC perpendicular, and AB parallel to the horizon. The velocity at A in falling down A Cis the same as it would be in falling perpendicularly down the height B C.

Put t=time in falling from C to A.

1=AC the length of the inclined plane. k = B C the height of ditto.

2P
f h

Let ADEB be a circle whose diameter A Bis perpendicular to the horizon. The times of a body falling down any chords A D, A E are equal, and equal to the time in falling vertically through

AB.

The Time of Oscillation of a Simple Pendulum.
Let A B the length of the pendulum = }
And 3-14159. de; g = 32-19

T= time in seconds oscillating from the point

Bto D.

B

Rule to find the Time of one Oscillation of a Simple Pendulum.

Divide the length of the pendulum by 32:1 by 3219; extract the square root of this quotient, and multiply the result by 3-1416, and the produet will be the time of oscillation in seconds.

If L be the length of a pendulum which oscillates in one second,

T

lagutininga od Pai G

The value of L for the 1

57

dlovsa ob gruzanų ploulez

latitude of London is 39.1386 inches. A

pendulum 935, 42, 27 inches long, v third, a quarter seconds respectively.

will oscillate in a half, a

If n be the number of oscillations made by a pendulum in one hour, then ferroe

The time of oscillation is not dependent on the weight of the bob. totulotos and 1» montary to gulbry wilt belle ai d son add sal Centrifugal Force.

Put V =

Let the weight W, placed at B, be connected with a cord, or wire, with the fixed point A Bround which it revolves with a uniform velocity.

velocity of rotation.

r = A B. the length of the cord in feet.

F= centrifugal force, or the force which is exerted to break the cord in the direction of its length.

F

Athadu s 5/32:19 × K.

If ʼn be the number of revolutions in one minute,

331

...F=

× Wrn2.

ads bellas at Q mud 1000000 for a melons sigte 227% senili If W be measured in tons, then F will be in tons also.

If w be the angular velocity,

Wr w2

If T be the time of the weight making a complete revolution,

shemalt parim eiz If there be several bodies at E

B, C, D, and revolving round the axis passing through A, and perpendicular to the plane ADB C,

f

angular velocity, W W W, &c. the weights at and r &c, the distances A B A C A D. &c. 1 weights at B and C be 80 and 90 lbs. respectively, revolving at a distance A B = 8 feet, AC 12 feet, with a velocity making 40 revolutions per minute. Find force, or the pressure on the axis passing through

Had a stallions, 12 x 40 od 4

w=

The moment of inertia.

3

the centrifugal

is radT

[{W, +W2+W; + &c.) k2 = W1 r1 + W2 r2 + W ̧ r2+ &c. Each side of this que equation is called the moment of inertia, and the distance k is called the radius of gyration of the revolving system.

Let a constant force. ce. Fact at a distance Afa from the axis

of motion.

Log a di The angular velocity at the end of a second

9 Fa

The angular velocity

ty at the end of one revolution schon in at

2. Ng Farol eli to zolijoub stb al babo

VW1 + W2+ W2 + &c. × k

2

3

If a point O be determined from the equation

nisati pap at Murloren to redans afted w

where G is the centre of gravity of the system, then O is called the centre of oscillation of wà modt, gnod gi bowaboni od V1I The values of k in Geometrical Solids. pan wit ed wH A rectangular parallelopipedon revolving about an axis passing through its centre of gravity, and parallel to either of its edges. rekulover blquno a gaʊlytow sill be smit of

where be are the length and breadth at right angles to the axis of revolution

An upright triangular prism about a vertical axis passing through its centre of gravity.

by (a)

The section of the prism perpendicular to the revolving axis is an isosceles triangle, the base being and the per pendicular upon it from the angle contained by the equal sides by (c).

In a cylinder, whose radius is (r), revolving about its axis, tal drode

gaivľovej agy voltergy to enhar

L* =
2.

In a hollow eylinder, whose internal and external radii area and respectively, revolving about its laxisang taout or brode kovių ste a2 + b2

In a cylinder, whose radius is r and length &, revolving round a line at right angles to its axis, and passing through its middle,

k

12

In a sphere, whose radius is r, revolving about its diameter,

In a hollow sphere, whose internal and external radii are (a) and (a) respectively, revolving about its diameter,

In a cone, whose radius of base is r and height, revolving about a line at right angles to its axis, and passing through its centre of gravity,

3 (4y2+43)
80

The square of the radius of gyration about any

line in a revolv

ing system, is equal to the square of the radius of gyration about a line parallel to it passing through the centre of gravity and the square of the distance from the centre of gravity to the line about. which the system revolves.

Let G be the centre of gravity of any body; draw AB any line about which the system revolves. Let CD be parallel to AB, and draw GH perpendicular to A B.

Let Kradius of gyration when revolving
about A B

kradius of gyration when revolving
about CD.

: K2 = L2 + G H2.

This important theorem will readily extend the theorems which are given above to most practical cases.

The Centre of Gyration

is that part of a body revolving about an axis, into which, if the whole quantity of matter were collected, the same moving force would generate the same angular velocity.

To find the centre of Gyration, multiply the weight of the several particles by the squares of their distances from the centre of motion, and divide the sum of the products by the weight of the whole mass; the square r root of the quotient will be the distance of the centre of gyration, from the centre of motion.

The distances of the centre of gyration from the centre of motion, of different revolving bodies, are as follows:

In a straight rod revolving about one end, the length × 5773. In a circular plate, revolving on its centre, the radius × 7071. In a circular plate, revolving about one diameter, the radius × 5. In a thin circular ring, revolving about one diameter, radius * '7071.

In a solid sphere, revolving about one diameter, the radius

× 6325.

In a thin hollow sphere, revolving about one diameter, the radius

x 8164.

In a cone, revolving about its axis, the radius of the base ×

*5477.

In a right-angled cone, revolving about its vertex, the height X4886.98

In a paraboloid, revolving about its axis, the radius of the base × 5773.

The Centre of Percussion

C

is that point in a body revolving about a fixed axis, into which the whole of the force or motion is collected

3

It is, therefore, 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 the centre of percussion.