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Ch5.1 - Centripetal Force. Top View. Centripetal Force. INERTIA (velocity). Top View. Centripetal Force. INERTIA (velocity). There is no force pushing outward! Inertia wants object to fly off tangent to circle.

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## Ch5.1 - Centripetal Force

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**Ch5.1 - Centripetal Force**Top View Centripetal Force INERTIA (velocity)**Top View**Centripetal Force INERTIA (velocity) There is no force pushing outward! Inertia wants object to fly off tangent to circle. (Centrifugal Force is a fake force we think we feel throwing us outward.) Centripetal Force – real force acting towards center of circle.**Ex 1) What agent exerts the centripetal force in each?**m earth**Ex 1) What agent exerts the centripetal force in each?**m Fg Fnet = Fg earth**Ex 1) What agent exerts the centripetal force in each?**m Fg Fnet = Fg m.ac = Fg earth**Ex 1) What agent exerts the centripetal force in each?**FT Fnet = FT**Ex 1) what agent exerts the centripetal force in each?**FT Fnet = FT m.ac = FT**Ex 1) What agent exerts the centripetal force in each?**(Top view of record player) (coin) .**Ex 1) What agent exerts the centripetal force in each?**(Top view of record player) Fnet = Ff,s (coin) . Ff,s**Ex 1) What agent exerts the centripetal force in each?**(Top view of record player) Fnet = Ff,s m.ac = Ff,s (coin) . Ff,s**Centripetal Acceleration**Ex 2) The ride “Spin- Out” is a circular room, 5m in diameter. Once it gets to speed, its linear speed is 10 m/s. What is the centripetal acceleration?**Centripetal Acceleration**Ex 2) The ride “Spin- Out” is a circular room, 5m in diameter. Once it gets to speed, its linear speed is 10 m/s. What is the centripetal acceleration?**CENTRIPETAL FORCE**Ex 3) A 60 kg astronaut stands on a bathroom scale in a 2 km diameter rotating space station that spins every 62.8 sec. What does the scale read?**CENTRIPETAL FORCE**Ex 3) A 60 kg astronaut stands on a bathroom scale in a 2 km diameter rotating space station that spins ever 62.8 sec. What does the scale read? Fnet = FN Fn**CENTRIPETAL FORCE**Ex 3) A 60 kg astronaut stands on a bathroom scale in a 2 km diameter rotating space station that spins ever 62.8 sec. What does the scale read? Fnet = FN m.ac = FN Fn**CENTRIPETAL FORCE**Ex 3) A 60 kg astronaut stands on a bathroom scale in a 2 km diameter rotating space station that spins ever 62.8 sec. What does the scale read? Fnet = FN m.ac = FN**Ex 4) A sport car on a flat track rounds a corner with a**radius of 100m, at a maximum speed of 25 m/s before sliding out. What is the coefficient of static friction? Ch5 HW#1 1 – 5**Ex 4) A sport car on a flat track rounds a corner with a**radius of 100m, at a maximum speed of 25 m/s before sliding out. What is the coefficient of static friction? Fnet = Ff,s Ff,s**Ch5 HW#1 1 – 5**1. A youngster on a carousel horse 5.0 m from the center revolves at a constant rate, once around in 15.0 s. What is her acceleration? 2. A beetle standing on the edge of a 12-inch record (r = .152m) whirls around at 33.3 rotations per minute. What is its centripetal accl? What agent exerts the force? 3. A ‘foreign-made’ space station only has a radius of 3 meters. If it rotates around once every 1.15 sec, what is the centripetal accl at the floor? If an astronaut is 1.5 m tall, what is the centripetal accl at his head? Does anyone see any physiological implications of this? How could we remedy this?**A ‘foreign-made’ space station only has a radius of 3**meters. If it rotates around once every 1.15 sec, what is the centripetal accl at the floor? If an astronaut is 1.5 m tall, what is the centripetal accl at his head? Does anyone see any physiological implications of this? How could we remedy this? • At head: At feet:**A ‘foreign-made’ space station only has a radius of 3**meters. If it rotates around once every 1.15 sec, what is the centripetal accl at the floor? If an astronaut is 1.5 m tall, what is the centripetal accl at his head? Does anyone see any physiological implications of this? How could we remedy this? • At head: At feet:**4. The Talladega Speedway …1500kg cars complete turns of**radius 100m at a speed of 80m/s. What is net force on car? What agent exerts it? Fnet = Fn m.ac = Fn**5. 700kg flat track car rounds a corner of radius 30m at a**speed of 20m/s, without slipping. What is the minimum coeffiecient of static friction that can accomplish this? Fnet = Ff,s m.ac = μ.Fn**Ch5.2 - Gravity**All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law) Newton’s Law of Universal Gravitation**Ch5.2 - Gravity**All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law) Newton’s Law of Universal Gravitation - Universal gravitational constant, G = 6.67x 10-11 Ex1) Compute the gravitational attraction between 2 100kg uniform spheres by 1.00m. (Roughly 2 220-lb football players) Ex2) The mass of the moon is 7.35x 1022kg and its distance from the Earth is 3.84x 10 3 km. Taking the Earth’s mass to be 5.98x10 24 kg, what force keeps the moon in her orbit?