1 Gravititional Acceleration, Simple Harmonic and Rotational Motion

8/9/2019 1 Gravititional Acceleration, Simple Harmonic and Rotational Motion

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International Agriculture Section - Course in Physics - by Dr. Mohamed Fahmy Hussein

Gravitational Acceleration and Rotational Motion

Content of Lecture

1. The Scientific Methodology2. The simple harmonic motion versus the rotational motion3. Analysis concerning the circular pathway4. Analysis concerning the simple pendulum5. Derivation of an operational formula for determining the gravitational

acceleration using a simple pendulum6. Experimental work (see the lab session) (LINK)7. Remarks on the rotational movement

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The Scientific Method

Scienceis a way of acquiring knowledge. To do science, one must follow a specific universalmethodology. The central theme in this methodology is the testing ofhypotheses. The overall goalof science is to better understanding the Nature and our universe.

Various fields of study, like physics, chemistry, biology, and medicine have used science toexpand their knowledge base because science is neutral and unbiased in its operation. Learning andmaking decisions about the environment also requires similar qualities. As a result, science has

become the dominant form of inquiry when studying the environment and its problems.The broadest, most inclusive goal of science is to understand nature, or the match between

observed reality and some conceptual idea. Understanding encompasses all the other goals ofscience, many of which are quite specialized. Explanation is perhaps the next most important goalof science. Explanation consists of relating observed reality to a system of concepts, laws, orempirically based generalizations. Explanation may also relate observed phenomena to a networkof causes, or link them hierarchically to lower-level mechanisms.

Another general goal of science is generalization, which may be considered in two ways.

First, generalization may be considered as the condensation of a body of empirical fact to a simplestatement. In the process of such condensation, it is likely that some detail must be omitted and the

phenomenon abstracted. Generalization may also involve isolating the phenomenon from otheraspects of the system of interest. This constitutes idealization. A second view of generalization isthe unification of apparently unrelated phenomena in the same abstract or ideal system of concepts.

Francis Bacon (1561-1626), 17th century English philosopher, was the first individual tosuggest a universal methodology for science. He believed that scientific method required the

process ofinduction. Karl Popperlater refuted this idea in the 20th century. Popper suggested thatscience could only be done using a deductivemethodology.

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1. Simple Harmonic Motion versusRotational MotionLooking at the movement of a small ball on a circular track, we observe that the projections

(on a wall) of this movement will follow a simple harmonic motion (similar to that of a ball fixed atthe end of a simple pendulum).

We can use this relationship in order to understand both of these motions and to drive aformula for the second motion. We can also use the derived formula for the determination of the

gravitational acceleration (g).Note that all derivations are based on the assumption that the simple pendulum makes small

angle (Z) shifts (which are recommended to be


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Figure 1b. The driving force (-mg sin Z) of the simple harmonic motion.

Analysis concerning the circular pathway

1. There could be a tangential linear acceleration a tangential acting on the same direction as V1.When it is present, it is related to angular by (a tangential = angular r). This means that V1, ingeneral, is not constant. However, when the angular velocityis constant, V1 is also constant(this it is the case in a steady moving centrifuge and at the tip of an airplane propeller, as

supposed in Fig. 1). In this last case, there is no a tangential term. In Fig. 1 and in the followingderivations, note that same symbol, a, is used for the acceleration of a simple harmonic motion.DO NOT confuse it with and atangential.

2. In the case of non-constant angular velocity , the accelerations centripetal and tangential have a

vector(=). This vector will vanish ifis constant.3. For a steady moving centrifuge there is only a centripetal acceleration centripetal = 2r.

Figure 2. Derivation of the angular centripetal acceleration = 2r.

In a steady-moving centrifuge, despite the constant magnitude ofV1, the change of directionmeans that there is an angular acceleration (centripetal acceleration).

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Since the track is circular, the force (= mass of the ball moving on the circumferencemultiplied by the centripetal acceleration) acting on the ball should be pointing to the center of the

circle and perpendicular to the tangent V1.Note that cos 90 = zero. Consequently, there is no component of this force in the direction of

the tangent. This means that the velocity V1 on the tangent is constant; only the direction of the

tangent is permanently changing. Otherwise, the ball will not follow the circular pathway, andvelocity on the tangent would be changing in both the direction and the magnitude (which iscontradictory to the supposed steady circular movement).

