1. PHY 151: Lecture 6 6.3 Extending Particle in Uniform Circular Motion Model (Continued) 6.4 Nonuniform Circular Motion 6.5 Motion in Accelerated Frames 6.6 Motion in Presence of Resistive Forces

2. PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6.3 Extending Particle in Uniform Circular Motion Model (Continued)

3. Gravitron - 1 Circular room with a person pressed against the wall What is  s to keep the person from sliding down? Radius is 3 meters Rotating at 20 revolutions/minute

4. Gravitron - 2 v = 20 rpm = 20 x 2  r/60 m/s v = 20 rpm =2.09(3) = 6.28 m/s Friction force = weight  s N = mg Normal force is centripetal force F c = N = mv 2 /r  s mv 2 /r = mg  s = gr/v 2 = (9.8)(3)/6.28 2 = 0.76

5. Satellites in Uniform Circular Motion

6. Centripetal Force Satellites in Orbit - 1 Satellite of mass m is in orbit around earth Gravitational pull of earth provides centripetal force  F c = GM e m/r 2 = mv 2 /r  GM e /r = v 2  v = sqrt(GM e /r) Orbital velocity for a given radius is independent of mass of satellite

7. Centripetal Force Satellites in Orbit - 2 v = sqrt(GM e /r)(from prior slide) v = 2  r/T(definition of period) 2  r/T = sqrt(GM e /r) T 2 = 4  2 r 3 /Gm e r 3 = T 2 Gm e /4  2 Relationships between period, T, and orbital radius r

8. Low Earth Orbit - 1 Velocity of a satellite in low earth orbit M e = 5.98 x 10 24 kg r e = 6.37 x 10 6 m Height of satellite = 120 miles = 0.19 x 10 6 m r orbit = 6.56 x 10 6 m G = 6.67 x 10 -11 nm 2 /kg 2 v = sqrt(GM e /r) v = sqrt(6.67x10 -11 x 5.98 x 10 24 / 6.56 x 10 6 ) v = 7797 m/s = 17,446 mi/hour

9. Low Earth Orbit - 2 Period of satellite in low earth orbit T 2 = 4  2 r 3 /Gm e r e = 6.38 x 10 6 m Height of satellite = 120 miles = 0.19 x 10 6 m r orbit = 6.56 x 10 6 m m e = 5.98 x 10 24 kg T 2 = 4  2 (6.38 x 10 6 ) 3 /[6.673 x 10 -11 x 5.98 x 10 24 ] T 2 = 27941796 s 2 T = 5286 s = 88 minutes

10. Geosynchronous Orbit - 1 A satellite in geosynchronous orbit stays above a specific place on the earth For this to occur, the period of the orbit must be 24 hours = 86400 s r 3 = T 2 Gm e /4  2 r 3 = (86400) 2 x [6.673 x 10 -11 x 5.98 x 10 24 ] /4  2 r 3 = 7.55 x 10 22 m 3 r = 4.23 x 10 7 m = 26,268 miles This is distance from center of the earth

11. Geosynchronous Orbit - 2 r = 4.23 x 10 7 m = 26,268 miles This is distance from center of the earth Radius of earth is 6.38 x 10 6 m Distance from surface is 4.23 x 10 7 – 0.64 x 10 7 = 3.59 x 10 7 m 22,312 miles

12. Satellites in Orbit – Example 3 Earth orbits the sun as a satellite. What is sun’s mass? Distance from earth to sun, r se = 1.5 x 10 11 m Period of earth around sun  T = 365.24 days x 24 x 60 x 60 = 3.156 x 10 7 s d = vt v = d/t = 2  r se /T GM sun /r se = v 2 = 4  2 r se 2 /T 2 M sun = 4  2 r se 3 /GT 2 M sun = 4  2 (1.5 x 10 11 ) 3 /6.67x10 -11 /(3.156 x 10 7 ) 2 M sun = 2 x 10 30 kg M sun /M earth = 2 x 10 30 / 6 x 10 24 = 330,000

13. Satellites in Orbit – Example 4 Moon is satellite of Earth Io is satellite of Jupiter What is ratio of Jupiter’s mass to earth’s mass Period, T moon, of the moon is 27 days Period, T io, of Io is 1.5 days Radii of the moon and Io are approximately the same M earth = 4  2 r me 3 /GT moon 2 M jupiter = 4  2 r ij 3 /GT io 2 M jupiter /M earth = T moon 2 /T io 2 = (27/1.5) 2 = 324

14. Satellites in Orbit – Example 5 Earth and Jupiter are satellites of the sun What is ratio of Jupiter’s period to earth’s period 1 AU = 1.5 x 10 11 m, distance from earth to sun Earth’s distance from sun is 1 AU (astronomical unit) Jupiter’s distance from sun is 5.2 AU M sun = 4  2 r se 3 /GT earth 2 T earth 2 = 4  2 r se 3 /GM sun T jupiter 2 = 4  2 r sj 3 /GM sun T jupiter /T earth = sqrt(r sj 3 /r se 3) = sqrt(5.2 3 /1 3 ) = 11.86

15. Stars Orbiting Edge of Galaxy - 1 Mass of galaxy is 2 x 10 41 kg This is the mass of visible stars and dust A star orbits the galaxy at 2.7 x 10 20 m from the center of the galaxy This is close to the edge of the galaxy Assume 90% or more of the galactic mass is inside the star’s orbit Find a formula for the orbital velocity of star’s at the edge of the galaxy

16. Stars Orbiting Edge of Galaxy - 2

17. Graph

18. Actual Data Rotation curve of a typical spiral galaxy: predicted (A) and observed (B). The discrepancy between the curves is attributed to dark matter.

19. PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6.4 Nonuniform Circular Motion

20. Non-Uniform Circular Motion The acceleration and force have tangential components produces the centripetal acceleration produces the tangential acceleration The total force is

21. Vertical Circle with Non-Uniform Speed The gravitational force exerts a tangential force on the object –Look at the components of F g Model the sphere as a particle under a net force and moving in a circular path –Not uniform circular motion The tension at any point can be found

22. Top and Bottom of Circle The tension at the bottom is a maximum The tension at the top is a minimum If T top = 0, then

23. PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6.5 Motion in Accelerated Frames

24. Motion in Accelerated Frames A fictitious force results from an accelerated frame of reference –The fictitious force is due to observations made in an accelerated frame –A fictitious force appears to act on an object in the same way as a real force, but you cannot identify a second object for the fictitious force Remember that real forces are always interactions between two objects –Simple fictitious forces appear to act in the direction opposite that of the acceleration of the non-inertial frame

25. “Centrifugal” Force From the frame of the passenger (b), a force appears to push her toward the door From the frame of the Earth, the car applies a leftward force on the passenger The outward force is often called a centrifugal force –It is a fictitious force due to the centripetal acceleration associated with the car’s change in direction In actuality, friction supplies the force to allow the passenger to move with the car. –If the frictional force is not large enough, the passenger continues on her initial path according to Newton’s First Law

26. “Coriolis Force” This is an apparent force caused by changing the radial position of an object in a rotating coordinate system The result of the rotation is the curved path of the thrown ball From the catcher’s point of view, a sideways force caused the ball to follow a curved path

27. Fictitious Forces, examples Although fictitious forces are not real forces, they can have real effects Examples: –Objects in the car do slide –You feel pushed to the outside of a rotating platform –The Coriolis force is responsible for the rotation of weather systems, including hurricanes, and ocean currents

28. Fictitious Forces in Linear Systems The inertial observer models the sphere as a particle under a net force in the horizontal direction and a particle in equilibrium in the vertical direction The non-inertial observer models the sphere as a particle in equilibrium in both directions The inertial observer (a) at rest sees The non-inertial observer (b) sees These are equivalent if F fictiitous = ma

29. PHY 151: Lecture 6 Circular Motion Other Applications of Newton’s Laws 6.6 Motion in the Presence of Resistive Forces

30. Motion with Resistive Forces Motion can be through a medium –Either a liquid or a gas The medium exerts a resistive force,, on an object moving through the medium The magnitude of depends on the medium The direction of is opposite the direction of motion of the object relative to the medium –This direction may or may not be in the direction opposite the object’s velocity according to the observer nearly always increases with increasing speed

31. Motion with Resistive Forces, cont. The magnitude of can depend on the speed in complex ways We will discuss only two: – is proportional to v Good approximation for slow motions or small objects – is proportional to v 2 Good approximation for large objects

32. Resistive Force Proportional To Speed The resistive force can be expressed as b depends on the property of the medium, and on the shape and dimensions of the object The negative sign indicates is in the opposite direction to

33. Resistive Force Proportional To Speed, Example Assume a small sphere of mass m is released from rest in a liquid Forces acting on it are: –Resistive force –Gravitational force Analyzing the motion results in

34. Resistive Force Proportional To Speed, Example, cont. Initially, v = 0 and dv/dt = g As t increases, R increases and a decreases The acceleration approaches 0 when R  mg At this point, v approaches the terminal speed of the object

35. Terminal Speed To find the terminal speed, let a = 0 Solving the differential equation gives t is the time constant and t = m/b

36. Resistive Force Proportional To v 2 For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed. R = ½ D  Av 2 –D is a dimensionless empirical quantity called the drag coefficient –  is the density of air –A is the cross-sectional area of the object –v is the speed of the object

37. Resistive Force Proportional To v2, example Analysis of an object falling through air accounting for air resistance

38. Resistive Force Proportional To v2, Terminal Speed The terminal speed will occur when the acceleration goes to zero Solving the previous equation gives

39. Some Terminal Speeds

40. Example: Skysurfer Step from plane –Initial velocity is 0 –Gravity causes downward acceleration –Downward speed increases, but so does upward resistive force Eventually, downward force of gravity equals upward resistive force –Traveling at terminal speed

41. Skysurfer, cont. Open parachute –Some time after reaching terminal speed, the parachute is opened –Produces a drastic increase in the upward resistive force –Net force, and acceleration, are now upward The downward velocity decreases –Eventually a new, smaller, terminal speed is reached

42. Example: Coffee Filters A series of coffee filters is dropped and terminal speeds are measured The time constant is small –Coffee filters reach terminal speed quickly Parameters –m each = 1.64 g –Stacked so that front-facing surface area does not increase Model –Treat the filter as a particle in equilibrium

43. Coffee Filters, cont. Data obtained from experiment: At the terminal speed, the upward resistive force balances the downward gravitational force R = mg

44. Coffee Filters, Graphical Analysis Graph of resistive force and terminal speed does not produce a straight line The resistive force is not proportional to the object’s speed

45. Coffee Filters, Graphical Analysis 2 Graph of resistive force and terminal speed squared does produce a straight line The resistive force is proportional to the square of the object’s speed

46. Resistive Force on a Baseball – Example The object is moving horizontally through the air The resistive force causes the ball to slow down Gravity causes its trajectory to curve downward The ball can be modeled as a particle under a net force –Consider one instant of time, so not concerned about the acceleration Analyze to find D and R