CHAPTER 5 - Circular Motion; Universal Gravitation



Physics 3310

Chapter 5

NOTES

Circular Motion; Gravitation

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STUDENT NAME

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TEACHER’S NAME

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PERIOD

CHAPTER 5 - Circular Motion; Universal Gravitation

Circular Motion is a class of PERIODIC MOTION. Motion that repeats itself; that completes a cycle; that finishes where it starts; that oscillates; motion that repeats itself; that completes a cycle; that finishes where it starts; that oscillates; motion that repeats itself; that completes a cycle; that finishes where it starts; that oscillates…

Recall:

[pic]

Conversions:

[pic]

Uniform Circular Motion

• Uniform in this case means constant speed.

• Can an object moving with a constant speed have acceleration?

• If so, we physicists are going to want to find a value for it, but how?

NOTE: The magnitude of the velocity is not changing, but the direction is changing.

In Figure 2 triangle made by the two radii and Δl

will be similar to the triangle made by v1 ,v2 , Δv.

PROOF:

By definition:

The angle between the radius and the instantaneous velocity is always 90o. (See figure 1)

For this to remain true, the angle between r1 and r2 must be equal to the angle between v1 and v2. (See figure 2 (a) and (b))

Therefore, [pic]

And [pic]

If [pic]

Note: For small values of Δθ, Δl is ≈ s (arc length).

See Figure 2

Therefore since [pic], and [pic] is our definition of linear speed, v.

Then [pic] (Equation 1)

Hopefully, it is clear that this formula can only give positive answers, and therefore only can find the magnitude of the acceleration.

Acceleration is a VECTOR so… in what direction is the acceleration?

PROOF:

[pic] Centripetal vs. Centrifugal

Centri from the Latin Centrum: the middle point of a circle, the center.

Petal from the Latin Peto – to seek, strive after, endeavor to obtain.

Fugal from the Latin Fugito – to flee, to fly from, avoid or shun.

By looking at the direction of the change in velocity, you should be able to see the direction of the acceleration.

Is it Centripetal or Centrifugal?

[pic]

The acceleration of an object moving with Uniform Circular Motion is known as centripetal acceleration or radial acceleration.

To use equation # 1 we will need to know both the radius of the circle and the speed of the object.

[pic]

Problems:

1. (I) A jet plane traveling 1800 km/h (500 m/s) pulls out of a dive by moving in an arc of radius 6.00 km. What is the plane’s acceleration in g’s?

2. (I) Calculate the centripetal acceleration of the Earth in its orbit around the Sun and the net force exerted on the Earth. What exerts this force on the Earth? Assume that the Earth’s orbit is a circle of radius 1.50 x 1011 m.

3. Will the acceleration of a car be the same when it travels around a sharp curve at 60 km/h, as when it travels around a gentle curve at the same speed? Explain.

How do we find the speed of an object going in a circle?

• Measure the time it takes for the object to complete a certain number of revolutions.

EXAMPLE: We know an object makes 10 revolutions in 2 seconds.

With the information given we know the frequency of each revolution.

[pic]

The time needed to complete one revolution is known as the Period (T).

[pic]

How far does an object move in one revolution? (circumference)

1 rev =2πr

(Equation 2)

More Useful Formulas…

If we substitute Equation 2 into Equation 1…

[pic]

THEN SIMPLIFY

[pic] (Equation 3)

COMBINE THE OLD WITH THE NEW

Nonuniform Circular Motion

The total acceleration of a body is the tangential (linear) acceleration plus the radial (centripetal) acceleration.

[pic]

4. (I) A racing car starts from rest in the pit area and accelerates at a uniform rate to a speed of 30 m/s in 11 s, moving on a circular track of radius 500m. Determine the tangential and centripetal components of the total acceleration exerted on the car (by the ground).

EXAMPLE PROBLEM: Nonuniform circular motion

A particle moves clockwise in a circle of radius 1.0 m. It starts at rest at the origin at time, t=0. Its speed increases at a constant rate of 1.6 m/s2. See Figure below.

How long does it take to travel half way around the circle (point A to Point B)?

A. What is its speed at point B?

What is the direction of its velocity at that time?

What is the centripetal (radial) acceleration?

What is the Tangential (linear) acceleration?

What is the magnitude and direction of the total acceleration at point B?

DYNAMICS of Circular Motion

Revisiting Newton’s 2nd Law

Newton stated that if a body is accelerating, there must be a NET force acting on it, in the direction of the acceleration.

The NET force the cause circular motion is called the CENTRIPETAL force.

Note: This is not a new type of force, we will treat it very much the same as we did any NET force.

[pic]

A centripetal (center seeking) force causes circular motion NOT a centrifugal (center fleeing) force.

• The force that is acting on the ball causing circular motion is inward (centripetal).

• The force acting on the hand is outward, and NOT causing circular motion.

