1. Circular Motion and Gravity A Special Kind of Force Problem

2. Rotation Motion about an axis inside a body Linear(tangential) speed = radius x angular(rotational) speed Units: meters/sec = meters x radians/sec(also revolutions per minute) All parts of a rigid rotating body have same angular speed but different linear speed.

3. Acceleration is Centripetal An object moving at steady speed in a circle of radius r is accelerated toward the center with a = v 2 r v Lets see why… a

4. What if a String Held the Object? It would be in tension The direction of this force could only be along the string Thus force and acceleration must be toward the center of the circle Velocity of object is always perpendicular to string…and vector a toward center of circle.

5. v 2 /r derivation s is arc length by definition of radians or similar triangles Courtesy University of Pennsylvania Physics Dept.

7. Linear velocity and angular velocity Angle Theta  Angular velocity (radians per second)  = d  dt Linear velocity v = r  s  r  ds/dt = r d  /dt v = r d  /dt = r 

8. Force On Object is Inward From non-moving rest frame observing the object, force on it is inward Called centripetal force In rest frame of object (accelerated) acceleration and force are experienced as outward (centrifugal) pseudoforce Be CAREFUL about the rest frame It’s probably better to forget about centrifugal force altogether

9. Centripetal Force is Not a Real Force Real forces such as gravity, tension in string, friction PROVIDE centripetal force Centripetal force is algebraically equal to the RIGHT side of Newton’s F = ma Real forces are on the LEFT side Some people call centripetal force a “pseudoforce.”

10. Centripetal force can be provided by--- Normal force Static friction Tension Gravity Electrical force Magnetic force

11. F = ma c = mv 2 /r problems A 50 gram mass makes a horizontal circle at uniform speed in 1.0 sec at the end of a 0.5m string. What is the force on it? v = 2  r/1.0 sec =3.14 m/s F = mv 2 /r = 0.05Kg x 3.14 2 /0.5 m = 0.9869 N = 0.99 N

12. Period and Frequency Period = T = time to go around once Frequency = number of rotations or revolutions per second T = 1/f f = 1/T

13. Ferris Wheel What is the upward force of the seat of a 50 Kg rider at the bottom of a 100m diameter Ferris Wheel if it rotates at 2.0 rev/min?  F = mv 2 /r F N – mg = mv 2 /r F N = mv 2 /r + mg v = 2  r/30s = 10.47 m/s Continued… mg FNFN

14. Ferris Wheel, continued F N = mv 2 /r + mg v = 10.47 m/s F N = 50Kg(10.47) 2 /50m + 490N F N = 110N + 490N = 600N up a=v 2 /r = (10.47) 2 /50m = 2.19 m/s 2

15. Parking Lot Wheelies A car makes a circle radius 40m while traveling 20m/s. What coefficient of static friction is required to keep the car in the circle? Static friction directed toward the center of the car’s circle provides the necessary centripetal acceleration to keep the car moving in a circle Continued.. FCFC

16. Parking Lot, continued F =ma F = mv 2 /r f S = mv 2 /r  s F N = mv 2 /r  s mg = mv 2 /r  s  v 2 /rg = 20 2 /40x9.8  s  fSfS mg FNFN

17. Banked Turn (no friction) What angle of banking is required to keep a car traveling 30m/s around a curve of radius 100m from slipping out of the turn? FNFN F Nx = F N sin  F Ny = F N cos  x y mg 

18. Banked Turn, continued F N cos  mg = 0 F N cos  mg F N sin  mv 2 /r Divide tan  v 2 /rg = 30 2 /(100 x 9.8) tan  = 0.918  = 42.6 0 = 43 0

19. Blue Angel Loops Blue Angels fly 150 m/s in a vertical circle of radius 1.0 km. Find a c and F n at top and bottom of loop. Pilot is 80 Kg.

