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diagram of forces as in the figure, it is clear that F: mg::r: h; therefore

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Consequently if we have two masses m and m' hung by cords of different lengths, they will have the same period of rotation if the height h is the same in both cases.

The more rapid the rotation the higher the mass rises, and in the steam-engine governor the rising of the weights operates a lever through which steam is cut off and the speed of the engine decreased.

Problems.

1. How much weight can a cord sustain by which a mass of 100 grms. can be whirled in a circle of 1 meter radius making 2 turns per second, neglecting the effect of gravity in the circular motion?

2. A mass of 50 grms. has a velocity of 750 cms. per sec. in a circle of radius 60 cms. Find the acceleration in amount and direction and centripetal force in dynes. Also find angular velocity. 3. A 100-ton locomotive rounds a curve at a uniform speed of 40 miles per hour. Find the acceleration if the radius of curvature of the track is 1000 ft. Also find the horizontal force exerted against the rails.

4. In case of the last problem, how much higher must the outer rail be than the inner, in order that the resultant force due both to the weight of the locomotive and its centrifugal force, may be perpendicular to the road bed?

5. A mass of 1 lb. is whirled in a circle of 2 ft. radius on a smooth level table, being held in the circle by a cord which passes without friction through a hole in the center of the table and supports a 2 lb. weight. Find the angular velocity and revolutions per sec. of the 1-lb. mass necessary to support the weight.

6. A 200-gram mass is whirled in a vertical circle of radius 80 cms. with a uniform angular velocity 8 radians per sec. Find the period of revolution and the acceleration. Also what is the tension on the cord in grams when the mass is at the top of the circle and what when it is at the bottom?

7. A weight of 2 lbs. is whirled in a vertical circle. If its velocity is 100 cms. per sec. at the top of the circle, what will be its velocity at the bottom, the gain being due to the acceleration of gravity as it falls, just as in an inclined plane (see §103)? Radius 80 cms. 8. A 10-lb. mass is hung as a pendulum by a cord 4 ft. long. How high must it swing in order that the tension on the cord at the lowest point of its swing may be double the tension when at rest?

9. In case of "looping the loop," how high above the level of the topof the circle must the car start that it may just have speed enough to keep to the circle, neglecting friction? Circle 30 ft. in diameter. 10. Find the angular velocity and period of a conical pendulum hung by a cord one meter long and swinging around in a circle of 60 cms. radius.

II. A stream of water from a horizontal nozzle falls 3 ft. below the level of the nozzle in a distance of 20 ft., measured horizontally. Find the velocity of the escaping jet.

VIBRATORY MOTION.

120. Simple Harmonic Motion. If a body moving with. constant speed in a circular path is observed from a distant point in the plane of the circle, it appears to oscillate back and forward in a straight line.

The kind of vibratory or oscillatory motion that the particle appears to have in this case is known as simple harmonic motion, it may be defined as the projection upon a straight line of uniform motion in a circle.

There are other kinds of vibratory motion that are not simple harmonic, such, for example, as the particle would appear to have in the above instance if it moved around the circle in any manner whatever except with constant speed. Simple harmonic vibration is therefore one particular mode of oscillation; but it is by far the most important, for it is the most common of all, and all other modes of vibration may be expressed as the resultant of a sum of simple harmonic vibrations, as was shown by the French mathematician Fourier.

:

A simple mechanical device illustrating this kind of motion. is shown in figure 61. A pin P projects from the face of a rotating disc D and fits in a slot in a cross head which is attached to rods that can slide back and forth in the bearings B B'. When the disc rotates with uniform speed every point in the rods and cross head will move back and forth with simple harmonic motion.

The amplitude of the vibration is the distance that the vibrating body moves on each : side away from its central or mean position.

B

FIG. 61.

121. Velocity in Simple Harmonic Motion.-Let a particle A move around the circle (Fig. 62) with constant speed, and let another particle B move back and forth along a diameter D C in such a way that the line joining A and B is always perpendicular to D C. Then B oscillates with simple harmonic motion. Let vo represent the velocity of A. It may be resolved into two components, as shown in the diagram, one at right angles to the direction in which B moves and the other parallel with B's motion. Since B always keeps abreast of A, the velocity of B at any point must be equal to that component of A's velocity which is parallel to DC, namely to the component v.

D

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FIG. 62.

B

Letting

r represent the radius of the circle and y the distance AB, we have by similar triangles

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The velocity of B is therefore zero at the ends of its path at C and D, for there y=0. While at the center y=r and the velocity of B is equal to v。, its maximum value.

The complete period of an oscillation of B is evidently the same as the time in which A goes completely around the circle. Let P represent this period, and the velocity of A is

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which also expresses the velocity of B at its middle point.

122. Acceleration in Simple Harmonic Motion.-Since A and B have exactly the same motion in the direction D C, the acceleration of B must be the same as that component of the acceleration of A which is parallel to D C. The acceleration a。 of A, moving with uniform speed in a circle, is directed toward the 4 πο center of the circle and is equal to r

a. a

D

Χ

B

FIG. 63.

C

P2

(§ 117). Resolving the acceleration a。 into two components and letting a represent the component parallel to DC and b that perpendicular to it, we have by similar triangles,

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The acceleration of B is therefore proportional to its distance from the center, it is greatest when x=r and is zero when B is at the center. It will also be noticed that the acceleration of B is always directed toward the center; that is, B is always losing velocity as it moves away from the center and gaining velocity as it moves toward the center, and consequently its velocity is greatest at the center as we have already seen.

123. Force in Simple Harmonic Motion.-The fundamental dynamical equation F=ma enables us to express at once the

force in simple harmonic motion. When the mass of the oscillating particle is m and its complete period of oscillation is P, the acceleration at the instant when the particle is a distance x from its central position has just been shown to be

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and is directed always toward the center, or equilibrium position.

Therefore when a mass in equilibrium is so situated that if displaced it is always urged back toward its equilibrium position by a force which is proportional to the displacement, it will, on being displaced and then set free, oscillate with simple harmonic motion about its position of equilibrium as a center.

Now the force required to cause a small strain in almost any elastic body is proportional to the amount that the body is strained, whether the body is bent or stretched or twisted (Hooke's Law, $238), hence when such bodies are strained and then let go they oscillate to and fro in simple harmonic motion, as in case of the small vibrations of a tuning fork.

124. Problem.-Let a mass of 1 kilogram be supported by a steel spring of such stiffness that an additional weight of 100 grams will stretch it just 1 centimeter. It is required to find the period of oscillation of the weight if disturbed, neglecting the mass of the spring.

If the kilogram weight is pushed up or pulled down as it hangs on the spring, it will move through a distance which is proportional to the force used, a force of 100 grms. being required to displace it 1 cm. To produce a displacement of x centimeters the force required is 100 x grams or 100 xg dynes. But from equation (1) above we have, since m = 1000,

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