CIRCULAR MOTION - GRAVITATION · Circular Motion An object moving in a circle must have a force...

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1 CHAPTER 5 CIRCULAR MOTION - GRAVITATION http://www.physicsclassroom.com/Class/circles/ circtoc.html Units Kinematics of Uniform Circular Motion Dynamics of Uniform Circular Motion Highway Curves, Banked and Unbanked Nonuniform Circular Motion Centrifugation Newton’s Law of Universal Gravitation Gravity Near the Earth’s Surface; Geophysical Applications Satellites and “Weightlessness” Kepler’s Laws and Newton’s Synthesis Circular Motion An object moving in a circle must have a force acting on it; otherwise it would move in a straight line. The direction of the force is towards the center of the circle. Kinematics of Uniform Circular Motion Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that 2 R v a r

Transcript of CIRCULAR MOTION - GRAVITATION · Circular Motion An object moving in a circle must have a force...

Page 1: CIRCULAR MOTION - GRAVITATION · Circular Motion An object moving in a circle must have a force acting on it; otherwise it would move in a straight line. The direction of the force

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CHAPTER 5 CIRCULAR MOTION - GRAVITATION

http://www.physicsclassroom.com/Class/circles/

circtoc.html

Units

• Kinematics of Uniform Circular Motion • Dynamics of Uniform Circular Motion • Highway Curves, Banked and Unbanked • Nonuniform Circular Motion • Centrifugation • Newton’s Law of Universal Gravitation • Gravity Near the Earth’s Surface; Geophysical Applications • Satellites and “Weightlessness” • Kepler’s Laws and Newton’s Synthesis

Circular Motion

An object moving in a circle must have a force acting on it; otherwise it would move in a straight line. The direction of the force is towards the center of the circle.

Kinematics of Uniform Circular Motion

Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that

2

R

va

r

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This acceleration is called the centripetal, or radial, acceleration, and it points towards the center of the circle.

Forces in Accelerating Reference Frames

• Distinguish real forces from fictitious forces • “Centrifugal” force is a fictitious force • Real forces always represent interactions between objects

Example 1: A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

2 2(3.14)(0.600 )

7.54 /(0.500 )

r mv m s

T s

Example 2: The Moon’s orbit about the Earth has a radius of 384,000 km and a period T of 27.3 days. Determine the acceleration of the Moon toward the Earth.

In terms of g ( 29.80 /m s ):

Dynamics of Uniform Circular Motion

There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome. If the centripetal force vanishes, the object flies off tangent to the circle.

2 22(7.54 / )

94.7 /(0.600 )

R

v m sa m s

r m

2 2 2 2 83 2

2 2 6 2

(2 ) 4 4 (3.84 10 )2.72 10 /

(2.36 10 )R

v r r x ma x m s

r T r T x s

3 2 4

22.72 10 / 2.78 10

9.80 /

ga x m s x g

m s

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This force may be provided by the tension in a string, the normal force, or friction, among others.

Example 3: Estimate the force a person exerts on a string attached to a 0.150-kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second (T = 0.50s)

Vertical Circle

• Look at the forces at the top of the circle • The minimum speed at the top of the circle can be found

Example 4: A 0.150-kg ball on the end of a 1.1-m-long cord is swung in a vertical circle. (a) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle.

2 / 2 /RF ma a v r v r T

2 (0.600 ) /(0.500 ) 7.54 /v m s m s

2 2(7.54 / )(0.150 ) 14

(0.600 )T

v m sF m kg N

r m

topv gR

v gr

2(9.80 / )(1.10 ) 3.28 /v m s m m s

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(b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (a).

2

2(6.56 / )(0.150 ) (0.150 )(9.80 / ) 7.34

(1.10 )T

m sF kg kg m s N

m

Highway Curves, Banked and Unbanked

When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show. As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways:

1. The kinetic frictional force is smaller than the static.

2. The static frictional force can point towards the center of the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.

Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the horizontal component of the normal force, and no friction is required. This occurs when:

2

T

vF m mg

r

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Example 5: A 1000-kg car rounds a curve on a flat road of radius 50 m at a speed of 14 m/s. Will the car follow the curve, or will it skid? (a) Assume the pavement is dry and the coefficient of static friction is

0.60s .

NF on the car is equal to the weight mg since the road is flat.

The net horizontal force required to keep the car moving in a circle around the curve is

The maximum friction force attainable is

Since the force of only 3900 N is needed the car can follow the curve.

(B) If the pavement was icy and 0.25S then the maximum static friction force possible is

The car will skid because the ground cannot exert sufficient force (3900N) to keep it moving in a curve.

Example 6: What is the banking angle needed for an expressway off-ramp with a curve radius of 50 m at a design speed of 50 km/h?

2

2

(14 / )tan 0.40 21.8

(50 )(9.80 / )

om s

m m s

Nonuniform Circular Motion

If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one. This concept can be used for an object moving along any curved path, as a small segment of the path will be approximately circular.

