DYNAMICS.

DYNAMICS is the investigation of body, force, velocity, space, and

time.

Let them be represented by their initial letters bfvst, gravity by g, and momentum or quantity of motion by m; this is the effect produced by a body in motion.

Force is motive, and accelerative or retardative.

Motive force, or momentum, is the absolute force of a body in motion, and is the product of the weight or mass of matter in the body, multiplied by its velocity.

Accelerative or retardative force is that which respects the velocity of the motion only, accelerating or retarding it, and is found by the force being divided by the mass or weight of the body. Thus,

if a body of 4 lbs. be acted upon by a force of 40 lbs., the accelerating force is 10 lbs. ; but if the same force of 40 act upon another body of 8 lbs., the accelerating force then is 5 lbs., only half the former, and will produce only half the velocity.

Uniform Motion.

The space described by a body moving uniformly is represented by the product of the velocity into the time.

With momenta, m varies as bv.

EXAMPLE.-Two bodies, one of 20, the other of 10 lbs., are impelled by the same momentum, say 60. They move uniformly, the first for 8 seconds, the second for 6; what are the spaces described by both?

Then tv 3×8 = 24=s, and 6×6=36=s.

Thus the spaces are 24 and 36 respectively.

Motion Uniformly Accelerated.

- In this motion, the velocity acquired at the end of any time whatever, is equal to the product of the accelerating force into the time, and the space described is equal to the product of half the accelerating force into the square of the time.

The spaces described in successive seconds of time are as the odd numbers, 1, 3, 5, 7, 9, &c.

Gravity is a constant force, and its effect upon a body falling freely is represented by g.

The following theorems are applicable to all cases of motion uniformly accelerated by any constant force:

When gravity acts alone, as when a body falls in a vertical line, F is omitted

and we have,

NOTE. g is obviously 32.166 from what has been given in rules for Gravitation, and is the force of gravity.

If, instead of a heavy body falling freely, it be propelled vertically upward or downward with a given velocity, v, then

s=tv

>

an expression in which must be taken when the projection is upward, and + when it is downward.

Motion over a Fixed Pulley.

Let the two weights which are connected by the cord that goes over the pulley be represented by W and w; then

Or, if the resistance of the friction and inertia of the pulley be represented by r1 then

EXAMPLE.-If by experiment it is ascertained that two weights of 5 and 3 lbs. over a pulley, the heavier weight descended only 50 feet in 4 seconds, what is the measure of r?

If r is not considered, the heavier weight would fall 64 feet.

TABLE of the Effects of a Force of Traction of 100 lbs. at different Velocities, on Canals, Railroads, and Turnpikes.

13.20

4.282 3.040

14.400 10 14.66 3.468 2.462 14.400 13.5 19.9 1.900 1.350 14.400

The load carried, added to the weight of the vessel or carriage which contains it, forms the total mass moved, and the useful effect is the load.

The force of traction on a canal varies as the square of the velocity; on a railroad or turnpike the force of traction is constant, but the mechanical power necessary to move the carriage increases as the velocity.

10.800 1.800

1.350

10.800

1.800 1.350

10.800

1.800 1.350

PENDULUMS.

THE Vibrations of Pendulums are as the square roots of their lengths. The length of one vibrating seconds in New-York at the level of the sea is 39.1013 inches.

To find the Length of a Pendulum for any Given Number of Vibrations in a Minute.

RULE.-As the number of vibrations given is to 60, so is the square root of 39.1013 (the length of the pendulum that vibrates seconds) to the square root of the length of the pendulum required. EXAMPLE.-What is the length of a pendulum that will make 80 vibrations in a minute?

As √39.1013×60=375, a constant number,

Then

375 80

=

= 4.6875, and 4.68752 21.97 inches, Ans.

The lengths of pendulums for less or greater times is as the square of the times; thus, for a second it would be the square of 39.1013 = 9.7753 inches, the length of a second pendulum

, or 4 at New-York.

To find the Number of Vibrations in a Minute, the Length of the Pendulum being given.