20. Top mg + F n = mv 2 /r F n = mv 2 /r - mg = m(v 2 /r - g) = 80 kg( 150 2 /1000m – 9.8) = 1016 N

21. Bottom F n – mg = mv 2 /r F n = m (g + v 2 /r) = 2584 N v 2 /r = 150 2 / 1000m = 22.5 m/s 2 = 2.3 g’s Smaller loop at same speed dangerous because you black out about 6 g’s

22. Gravity… …was a mystery until Isaac Newton developed his theory of universal gravitation. He tried to understand why objects fell to earth and why the planets moved in the sky as they did Supposedly an apple fell on his head inspiring him to conclude that the two mysteries has a common solution

23. Sources for Review Physics Classroom – Glenbrook South Hyperphysics FizzicsFizzle Simulation

24. “Gravity is a Universal Force” says Isaac Newton All masses attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them F = G mM/r 2 m M r G is not the same as g !!!

25. Inverse Square Law Courtesy “Hyperphysics,” Georgia State University

26. Finding G in F = G mM/r 2 Newton did not know the value of G Difficult to measure because small Gravitational forces between objects of everyday size is undetected We only are aware of gravitation when one body in the law is astronomical in size

27. G is Gravitational Constant Same in all places in Universe F = G mM/r 2 Same at all times Measured by Cavendish 100 years after Newton G = 6.67 x 10 -11 N/Kg 2 -m 2 Don’t confuse G with g, the acceleration of an object due to gravity at surface of a planet, which is 9.8 m/s 2

28. Measuring G - Cavendish Courtesy The Physics Classroom System rotates when lead balls are brought near smaller balls Pasco scientific gravitational torsion balance F = G m 1 m 2 /d 2

29. Mass of the Earth F = G mM/r 2 Weight of object mg = G mM/r 2 g = G m E /r E 2 m E = gr E 2 /G = (9.80 m/s 2 )(6.38x 10 6 m) 2 ÷ (6.67x10- 11 Nm 2 /kg 2 ) = 5.98x10 24 kg Still the only way we know mass of Earth

30. Applying Newton’s Gravity Law to Orbits Uses a c = v 2 /r Gravity is only force acting between two astronomical bodies F = G mM/r 2 = mv 2 /r Io Jupiter and Io, Courtesy of NASA Jet Propulsion Lab G = 6.67 x 10-11 m 2 N/Kg 2 Universal Gravitation Constant

31. Velocity of a Satellite G mM/r 2 = mv 2 /r ( m is mass of satellite) G M/r 2 = v 2 /r G M/r = v 2 v = (GM/r) 1/2 Implications Mass of a planet does not affect its speed Planet 4 x away goes half the speed

32. Example Planet X of a star orbits at half the distance of Planet Y and is twice as massive. Compare the speeds of the planets. Answer: Planet X is faster v =v =

33. Kepler’s Three Laws 1. Planets move in ellipses with the sun at one focus 2. Planets move so that a line drawn from the Sun to each planet sweeps out equal areas in equal times 3. The ratio of the squares of the periods of any two planets revolving around the sun equals the ratio of the cubes of their semimajor axes (half the long axis) ie (T 1 /T 2 ) 2 = (s 1 /s 2 ) 3 or s 3 /T 2 = constant for all planets

34. Deriving Kepler’s Laws Only Algebra needed for Third Law Need Calculus for laws 1 and 2 Link to derivations

35. Kepler’s Third Law Square of period of planet proportional to cube of average distance to sun. F = ma G mM/r 2 = mv 2 /r GM/r = v 2 = (2  r/T) 2 =4  2 r 2 /T 2 r 3 /T 2 = GM/4  2 = constant M = 4  2 /G x r 3 /T 2 (can use to find mass of star or planet)

36. Gravitational field mg = GmM/r 2 g = GM/r 2 is sometimes called “strength of gravitational field.’ Example. Consider two planets X and Y. Betty weighs 600N on planet X. If planet Y is half the mass and twice the diameter of X, what will Betty weigh on Planet Y? Answer: 75 N