2(1000 )(9.8 / ) 9800NF mg kg m s N

2 2(14 / )(1000 ) 3900

(50 )R R

v m sF ma m kg N

r m

(0.60)(9800 ) 5900fr S NF F N N

(0.25)(9800 ) 2500fr S NF F N N

2

tanv

rg

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Centrifugation

A centrifuge works by spinning very fast. This means there must be a very large centripetal force. The object at A would go in a straight line but for this force; as it is, it winds up at B.

Newton’s Law of Universal Gravitation

If the force of gravity is being exerted on objects on Earth, what is the origin of that force? Newton’s realization was that the force must come from the Earth. He further realized that this force must be what keeps the Moon in its orbit.

• Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

The gravitational force on you is one-half of a Third Law pair: the Earth exerts a downward force on you, and you exert an upward force on the Earth.

When there is such a disparity in masses, the reaction force is undetectable, but for bodies more equal in mass it can be significant. Therefore, the gravitational force must be proportional to both masses. By observing planetary orbits, Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses.

1 2

2

m mF G

r

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Gravitation Constant

• Determined experimentally • Henry Cavendish

– 1798 • The light beam and mirror serve to amplify the motion

Example 6: A 50-kg person and a 75-kg person are sitting next to each other (0.5m) in the classroom. Estimate the magnitude of the gravitational force each exerts on the other.

Example 7: Three 0.300-kg pool balls are placed on a table at the corners of a right triangle. Find the net gravitational force on the cue ball.

Find the net gravitational force on the cue ball.

21F on m2 on m1

31F by m3 on m1

11 113 131 2 2

31

(0.3 )(0.3 )(6.67 10 ) 6.67 10

(0.3 )

m m kg kgF G x x N

r m

2 2 2 2 11 11(6.67) (3.75) 10 7.65 10X YF F F X N x N

11 2

1 2

2 2

(6.67 10 )(50 )(75 )

(0.5 )

m m x N m kg kgF G

r m

61 10x N

2 121 2

21

m mF G

r

11 11

2

0.3 )(0.3 )(6.67 10 ) 3.75 10

(0.4 )

kg kgx x N

m

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Example 8: An astronaut standing on the surface of Ceres, the largest asteroid drops a rock from a height of 10.0 m. It takes 8.06 s to hit the ground. (a) Calculate the acceleration of gravity on Ceres:

(b) Find the mass of Ceres, given that the radius of Ceres is 510 km.

Gravity near the Earth’s Surface; Geophysical Applications

Now we can relate the gravitational constant to the local acceleration of gravity. We know that, on the surface of the Earth:

Solving for g gives:

Now, knowing g and the radius of the Earth, the mass of the Earth can be calculated:

Kg

Gravity Near the Earth’s Surface; Geophysical Applications

The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.

Example 9: Estimate the effective value of g on top of Mt. Everest, 8850 m above sea level.

r = 6380 km + 8.9 km = 6389 km

21

2ox v t at 2 21

10.0 0 [ (8.06 ) ] 0.308 /2

Cm g s m s

2

CC

C

M mmg G

R

2211.20 10C C

C

g RM x kg

G

11 2 2 242

2 6 2

(6.67 10 / )(5.98 10 )9.77 /

(6.389 10 )

Em x N m kg x kgg G m s

r x

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Satellites and “Weightlessness”

Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth’s gravity altogether.

The satellite is kept in orbit by its speed – it is continually falling, but the Earth curves from underneath it. Objects in orbit are said to experience weightlessness. They do have a gravitational force acting on them, though! The satellite and all its contents are in free fall, so there is no normal force. This is what leads to the experience of weightlessness.

Example 10: A geosynchronous satellite is one that stays above the same point on the Earth, which is possible only if it is above a point on the equator. Determine the height above the Earth’s surface such a satellite must orbit. T = 1 day = (24h)(3600s/h)=86,400s

The satellites speed has to be

Escape Speed

• The escape speed is the speed needed for an object to soar off into space and not return

23

24

EGm Tr

2 11 24 23 22 3

2 2

(6.67 10 )(5.98 10 )(86,400)7.54 10

4 4

EGm T x xr x m

74.23 10r x m 42,300 6380 36,000or km km km

11 24

7

(6.67 10 )(5.98 10 )3070 /

(4.23 10 )

EGm x xv m s

r x

2 Eesc

E

GMv

R

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• For the earth, vesc is about 11.2 km/s

• Note, v is independent of the mass of the object

Various Escape Speeds

• The escape speeds for various members of the solar system

• Escape speed is one factor that determines a planet’s atmosphere

Johannes Kepler (1571 – 1630) was a German mathematician, astronomer and astrologer,

and a key figure in the 17th

century astronomical revolution. He is best known for his laws of planetary motion.

Kepler’s Laws and Newton's Synthesis

Kepler’s laws describe planetary motion. 1. The orbit of each planet is an ellipse, with

the Sun at one focus.

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2. An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.

3. The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.

– For orbit around the Sun, K = KS = 2.97x10-19 s2/m3

– K is independent of the mass of the planet Example 11: Mar’s period was noted by Kepler to be about 687 days (Earth days), which is (687d/365d) = 1.88yr. Determine the distance of Mars from the Sun using the Earth as a reference.