Analysis concerning the pendulum

The ball of thependulum moves with a linear velocityV2. This velocity decreases to zero atboth ends of the track, and has a maximum value at the middle. Since this velocity changes in

magnitude, there is a decelerationa (= a negative acceleration) for the simple harmonic motion.From Fig. 1, there is a relationship between the linear velocity of the simple harmonic motion

V2 and the linear velocity tangent to the circular pathwayV1, This relationship is determined bycos . V2 vanishes at = 90.V2 = V1 cos

In addition to the expression of the velocity on the circular pathway by a tangent linear

velocityV1, the circular motion can also be expressed by dividing the angular displacement (rad)on the circumference by time, t. The result is known as angular velocity,, (also known as theradian velocity, since measured in radians/sec) which can be related to V1 as the following:Angular velocity =

However, the angular displacementrad =Consequently, angular velocity = *Multiply both sides by rr = r =r = V1So the relationV2 = V1 cos will now read V2 = rcos

Also, in the shadowed right-angle triangle inside the circle in Fig. 1,

cos = i.e. vertical = rcos Howerver, vertical =Consequently, V2 =

Furthermore, from Fig. 1, there is a relationship between the deceleration (a) of the simplependulum and the centripetal acceleration () governed by cos (90-) i.e. sin :

a = - cos (90- ) or a = - sin

In Figure 2, the centripetal acceleration can be related to the angular velocity through themethod known as the closed triangle as following (Fig. 2).

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In order to calculate the change in the linear velocity (V1) on the circular pathway (this is adirectional change, not a change in magnitude, remember!), when the angle is small (this is animportant condition and assumption, remember!), we have:

The arc AB the segment or the hypotenuse ABAnd from the two similar triangles

= =The segment AB is a distance. So it can be replaced by the product V t. Consequently,

=Multiply by V

V =Divide by t

V / t =

=However, we have previously shown that:r = V1,

Consequently,

= = 2r

2. Derivation of an Operational Formula for Determining the

Gravitational Acceleration Using a Simple Pendulum

By changing the length (L) of the pendulum string, the square of the time of one oscillation

(t2) will proportionally change. A formula that is based on this relation can be found and used todetermine the gravitational acceleration (g).

From deriving the relationship of the circular movement to the simple harmonic motion, we

have already learned that the acceleration (a) of the simple harmonic motion is related to thecentripetal acceleration() by:

a = - sin

In addition, we have previously shown that:

= 2rConsequently,

a = - 2rsin

a = - 2ra = - 2y

However, we know that the force acting on the ball of simple pendulum is given by:F = m * acceleration

= - m * g sin ZConsequently, this negative acceleration (deceleration) of the pendulum is

Acceleration = - g sin Z6


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Acceleration = - gComparing the relation a = - 2y with Acceleration = - gIt can be noticed that:

2y = g or 2 = i.e. = and g = 2LHowever, we already know that:

= and for a circle rad = 2Consequently, for one turn on a circle (corresponding to one oscillation of a simple

pendulum):

=This means that:

==

g = 42

It is obvious that this relation can be used to calculate g by one reading forL and t2

.However, it is interesting to write this relation in the following ready to plot form:

t2 = * LThis is the form of a linear equation, where t2 is the dependent variable (to be plotted on the

ordinate. i.e. on the vertical Y-axis). This variable proportionally changes with the change of

pendulum length L (to be plotted on the abscissa. i.e. on the horizontal X-axis) as the independentvariable.

This linear relationship is the basis for plotting your set of observations (t2 in function ofL).From the plot, you get the slope = and its inverse = .

Note that if you multiply this inverse by 42(=39.48) by, we get 42.

This means that g can directly be calculated from the plot by the product 42 .g = 42

This is the operational equation that we were looking-for!