• A centrifugal force is a “fictitious” or “pseudo” force caused by viewing the situation from an accelerating reference frame.

A “fictitious” force is real to an observer in an accelerating frame of reference. SEE CHAPTER 4 NOTES

EXAMPLE: Person in a rotating reference frame.

THE ROTOR RIDE

• In the accelerating reference frame (i.e. the person rotating around with the wall) the person is at rest and not moving. This person can’t use Newton’s Laws to explain why he is at rest. He must introduce a “fictitious” force (centrifugal force) which is quite real to him. He is in a non-inertial (accelerating) reference frame.

Why is it said that centrifugal “fictitious” force is a misconception, and NOT real?

• Because to an observer in an inertial reference frame “fictitious” force does not exist.

• Remember, Newton’s laws only apply in inertial reference frames.

• Centrifugal (fictitious) force is a non-Newtonian, and is caused by acceleration and is only real to an observer in the accelerating reference frame.

• In the picture above (a) represents what would happen if centrifugal forces were real. (b) Shows what does happen in an inertial frame of reference.

Problems:

4. (I) A horizontal force of 280 N is exerted on a 2.0-kg discus as it is rotated uniformly in a horizontal circle (at arms length) of radius 1.00 m. Calculate the speed of the discus.

5. (II) A flat puck (mass M) is rotated in a circle on a frictionless air hockey tabletop, and is held in this orbit by a light cord which is connected to a dangling mass (mass m) through the central hole as shown in Fig. 5-32. Show that the speed of the puck is given by [pic].

6. (II) A 0.40-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 60 N, what is the maximum speed the ball can have? How would your answer be affected if there were friction?

7. (II) A ball on the end of a string is cleverly revolved at a uniform rate in a vertical circle of radius 85.0 cm, as shown in Fig. 5-33. If its speed is 4.15 m/s and its mass is 0.300 kg, calculate the tension in the string when the ball is (a) at the top of its path, and (b) at the bottom of its path.

8. (II) A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 10.0 m. If the force felt by the trainee is 7.75 times her own weight, how fast is she rotating? Express your answer in both m/s and rev/s.

9. (II) How many revolutions per minute would a 15-m-diameter Ferris wheel need to make for the passengers to feel “weightless” at the topmost point of the trip?

10. (II) In a “Rotor-ride” at a carnival, people pay money to be rotated in a vertical cylindrically walled “room.” (Refer to note packet for picture) If the room radius is 5.0 m, and the rotation frequency is 0.50 revolutions per second when the floor drops out, what is the minimum coefficient of static friction so that the people will not slip down? People describe this ride by saying they were being “pressed against the wall.” Is this true? Is there really an outward force pressing them against the wall? If so, what is its source? If not, what is the proper description of their situation (besides “scary”)?

[Hint: First draw the free-body diagram for a person.]

Conical Pendulum – (horizontal circular motion)

Constant speed

A particle of mass, m, is suspended from a string of length, L, and travels at a constant speed, v in a horizontal circle if radius, r. The string makes an angle, θ, given by sin θ = r/L, as shown in the figure.

a. Find the tension in the string and the speed of the particle.

EXAMPLE 5-7 (from book)

A Car rounding a bend

A 1000 kg car rounds a curve on a flat road of radius 50 m at a speed of 50 km/hr (14 m/s). Will the car make the turn, or will it skid, if: (a) the pavement is dry and the coefficient of static friction is μs = 0.60; (b) the pavement is icy and μs = 0.25?

11. (II) What is the maximum speed with which a 1050-kg car can round a turn of radius 70 m on a flat road if the coefficient of friction between tires and road is 0.80? Is this result independent of the mass of the car?

A Car on incline rounding a bend

For a car traveling with a speed, v, around a curve of radius, r.

a. Determine a formula for the angle at which a road should be banked so that no friction is required to keep the car on the road.

b. What is this angle for a curve of radius 50 m at a designed speed of 50 km/h?

c. If friction were introduced between the tires and the road, how would the problem change?

Newton’s Gravitational Theory (Universal Gravitation)

But First, some History

(Pre 600 BC) – Ancient Greek Model

• Flat earth as the center of the universe

• Stars rotate at 15o/hr

• Sun moves 1o/day

• Stars are fastened to a gigantic crystal sphere (celestial sphere)

• Sun’s path is called the ecliptic

• If the plane of the sun is tilted 23o from the equatorial plane, this accounts for why the sun is high in the summer and low in the winter.

(582-500 BC) – Pythagorus

• Recognized the earth was a sphere. He simplified the Greek earth centered model. It was no longer necessary to adjust the axis of the celestial sphere to account for the variation of the observed height of the pole star with latitude (north-south).

• The Greeks noticed that most stars rotate at the same rate, but some rotated at rates slower than the other stars. They called these wandering stars or planets.