So Mars is 1.52 times the Earth’s distance from the sun or 112.28 10x m .

2 3T Kr

111.50 10ESr x m2 2

3 31.881.52

1

MS M

ES E

r T yr

r T yr

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CHAPTER 5

CIRCULAR MOTION & GRAVITATION CONCEPTS

1. Two masses, A and B, move in circular paths as shown to the right. The centripetal acceleration of mass A, compared to that of mass B, is one-half as great. 2. The diagram to the right represents a 4.0 x 102 kg satellite, S, in a circular orbit at an altitude of 5.6 x 106 m. The orbital speed of the satellite is 5.7 x 103 m/s and the radius of the Earth, R, is 6.4 x 106 m. If the altitude decreased, its centripetal acceleration would increase. 3. The diagram to the right represents a ball undergoing uniform circular motion as it travels clockwise on a string. At the moment shown in the diagram, the correct directions of both the velocity and centripetal acceleration of the ball is B.

4. A convertible car with its top down is traveling at constant speed around a circular track. When the car is at point A, a passenger in the car throws a ball straight up, the ball could land at point C. 5. A car going around a curve is acted upon by a centripetal force, F, if the speed of the car were twice as great, the centripetal force necessary to keep it moving in the same path would be 4F. Concepts 6 – 8 refer to the following: The diagram on the right shows an object with a mass of 1.0 kg attached to a string 0.50 m long. The object is moving at a constant speed of 5.0 m/s in a horizontal circular path with center at point O.

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6. While the object is undergoing uniform circular motion, its acceleration is directed toward the center of the circle. 7. If the string is lengthened while the speed of the object remains constant, the centripetal acceleration of the object will decrease. 8. If the string is cut when the object is at position shown, the path the object will travel from this position will be a straight line tangent to the circle. Concepts 9 – 13 refer to the following: The diagram to the right represents a ball of mass M attached to a string. The ball moves at a constant speed around a flat horizontal circle of radius R. 9. The centripetal acceleration of the ball is constant in magnitude, but changing in direction. 10. If the velocity of the ball is doubled, the centripetal accelration is quadrupled. 11. If the string is shortened while the speed of the ball remains the same, the centripetal acceleration will increase. 12. When the ball is in the position shown, the directioin of the centripetal force is toward point D. 13. If the mass of the ball is decreased, the centripetal force required to keep it moving in the same circle at the same speed decreases. Concepts 14 & 15 refers to the following: A 60-kg adult and a 30-kg child are passengers on a rotor ride. When the rotating hollow cylinder reaches a certain constant speed, v, the floor moves downward. Both passenger stay “pinned” against the wall of the rotor, as shown to the right. 14. The magnitude of the frictional force between the adult and the wall of the spinning rotor is F. The magnitude of the frictional force between the child and the mass of the spinning rotor is 1/2 F. 15. Compared to the magnitude of the acceleration of the adult, the magnitude of the acceleration of the child is the same. 16. The best description of the force keeping the ball in the circular path is that it is perpendicular to the circle and directed toward the center of the circle. 17. The shape of the path of the Earth about the Sun is an ellipse with the Sun at one focus. 18. The shapes of the paths of the planets about the Sun are all ellipses with the Sun at one focus.

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19. The increase in a planet’s speed as it approaches the Sun is described by Kepler’s Second Law. The best explanation for this empirical law is that the gravitational potential energy lost as the planet nears the Sun becomes kinetic energy. 20. As the distance of a satellite from Earth’s surface increases, the time the satellite takes to make one revolution around the Earth increases. 21. The condition required for a satellite to be in a geosynchronous orbit around the Earth’s is that the period of revolution of the satellite must be the same as the rotational period of the Earth. 22. The banking angle in a turn on the Olympic bobsled track is not constant, But increases upward from the horizontal. Coming around a turn, the bobsled team will intentially “climb the wall”, then go lower coming out of the turn. They do this to take the turn at a faster speed. 23. A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with the track at the top of the

loop, the minimum value for its speed at that point is rg

24. A pilot executes a vertical dive, and then follows a semi-circular arc until it is going straight up. Just as the plane is at its lowest point, the force on him is more than g, and pointing up.

25. A spaceship is traveling to the moon. It is never beyond the pull of Earth’s gravity.

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26. An astronaut is inside a space capsule in orbit around the Earth. He is able to float inside the capsule because he and his capsule move with the same constant acceleration. 27. A car goes around a curve of radius r at a constant speed v. The direction of the net force on the car is toward the curve’s center. 28. Two objects attract each other gravitationally. If the distance between their centers doubles, the gravitational force is cut a fourth. 29. Two objects gravitationally attract each other. If the distance between their centers is cut in half, the gravitational force between them quadruples. 30. The acceleration of gravity on the moon is one-sixth what it is on Earth. An object of mass 72 kg is taken to the moon. Its mass there is 72 kg.