3. Experimental Work1. The needed equipment is fairly simple: a simple pendulum that you can arrange anywhere (by

fixing a string filament to a point on the top of metal bar fitted with a hanging block, and in thelower end of the filament, fix a small copper or steel ball).2. Use a stopwatch to record the time of 20 oscillations (one oscillation is the movement of the

pendulum ball from a start point forth, then back to the start point, this is equivalent to onecomplete turn on the circumference of a circle, Fig. 1). Divide by 20 to get the time of oneoscillation.

3. Record the time readings. Start at short pendulum lengths. Then gradually increase the length.

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OBSERVATIONS

L time of time of T2 X Y

cm 20 oscillations 1 oscillation sec2

L T2

120 42.0 2.10 4.41 120 4.41

110 40.0 2.00 4.00 110 4.00

100 39.0 1.95 3.80 100 3.80

90 38.0 1.90 3.61 90 3.61

80 36.0 1.80 3.24 80 3.24

70 33.8 1.69 2.86 70 2.86

60 30.9 1.55 2.39 60 2.39

50 28.5 1.43 2.03 50 2.03

40 25.5 1.28 1.63 40 1.63

30 22.5 1.13 1.27 30 1.27

20 18.9 0.95 0.90 20 0.90

15 15.3 0.76 0.58 15 0.58

10 14.0 0.70 0.49 10 0.49

5 11.0 0.55 0.30 5 0.30

0 0.0 0.00 0.00 0 0.00

4. Plot the relationship of L and T2 as shown on Figure 3. Calculate g using the practicalrelation:

g = 42m/sec2 = dimensionless term or rad2 * m/sec2

Example:To calculate the time needed for a simple pendulum to move with oscillation time of one

second (the time of one oscillation = 1 second), the needed length, L, can be calculated by:

= = = 0.253 m = 25.3 cmNotes:

1. Experimental errors are unavoidable, however, be aware of recording observations quitcarefully and in well-organized manner. Otherwise, you will get false results (and poorscore!)

2. You must record units (or dimensions) for each term in any equation. No physical

quantity is acceptable if it is void of units (even dimensionless terms must be indicatedas such!).

3. Do not use too short (or too long!) filament lengths and do not let the ball movethrough a great angle, nor to turn around itself while moving back and forth.

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g = gravitational acceleration

0

1

2

3

4

5

6

0 40 80 120

L, cm (abscissa)

T2

sec

2(

ordinate)

student data

general best fit

zero intercept best

fit

Assistant data

short statistical fit

calculate slope on

gaphic, take its

inverse, this givesL/T

2to be

used in the

equation:

g = 4 2 L/T2

Figure 3: Get the inverse of the slope, substitute in the equation. That is all !

ideal pendulum length, L= 30 - 100 cm

-10

-5

0

5

10

15

20

0 20 40 60 80 100 120

L = Length, cm

T2,

deviation%

data

Figure 4. Too short and too long cords relatively give large errors.

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0

0.1

0.2

0.3

0.4

0.5

0.6

0102030405060708090

V2

m s-1

of simplependulum

Angle of corresponding rotational motion

-0.03

-0.02

-0.01

0.00

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

the force

component of

the simple

harmonic

motion

- mg cos (90-Z)

=

- mg SIN Z

Newton

Angle, Z

Figure 5. Velocity and acceleration change during motion of the simple pendulum and thecorresponding rotational movement.

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4. Remarks on the rotational motion

The reaction of the centripetal force is the centrifugation force.

An angle in radians rad =. (Note that the circumference = 2 r) expressed in rad = = = .It is equal to 3.14rad. Its equivalent in degrees is 180.

For a circle, the circle arc=2r, so circle angle in rad = 2= (2 * 3.14) rad = 6.28 rad.

Transformations:

Angle in rad = Angle in degrees*

= Angle in degrees*

= Angle in degrees* 0.0174

Angle in degrees = Angle in rad*

= Angle in rad*

= Angle in rad* 57.2968

One radin degrees = 1* = = 57.3

example: 2rad = 2*() = 2*() = 2*(57.3) = 114.6

example: a circle is 360(= 2* in rad * 57.3 = 2* 3.14 * 57.3 = 360)

example: 360 in rad = (360* = 2 = 2 * 3.1416 = 6.2832 rad)2 = (i.e. is the ratio: ) is the ratio:) i.e. is the ratio )=

in addition, is the ratio between the angle 180 and the angle 1 radwhere rad should only be expressed as (180 / ), not in degrees.

example: for180, the equivalent in rad = 180* = = 3.14

example: for 60, the equivalent in rad = 60* = =

Angular velocity: = angle in rad / time t in seconds for example

Definition and calculation of:

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Figure 6. Division en 90/n angles.