• The planets rotate west to east but at certain times the direction is reversed. The looped part of the path is called the epicycle, and the reverse part is called the retrograde motion.

Hellenistic Age - Library in Alexandria

(276-196 BC) Eratosthenes

• Measured the circumference of the earth!!!

• He measured the angles the suns rays make with the vertical at noon on the first day of summer in Alexandria and Syene.

(312-230 BC) – Aristarchus of Samos

• Developed a method to determine the size of the moon and sun and to find the distances to the moon and sun. His method was sound, but his measurements were incorrect.

Step 1

Determine the ratio of the distance to the moon divided by the diameter of the moon.

Step 2

Get Dm by comparing the width of the earth’s shadow to the moon during an eclipse of the moon.

(312-230 BC) – Aristarchus of Samos (continued)

• Important historic note; until Aristarchus, all models of our solar system were geocentric.

• Aristarchus was the first to introduce a heliocentric model of the known universe.

• Aristarchus also proposed that while the earth travels around the sun it spins on its axis.

• Nicolaus Copernicus would reinvent this theory eighteen hundred years later.

• The Aristarchus-Copernican model gives the simplest explanation of the motion of the planets.

• Unfortunately, the Aristarchus-Copernican model was not generally accepted.

(190 –120 BC) – Hipparchus

• Using Aristarchus’ technique for finding the distance to the moon and sun, he redid Aristachus’ measurements and found values for the distance to the moon very similar to the accepted values we use today, while his sun measurements were still inaccurate.

• Using centuries of celestial data he determined the axis of the earth is not fixed, but slowly precesses. He determined the time for each precession is 25,000 years.

MUCH LATER…

(150 AD) Ptolemy

• Argued against any model in which the earth moved because he thought the atmosphere would be left behind.

• Ptolemy developed a very complicated but accurate earth-centered model.

• Ptolemy’s famous book, the “Almagest” provided a record of all the Greek models.

• The Almagest was used for more than 1000 years as the manual for determining the motion of the stars & planets.

HISTORY OF THE WORLD – PART II

The beginning of modern science

(1473 – 1543) Nicolaus Copernicus – Polish Monk

• Reinvented Aristarchus’ Model of the heliocentric frame of reference for the solar system.

• He knew that the church had adopted Ptolemy’s model as part of its dogma, and therefore would not accept this radical change.

• He intentionally delayed the publication of his great book “De Revolutionibus Orbium Coelestium” until the year he died, 1543.

(1564 –1643) Galileo Galilei – Italian

• Supported Copernicus model and used his newly invented “telescope” to help verify the model by observing the phases and apparent changes in size of Venus.

(1546 – 1601) Tycho Brahe – Danish nobleman

• Took years and years of very accurate data (pre-telescope) that he made with huge 20 ft protractors that he had built for sighting the directions to the heavenly bodies.

• He made records of his measurements on equally large spheres.

(1571 – 1630) Johannes Kepler – German

“The wandering mathematician”

• Kepler had the mathematical talent to use Tycho Brahe’s data, and put it to good use. However, it is unclear if Kepler was given Tycho’s data or whether he stole it.

• In either case he used it well, and the rest is history.

KEPLER’S THREE LAW’S OF PLANETARY MOTION

1st – The orbits of the planets are ellipses, with the sun at one focal point.

2nd – The line joining the sun and a planet (its radius vector) sweeps over equal areas in equal times.

3rd – The squares of the periods of the planets’ motions are proportional to the cubes of the semimajor axes of their elliptical paths; that is c3/T2 is the same for all the planets, where T is the time for the planet to complete one orbit about the sun and c is defined in the figure at the bottom of the last page.

• Galileo used his telescopes to discover the four moons of Jupiter and times their periods of orbit.

• He found that for all four moons Kepler’s ratio of c3/T2 was constant.

• This convinced most scientists that Kelper’s laws were not an accidental fit, and Copernicus’ heliocentric universe model must be correct.

• The Catholic Church was not convinced and Galileo was arrested for his teachings and excommunicated from the church. He has since been re-instated.

Kepler and Galileo had worked to show how the planets behaved, and how the motions of the moons of Jupiter and the planets of our sun obeyed the same rules. It was Newton’s turn to explain what causes planets to obey the same rules.

Newton’s Analysis of Planetary Motion

(1642 – 1727) Isaac Newton

• 1666 – Using mathematics, Newton showed that for planets to move in elliptical paths.

Newton proposed the following proportionality between the force, F and the distance between the centers of the planet and the center of the sun, d.

Deriving Newton’s Universal Law

HERE IT COMES!!!

ANOTHER STROKE OF GENIUS FROM NEWTON

Newton sees symmetry between these forces.

• If FS,P is proportional to the mass of planet, then the FP,S must be proportional to the mass of the sun, and therefore sets

where he called G the Universal Gravitational Constant.