RULE.As the square root of the length of the pendulum is to the square root of 39.1013, so is 60 to the number of vibrations required. EXAMPLE.-How many vibrations will a pendulum of 49 inches long make in a minute?

√49:39.1013: 60 number of vibrations.

Or,

375

√49

53.57 vibrations, Ans.

To find the Length of a Pendulum, the Vibrations of which will be the same Number as the Inches in its Length.

RULE.-Square the cube root of *375, and the product is the an

swer.

EXAMPLE.-3757:211247, and 7.2112472 — 52.002, Ans.

The Length of a Pendulum being given, to find the Space through which a Body will fall in the Time that the Pendulum makes one Vibration.

RULE.--Multiply the length of the pendulum by 4.93482528, and it will give the answer.

* 375 is the constant for the latitude of New-York; in any other place, multiply the square root of the length of the pendulum at that place by 60.

EXAMPLE. The length of the pendulum is 39.1013 inches; what is the distance a body will fall in one vibration of it?

39.1013×4.9348192.9578 inches, or 16.8298 feet, Ans.

All vibrations of the same pendulum, whether great or small, are performed very nearly in the same time.

In a Simple Pendulum, which is, as a ball, suspended by a rod or line, supposed to be inflexible, and without weight, the length of the pendulum is the distance from its centre of gravity to its point of suspension. Otherwise, the length of the pendulum is the distance from the point of suspension to the Centre of Oscillation,* which does not coincide with the centre of gravity of the ball or bob.

CENTRE OF GYRATION.

THE Centre of Gyration is the point in any revolving body, or system of bodies, that, if the whole quantity of matter were collected in it, the angular velocity would be the same; that is, the momentum of the body or system of bodies is centred at this point.

If a straight bar, equally thick, was struck at this point, the stroke would communicate the same angular velocity to the bar as if the whole bar was collected at that point.

To find the Centre of Gyration.

RULE 1-Multiply the weight of the several particles by the squares of their distances in feet from the centre of motion, and divide the sum of the products by the weight of the entire mass; the square root of the quotient will be the distance of the centre of gyration from the centre of motion.

EXAMPLE. If two weights of 3 and 4 lbs. respectively be laid upon a lever (which is here assumed to be without weight) at the respective distances of 1 and 2 feet, what is the distance of the centre of gyration from the centre of motion (the fulcrum)?

That is, a single weight of 7 lbs., placed at 1.64 feet from the fulcrum, and revolving in the same time, would have the same impetus as the two weights in their respective places.

*See Centre of Oscillation.

RULE 2.--Multiply the distance of the centre of oscillation, from the centre or point of suspension, by the distance of the centre of gravity from the saine point, and the square root of the product will be the answer.

EXAMPLE.--The centre of oscillation is 9 feet, and that of gravity is 4 feet from the centre of the system, or point of suspension; at what distance from this point is the centre of gyration?

=

9X4 36, and √36 = 6 feet, Ans.

The following are the distances of the centres of gyration from the centre of motion in various revolving bodies, as given by Mr. Farey :

In a straight, uniform Rod, revolving about one end; length of rod X.5773. In a circular Plate, revolving on its centre; the radius of the circle X.7071. In a circular Plate, revolving about one of its diameters as an axis; the radius X.5.

In a Wheel of uniform thickness, or in a Cylinder revolving about the axis; the radius X.7071.

In a solid Sphere, revolving about one of its diameters as an axis; the radius X.6325.

In a thin, hollow Sphere, revolving about one of its diameters as an axis; the radius X.8164.

In a Cone, revolving about its axis; the radius of the circular base X.5477. In a right-angled Cone, revolving about its vertex; the height of the cone X.866. In a Paraboloid, revolving about its axis; the radius of the circular base X.5773. In a straight Lever, the arms being R and r, the distance of the centre of gyraR3+r3 tion from the centre of motion; 3(R-)

NOTE. The weight of the revolving body, multiplied into the height due to the velocity with which the centre of gyration moves in its circle, is the energy of the body, or the mechanical power which must be communicated to it to give it that motion.

EXAMPLE. In a solid sphere revolving about its diameter, the diameter being 2 feet, the distance of the centre of gyration is 12X.63257.59 inches.