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Physics 203 College Physics I Fall 2012

Physics 203 College Physics I Fall 2012. S. A. Yost. Chapter 5 Part 2 . Circular Motion, Universal Gravitation. Announcements. Chapter 5’s homework, HW05, is due next Tuesday.

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Physics 203 College Physics I Fall 2012

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  1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 5 Part 2 Circular Motion, Universal Gravitation
  2. Announcements Chapter 5’s homework, HW05, is due next Tuesday. You should have finished reading the chapter, except for “Nonuniform Circular Motion”, which we will skip (sec. 5-4). Sec. 5-5 is optional, but an interesting application. Today: circular motion, banked turns, and universal gravitation (orbits).
  3. Quiz: Question 1 The net force vector on an object in uniform circular motion points (A) away from the center of the circular path. (B) in the direction opposite to the motion of the object. (C) toward the center of the circular path. (D) in the same direction as the motion of the object. (E) nowhere. The net force is zero because the object is moving with a constant speed.
  4. Quiz: Question 2 Two objects attract each other gravitationally. If the distance between them decreases by a factor of 2, what happens to the gravitational force? (A) It increases by a factor of 2. (B) It decreases by a factor of 2. (C) It increases by a factor of 4. (D) It decreases by a factor of 4. (E) It doesn’t change.
  5. Quiz: Question 3 The orbits of planets are (A) circular with the sun at the center. (B) circular with the sun off-center. (C) elliptical with the sun at the center. (D) elliptical with the sun off-center. (E) something else I haven’t thought of.
  6. Quiz: Question 4 Why does a satellite in a circular orbit travel at a constant speed? (A) The net force acting on the satellite is zero. (B) There is a force acting opposite to the direction of the motion of the satellite. (C) The gravitational force acting on the satellite is balanced by the centrifugal force acting on the satellite. (D) There is no component of force acting along the direction of motion of the satellite.
  7. Vertical Circular Motion Question A mass (red) is rotated at a constant speed in a vertical circle on the end of the rod. Which vector (green) could possibly show the force of the rod on the mass? A B C D
  8. Fg Fc Vertical Circular Motion Question The forces could look like this… Fr q
  9. The Rotor People ride in a Rotor ride at a fair. When it spins fast enough, the floor drops out, and the people feel like they are pressed against the wall.
  10. The Rotor Is there really an outward force pushing the people against the wall? A Yes B No
  11. The Rotor Which picture shows the correct forces on a passenger in the Rotor? A B C D E What is the name of each of the three forces? Ff FN mg
  12. The Rotor If the barrel’s radius is 3.0 m and the coefficient of friction is 1.2, how many RPM must the barrel make to keep the woman from sliding down?
  13. Banked Curve Suppose a car is going around a track with radius 200 m banked at an angle of 25o. What is the minimum coefficient of friction for the tires if the car can go around the track at 50 m/swithout skidding? v R 200 m
  14. Universal Gravitation Newton’s Law of Universal Gravitation: Any two masses exert an attractive force on each other in proportion to their masses and inversely proportional to the distance between them: F12 = G m1 m2 /R2 Newton’s Gravitational Constant: G = 6.67 x 10-11 N m2/kg2
  15. Orbital Velocity Example We can use the inverse square law to find the orbital velocity of a spaceship at a height 2Reabove the Earth’s surface. (Re = 6380 km) 2Re Re
  16. Orbital Velocity Example The spaceship is 3 times as far from the center of the Earth as the ground is, so at this height, a = g/9 = 1.1 m/s2 2Re Re I don’t need to use G or Me!
  17. Orbital Velocity Example Setting the gravitational acceleration equal to the centripetal acceleration gives a = g/9 = v2/R = v2 / (3Re). 2Re Re
  18. Orbital Velocity Example Setting these equal, g/9 = v2 / (3Re) which implies v = √ gRe/3 = √ (9.8 m/s2)(6380 km)/3 2Re Re = 144 m/s
  19. Newton’s Law and Orbits Determine the mass of the sun using the properties of Earth’s orbit. Newton’s Gravitational law: F = G Ms Me / R2. Newton’s 2nd Law: F = Mea = Mev2/R v = 2pR / T G Ms Me / R2 = Me (2p/T)2 R Ms = (2p/T)2 R3 / G
  20. Newton’s Law and Orbits Ms = 4p2 R3/(GT2) R = 1.50 x 1011 m T = 1 year = 3.16 x 107 s G = 6.67 x 10–11 Nm2/kg2 The numbers give Ms = 2.00 x 1030 kg
  21. Kepler’s Laws Law 1: All planets move in elliptical orbits. P = perihelion A = aphelion
  22. Kepler’s Laws Law 2: A line joining any planet to the sun sweeps out equal area in equal times.
  23. Kepler’s Laws Kepler’s Third Law: The square of a planet’s year, divided by the cube of its mean distance from the sun, is the same for all planets: T2/R3 = TE2/RE3 for Earth
  24. Kepler’s Third Law R3 / T2 = constant If R is measured in astronomical units (1 AU = the distance from the Earth to the sun) and T in Earth years, then R = 1 and T = 1 for the Earth. 2R
  25. Kepler’s Third Law Therefore, R3 / T2 = 1 if R is in AU and T in Earth years. 1 AU = 1.5 x 1011 m 1 year = 3.16 x 107 s 2R
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