With one last substitution Newton finds…

Realize Newton still did not know G or MS

But, he did know GMS =

Newton assumed that his law should work for any two objects not just planets. Newton’s Law of Universal Gravitation for any two objects is stated as follows…

The direction of the force is along the line joining the two objects and is always an attractive force.

THE CAVENDISH EXPERIMENT (1798)

OR

The torsion balance and the measurement of the gravitational constant, “G”.

Torsion Balance

Principle of Operation

1. Two small masses are placed at the end of a light rod.

2. The force required to rotate the fiber from its untwisted position (a), is measured.

3. The force required to twist the fiber is calibrated as a function of the angle θ, [pic]

4. The light beam and scale are used to magnify θ.

5. When the mass M, is brought near to mass m it applies a gravitational force onto the fiber. To check for symmetry it is placed in positions AA and BB.

6. Knowing that the force causing the twist was caused by gravity Cavendish now used The Law of Universal Gravitation.[pic]

7. Cavendish was now able to isolate G, and calculated it as…

[pic]

EXAMPLE PROBLEM: Assume the earth is moving in a circular orbit around the sun. Using the following data calculate the speed of the earth in its orbit in miles/hr.

Mean Radius of Orbit = 1.5 x 1011 m 1 mile = 1.61 km

Mass of Sun = 1.99 x 1030 kg G = 6.67 x 10-11[pic]

12. (I) Calculate the force of gravity on a spacecraft 12,800 km (2 earth radii) above the Earth’s surface if its mass is 1400 kg.

13. (I) A hypothetical planet has a mass 2.5 times that of Earth, but the same radius. What is g near its surface?

Satellite Motion:

Imagine placing a large cannon on top of Mount Everest.

If we were to load the cannon with enough gun powder and fire it HORIZONTALLY to the surface of the Earth below, we may hit a town far away.

If we were to load the cannon with more gun powder and fire it HORIZONTALLY to the surface of the Earth below, we may even hit a different country far away.

Let’s say we were to load the cannon with just enough gun powder and fire it HORIZONTALLY to the surface of the Earth below, we may even hit a country halfway around the world.

Well, let’s say we were to load the cannon with perfect amount of gun powder and fire it HORIZONTALLY to the surface of the Earth below, so that it just keeps falling for ever as the Earth curves away from its fall. If this cannon ball does not hit the back of the cannon which it was fired from, then this cannon ball will continuously go around the world; falling towards the Earth… FOREVER!!!!!

This is exactly what is happening to our moon with Earth and all the planets with our Sun.

[pic]

Orbiting Velocity and Escape Velocity:

RECALL:

[pic]

And

[pic]

[pic]

[pic]

[pic]

A 5000.0 kg satellite is moving in a stable circular orbit at altitude of 12,800 km above the earth's surface.

Rearth= 6.38 x 106 m Mearth= 5.98 x 1024 kg

G= 6.67 x 10-11 N*m2/kg2

a. Please calculate the orbiting velocity of the satellite.

a:__________________

b. What is its period (time to make one orbit) in hours.

b:_________________

A Geosynchronous satellite is a satellite that has a period of 1 Earth day so that it is stationary to one fixed location on the surface of the earth. What must be the altitude in terms of radius of the Earth and the orbiting velocity for this satellite in miles per hour?

Please show your work…

Altitude = ________________ Rearth

velocity =_________________miles/hour

14. (II) You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn’t so by calculating how much weaker gravity is 300 km above the Earth’s surface.

15. A projected space station consists of a circular tube that is set rotating about its center (like a tubular bicycle tire) (Fig. 5-42). The circle formed by the tube has a diameter of about 1.1 km. (a) On which wall inside the tube will people be able to walk? (b) What must be the rotation speed (revolutions per day) if an effect equal to gravity at the surface of the Earth (1.0 g) is to be felt?

[pic]

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[pic]

qð

Figure 1 a small object moving in a circle. Note the instantaneous velocity is always tangent to the circular path.

Figθ

Figure 1 a small object moving in a circle. Note the instantaneous velocity is always tangent to the circular path.

Figure 2 Determining the change in the velocity, Δv, for a particle moving in a circle. The length Δl is the cord from A to B. As Δθ approaches zero, Δl approaches the arc length, s, from A to B.

1

2

total

r =1.0 m

B

A

Wahoo!

I’m stuck to the wall!

I feel like I’m being pressed against the wall.

Force on person due to the wall

Fictitious force to explain zero acceleration

θ

[pic]

[pic]

18 centuries

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Mirror

Scale

Light Beam

tf

ti

vf

vi

The direction of acceleration is also towards the center of the circle! THUS, centripetal!

Notice the direction of (v is towards the center of the circle!

a=(v/(t

(v

vf

-vi

[pic]

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