If the angle (90) between two perpendicularradius radials is divided into a great number (n) ofsmall angles, the value of any of these small angles will be.And taking the sinus, sin =

Nevertheless, since the angle is small, the opposite is approximately equal to the arc, and thehypotenuse is a radius in the circle. Consequently,sin =,

i.e. it is equal to the value of the small angle as expressed in rad

However, this is also (by definition) the value of any angle in rad.

This means that, in the case of small angle (up to 15),

the value of the angle in rad = the sin of the angle

In brief: value of the small angle in rad = arc / radius= opposite / hypotenuse= sin

Multiplying by n, we get: n (sin)= n

Obviously, this can be expressed as =

But we already know that = / 2

So, n(sin) = / 2And = 2 n(sin)

As n becomes a great number, the value of will be numerically stable. That is why it isexpressed by the following limit (this is a definition of):

= 2 l im n (n sin )

The kinetic energy of a body moving on a circular pathway = mV2

= m2

r2

(Where mr2 is the moment of inertia with respect to the rotation axis)

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The centripetal force F = mV2 / r= m2r

The periodic time t (the time needed to make a complete rotation)

t =

t =i.e., the angular velocity, =

If(n) is the number of rotations per minute (rpm),The total angles (in rad) rad/min = 2 nAnd, rad/sec = 2

If we know the value ofin rad/sec, we can calculate (n) byn rev/sec = in rad/sec * ()In addition, dividing rad/sec by 60 you get in rad/min.

Also, knowing n rpm and the distance rfrom the rotation axis we can calculate (which isthe unique acceleration in a centrifuge with a constant , since the constant makes the tangentialacceleration vanished) using the equation:

= 2r where, rad/sec = 2

Example

If r = 0.20 m(n rpm) = 6000(n rps) = 6000/60 = 100

(The time of 1 revolution = 0.01 sec)

= (2 3.14) () rps = 6.28 100 rps = 628 rad/sec2 =394784 rad2/sec2

= 2r=394784 * 0.20 = 78957 m/sec2, (note that rad2 is usually dropped)The centripetal acceleration can be calculated as times the gravitational acceleration (this is

called xg) g as following:Divide the value ofby the value ofg (which is about 10 m.sec-2). Here you get 7896 g.(i.e. is approximately 8000 times g).

Inverse Example:

In order to adjust to 45 m.sec-2 (i.e. about 4.5g), for a rotation axis, r, of5 m,will be rad/sec = rad/sec, i.e. will be 3rad/sec.

So, the value ofnrps can be calculated:

nrps = = rps = 0.477 rps,nrpm = (0.477)*60 = 28 rpm.

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By increasing rto 11.25 m, will be 2rad/sec, and nrpm = 19 rpm.(To keep xg the same on a larger radius centrifuge, nrpm must be lowered. Inversely, to keepxg the same on a smaller radius centrifuge, nrpm should be increased).

Note that theperiodic time

tin minutes is

t=

And frequencyfis the inverse of the periodic time, , and: f =

Note that the rigorous units of the angular velocityare rad.sec-1 . However, rad is usuallyomitted. Also in its relation to the angular acceleration, which is = 2r, the units forarerigorously m.rad2/sec2. However, rad2 is always omitted and we only retain m/sec2.

Retain the following formulae

Centripetal = 2r

Angular velocity =Angular displacement rad =For a circle rad = 2For the simple pendulum = =For1 revolution on a circle (corresponding to 1 oscillation of simple pendulum)

=Fornrevolutions on a circle (corresponding to n oscillations of pendulum) during I minute:

rad/min =Where n is the number ofrevolutions per minute (rpm), and the denominatort is 1 min.Hence: rad/sec = 2

rad/sec = 2 n rpsn rps =

To get, divide by 60

= mCentripetal = m2r = m

= =

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