**Ch5.2 - Gravity**All matter is attracted to all other matter. - More matter = larger force - Distance apart is important (Inverse square law) Newton’s Law of Universal Gravitation - Universal gravitational constant, G = 6.67x 10-11 Ex2) The mass of the moon is 7.35x 1022kg and its distance from the Earth is 3.84x 10 3 km. Taking the Earth’s mass to be 5.98x10 24 kg, what force keeps the moon in her orbit?**Ex3) 2 objects both of mass m are a distance r apart and**exert a force F on each other. How does the force change if: a) Both masses are doubled. b) Instead the distance is doubled? c) Distance made 3X smaller?**Ex4) How fast does an object have to travel, to stay in**circle? - what is the direction of the instantaneous velocity? - what would happen if it traveled slower? - what would happen if it traveled faster? Ch5 HW#2 1 – 4 m M**Ch5 HW#2 1 – 4**1. Using our equation for the universal law of gravitation: Explain how the force will change if: a)One mass doubles b) Both masses cut in half and the distance between them cut in half. c) The masses stay the same and the distance triples. d) The masses double and the distance between is cut in half.**Ch5 HW#2**1. Using our equation for the universal law of gravitation: Explain how the force will change if: a) Both masses double b) Both masses cut in half and the distance between them cut in half. c) The masses stay the same and the distance quadruples. d) The masses triple and the distance between is cut to one third.**I want you to calculate my weight 2 ways:**• (My mass = 75 kg; Earth’s mass = 6x1024kg; Earth’s radius = 6x106m) • 1) Fg = m.g • 2) • (Roughly the same with rounding error.) • What is the force of gravity on me (75 kg) on the surface of Jupiter? • MJ= 2x1027kg RJ= 7x107m • What is the gravitational attraction between the earth and the Sun? • Msun = 2x1030 kg DES= 1.5x1011m**I want you to calculate my weight 2 ways:**• (My mass = 75 kg; Earth’s mass = 6x1024kg; Earth’s radius = 6x106m) • 1) Fg = m.g = (75kg)(9.8m/s2) = 748.5N • 2) • (Roughly the same with rounding error.) • What is the force of gravity on me (75 kg) on the surface of Jupiter? • MJ= 2x1027kg RJ= 7x107m • What is the gravitational attraction between the earth and the Sun? • Msun = 2x1030 kg DES= 1.5x1011m**I want you to calculate my weight 2 ways:**• (My mass = 75 kg; Earth’s mass = 6x1024kg; Earth’s radius = 6x106m) • 1) Fg = m.g = (75kg)(9.8m/s2) = 748.5N • 2) • (Roughly the same with rounding error.) • What is the force of gravity on me (75 kg) on the surface of Jupiter? • MJ= 2x1027kg RJ= 7x107m • What is the gravitational attraction between the earth and the Sun? • Msun = 2x1030 kg DES= 1.5x1011m**Ch5.3 - Gravity Applications**Acceleration of gravity at surface of any object:**Apparent weightlessness**-What does the bathroom scale read in each accelerating elevator? (scale reads FN) (assume m= 100kg) a = 0 m/s2 a = 10 m/s2**Satellite Orbits**Orbital Speed:**Satellite Orbits**Orbital Speed:**Ex1) A satellite in geostationary orbit must be at a height**of 3.6x107m above the earth’s surface. What speed must it travel at to stay directly above good ol’ AV?**Kepler’s 3 Laws of Planetary Motion:**1) The planets move in elliptical orbits with the sun as 1 focus. 2) Each planet sweeps out equal (areas) in equal time intervals. 3) The ratio of the average distance from the Sun cubed to the period squared is the same constant value. Ch5 HW#3**Ch5 HW#3 p172 37,44,46,47, + Hard Bonus Problem**Hard Bonus Problem. (Year 2000 AP Test FRQ #1) Prove that the acceleration on the surface of Mars is 38% the acceleration on the surface of the Earth, given MMars = 0.11.MEarth, RMars = 0.53.REarth.**37. The acceleration due to gravity on the surface of mars**is 3.7 m/s2. If the planet’s diameter is 6.8x106m, determine the mass of the planet and compare it to Earth. Fnet = FG**37. The acceleration due to gravity on the surface of mars**is 3.7 m/s2. If the planet’s diameter is 6.8x106m, determine the mass of the planet and compare it to Earth. Fnet = FG Mm = 6.4x1023 kg**44. Locate the position of a spaceship on the Earth-Moon**center line, such that at that point, the tug of each celestial body exerted on it will cancel and the craft would literally be weightless. E m Fnet = FER – FRM**44. Locate the position of a spaceship on the Earth-Moon**center line, such that at that point, the tug of each celestial body exerted on it will cancel and the craft would literally be weightless. Fnet = FER – FRM E m**44. Locate the position of a spaceship on the Earth-Moon**center line, such that at that point, the tug of each celestial body exerted on it will cancel and the craft would literally be weightless. Fnet = FER – FRM E m RER = 3.46x108m (346,000km)**46. Three very small spheres of mass 2.50 kg, 5.00 kg, and**6.00 kg are located on a strait line in space awayfromeverything else. The first one is a point between the other two, 10.0 cm to the right of the second and 20.0 cm to the left of the third. Compute the net gravitational force on it. 5 6 2

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