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Bellringer 11/12 A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done on the.

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Presentation on theme: "Bellringer 11/12 A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done on the."— Presentation transcript:

1 Bellringer 11/12 A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done on the bucket?

2 Chapter 7: Circular Motion and Gravitation

3 Object is moving tangent to the circle - Direction of the velocity vector is the same direction of the object’s motion – the velocity vector is directed tangent to the circle Object moving in a circle is accelerating

4 Tangential speed (v t ) – speed of an object in circular motion  When v t is constant = uniform circular motion  Depends on distance

5 Centripetal Acceleration Centripetal Acceleration – acceleration directed toward the center of a circular path (center-seeking) Centripetal Acceleration a c = v t 2 /r Centripetal Acceleration = (tangential speed) 2 /radius of circular path

6 Example A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05m/s 2, what is the car’s tangential speed?

7 Tangential acceleration – acceleration due to the change in speed

8 Centripetal Force – net force directed toward the center of an object’s circular path Centripetal Force F c = mv t 2 /r Centripetal Force= mass x (tangential speed) 2 /radius of circular path Example: Gravitational Force – keeps moon in its orbit

9 Example A pilot is flying a small plane at 56.6m/s in a circular path with a radius of 188.5m. The centripetal force needed to maintain the plane’s circular motion is 1.89x10 4 N. What is the plane’s mass?

10 Example A car is negotiating a flat curve of radius 50. m with a speed of 20. m/s. If the centripetal force provided by friction is 1.2 x 10 4 N. A. What is the mass of the car? B. What is the coefficient of friction?

11 Centripetal vs Centrifugal Centrifugal – center fleeing – away from the center/outward  DOES NOT EXIST!!!! Fake force!

12 Bellringer 11/14 A building superintendent twirls a set of keys in a circle at the end of a cord. If the keys have a centripetal acceleration of 145 m/s 2 and the cord has a length of 0.34m, what is the tangential speed of the keys?

13 Gravitational Force – mutual force of attraction between particles of matter Newton’s Law of Universal Gravitation F g = G m 1 m 2 r 2 Gravitational Force= constant x mass1 x mass2 (distance between masses) 2 G = 6.673x Nm 2 kg 2

14 Example Find the distance between a 0.300kg billiard ball and a 0.400kg billiard ball if the magnitude of the gravitational force between them is 8.92x N.

15 Gravity’s Influence Tides – periodic rise and fall of water

16 Field Force Gravitational force is an interaction between a mass and the gravitational field created by other masses Gravitational Field Strength g = F g /m g = 9.81m/s 2 on Earth’s surface

17 Weight changes with location Weight = mass x free-fall acceleration or ● Weight = mass x gravitational field strength F g = Gmm e / r 2 g = F g /m = Gmm e /m r 2 = Gm e / r 2

18 Weight Gravitational field strength depends only on mass and distance – your distance increases, g decreases…your weight decreases

19 Bellringer 11/15 A 7.55x10 13 kg comet orbits the sun with a speed of 0.173km/s. If the centripetal force on the comet is 505N, how far is it from the sun?

20 Example Suppose the value of G has just been discovered. Use the value of G and an approximate value for Earth’s radius to find an approximation for Earth’s mass.

21 Example Earth has a mass of 5.97x10 24 kg and a radius of 6.38x10 6 m, while Saturn has a mass of 5.68x10 26 kg and a radius of 6.03x10 7 m. Find the weight of a 65.0kg person at the following locations a.On the surface of Earth b.1000km above the surface of Earth c.On the surface of Saturn d.1000 km above the surface of Saturn

22 Example A scam artist hopes to make a profit by buying and selling gold at different altitudes for the same price per weight. Should the scam artist buy or sell at the higher altitude? Explain.

23 Bellringer 11/18 What is the force of gravity between two 74.0kg physics students that are sitting 85.0cm apart?

24 Motion in Space Claudius Ptolemy  Thought Earth was the center of the universe Nicolaus Copernicus  Thought Earth orbits the sun in perfect circles

25 Johannes Kepler Kepler’s Laws of Planetary Motion - First Law: Each planets travels in an elliptical orbit around the sun, the sun is at one of the focal points

26 Kepler’s Laws of Planetary Motion - Second Law: An imaginary line drawn from the sun to any planet sweeps out equal areas in equal time intervals

27 Kepler’s Laws of Planetary Motion - Third Law: The square of a planet’s orbital period (T 2 ) is proportional to the cube of the average distance (r 3 ) between the planet and the sun Planet Period (s) Average Dist. (m) T 2 /R 3 (s 2 /m 3 ) Earth3.156 x 10 7 s x x Mars5.93 x 10 7 s2.278 x x

28 Example The moons orbiting Jupiter follow the same laws of motion as the planets orbiting the sun. One of the moons is called Io - its distance from Jupiter's center is 4.2 units and it orbits Jupiter in 1.8 Earth-days. Another moon is called Ganymede; it is 10.7 units from Jupiter's center. Make a prediction of the period of Ganymede using Kepler's law of harmonies. Answer: 7.32 days

29 Period and Speed of an object in circular motion T = 2π r 3 v t = G m Gm r T = Orbital period r = mean radius m = mass of central objectv t = orbital speed √ √ m is the mass of the central object. Mass of the planet/satellite that is in orbit does not affect the period or speed

30 Example During a spacecraft’s fifth orbit around Venus, it traveling at a mean altitude of 361km. If the orbit had been circular, what would the spacecraft’s period and speed have been?

31 Example At what distance above Earth would a satellite have a period of 125 minutes?

32 Bellringer 11/19 At the surface of a red giant star, the gravitational force on 1.00kg is only 2.19x10 -3 N. If its mass equals 3.98x10 31 kg, what is the star’s radius?

33 What is the correct answer? Astronauts on the orbiting space station are weightless because… a.There is no gravity in space and they do not weight anything b.Space is a vacuum and they is no gravity in a vacuum c.Space is a vacuum and there is no air resistance in a vacuum d.The astronauts are far from Earth’s surface at a location where gravitation has a minimal affect

34 Weight and Weightlessness Weightlessness – sensation when all contact forces are removed

35 Astronauts in orbit Astronauts experience apparent weightlessness  No normal force is acting on them

36 Example Otis’ mass is 80kg. a.What is the scale reading when Otis accelerates upward at 0.40m/s 2 b.What is the scale reading when Otis is traveling upward at a constant velocity at 2.0m/s c.Otis stops at the top floor and then accelerates downward at a rate of 0.40m/s 2


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