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:
1Bellringer 11/12A 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?
3Circular Motion – motion of an object about a single axis at a constant speed 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 circleObject moving in a circle is accelerating
4Tangential speed (vt) – speed of an object in circular motion When vt is constant = uniform circular motionDepends on distance
5Centripetal Acceleration Centripetal Acceleration – acceleration directed toward the center of a circular path (center-seeking)Centripetal Accelerationac = vt 2/rCentripetal Acceleration = (tangential speed)2 /radius of circular path
6ExampleA 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/s2, what is the car’s tangential speed?
7Tangential acceleration – acceleration due to the change in speed
8Centripetal Force= mass x (tangential speed)2 /radius of circular path Centripetal Force – net force directed toward the center of an object’s circular pathCentripetal ForceFc = mvt 2/rCentripetal Force= mass x (tangential speed)2 /radius of circular pathExample: Gravitational Force – keeps moon in its orbit
9ExampleA 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 1.89x104 N. What is the plane’s mass?
10ExampleA 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 104 N.A. What is the mass of the car?B. What is the coefficient of friction?
11Centripetal vs Centrifugal Centrifugal – center fleeing – away from the center/outwardDOES NOT EXIST!!!! Fake force!
12Bellringer 11/14A 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/s2 and the cord has a length of 0.34m, what is the tangential speed of the keys?
13Newton’s Law of Universal Gravitation (distance between masses) 2 Gravitational Force – mutual force of attraction between particles of matterNewton’s Law of Universal GravitationFg = G m1m2r2Gravitational Force= constant x mass1 x mass2(distance between masses) 2G = 6.673x N•m kg 2
14ExampleFind 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.
15Gravity’s InfluenceTides – periodic rise and fall of water
16Gravitational Field Strength Field ForceGravitational force is an interaction between a mass and the gravitational field created by other massesGravitational Field Strengthg = Fg /mGravitational field strength equals free fall accelerationMasses create a gravitational field in space….g = 9.81m/s2 on Earth’s surface
17Weight changes with location Weight = mass x free-fall accelerationor● Weight = mass x gravitational field strengthFg = Gmme / r2g = Fg /m = Gmme /m r2 = Gme / r2
18WeightGravitational field strength depends only on mass and distance – your distance increases, g decreases…your weight decreases
19Bellringer 11/15A 7.55x1013 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?
20ExampleSuppose 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.
21ExampleEarth has a mass of 5.97x1024kg and a radius of 6.38x106 m, while Saturn has a mass of 5.68x1026 kg and a radius of 6.03x107 m. Find the weight of a 65.0kg person at the following locationsOn the surface of Earth1000km above the surface of EarthOn the surface of Saturn1000 km above the surface of Saturn
22ExampleA 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.
23Bellringer 11/18What is the force of gravity between two 74.0kg physics students that are sitting 85.0cm apart?
24Motion in Space Claudius Ptolemy Nicolaus Copernicus Thought Earth was the center of the universeNicolaus CopernicusThought Earth orbits the sun in perfect circles
25Kepler’s Laws of Planetary Motion Johannes KeplerKepler’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
26Kepler’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
27Kepler’s Laws of Planetary Motion - Third Law: The square of a planet’s orbital period (T2 ) is proportional to the cube of the average distance (r3 ) between the planet and the sunPlanetPeriod(s)AverageDist. (m)T2/R3(s2/m3)Earth3.156 x 107 sx 10112.977 x 10-19Mars5.93 x 107 s2.278 x 10112.975 x 10-19
28ExampleThe 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
29Period and Speed of an object in circular motion T = 2π r3 vt = G mGm rT = Orbital period r = mean radiusm = mass of central object vt = 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
30ExampleDuring 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?
31ExampleAt what distance above Earth would a satellite have a period of 125 minutes?
32Bellringer 11/19At the surface of a red giant star, the gravitational force on 1.00kg is only 2.19x10-3 N. If its mass equals 3.98x1031 kg, what is the star’s radius?
33What is the correct answer? Astronauts on the orbiting space station are weightless because…There is no gravity in space and they do not weight anythingSpace is a vacuum and they is no gravity in a vacuumSpace is a vacuum and there is no air resistance in a vacuumThe astronauts are far from Earth’s surface at a location where gravitation has a minimal affect
34Weight and Weightlessness Weightlessness – sensation when all contact forces are removed
35Astronauts in orbit Astronauts experience apparent weightlessness No normal force is acting on them
36Example Otis’ mass is 80kg. What is the scale reading when Otis accelerates upward at 0.40m/s2What is the scale reading when Otis is traveling upward at a constant velocity at 2.0m/sOtis stops at the top floor and then accelerates downward at a rate of 0.40m/s2