1. PHY 101: Lecture 5 5.1 Uniform Circular Motion 5.2 Centripetal Acceleration 5.3 Centripetal Force 5.4 Banked Curves 5.5 Satellites in Circular Orbits 5.6 Apparent Weightlessness and Artificial Gravity 5.7 Vertical Circular Motion

2. PHY 101: Lecture 4 Dynamics of Uniform Circular Motion 5.1 Uniform Circular Motion

3. Accelerated Motion No Acceleration  Object is at rest  Object is moving in a straight line with constant speed Acceleration  Object is moving in a straight line with changing speed  Object is moving in a curve with constant speed  Object is moving in a curve with changing speed

4. Uniform Circular Motion Uniform circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path The magnitude of the object’s velocity in circular motion is constant Direction of the velocity changes in time

5. Uniform Circular Motion Period Period (T) –Time required to travel once around the circle –Time for one revolution, one lap, or one cycle speed = distance / time Let v be the speed (normally v is velocity) d is the circumference of the circle, 2  r v = d / t = 2  r / T

6. Period - Example The tangential speed of a particle on a rotating wheel is 3.0 m/s Particle is 0.20 m from the axis of rotation How long will it take for the particle to go through one revolution?  d = vt  2  r = vT  T = 2  r/v = 2  (0.20)/3.0 = 0.42 s

7. PHY 101: Lecture 4 Dynamics of Uniform Circular Motion 5.2 Centripetal Acceleration

8. Magnitude: –Centripetal acceleration of an object moving with a speed v on a circular path of radius r has a magnitude a c a c = v 2 /r = 4  2 r/T Direction: –Centripetal acceleration vector points toward the center of the circle and continually changes direction as the object moves

9. Centripetal Acceleration – Example 1 A racing car with a speed of 33.3 m/s goes around a level, circular track with a radius of 1000 m What is the centripetal acceleration of the car?  a c = v 2 /r = (33.3) 2 /1000 = 1.11 m/s 2

10. Centripetal Acceleration - Example 2 A car with a constant velocity of 23.1 m/s enters a circular flat curve with a radius of curvature of 400 m The friction between the road and the car’s tires can supply a centripetal acceleration of 1.25 m/s 2 Does the car negotiate the curve safely?  a c = v 2 /r = (23.06) 2 /400 = 1.33 m/s 2  Centripetal acceleration provided by friction is not enough to have the car stay on the curve

11. PHY 101: Lecture 4 Dynamics of Uniform Circular Motion 5.3 Centripetal Force

12. Centripetal Force is name given to the net force required to keep an object of mass m, moving at velocity v, on a circular path of radius r Magnitude: F c = ma c F c = mv 2 /r = m4  2 r/T 2 = mr  2 Direction: –Centripetal force vector always points toward the center of the circle and continually changes direction as the object moves

13. Sources of Centripetal Force What provides centripetal force depends on the physical situation  Moon in orbit around earth – gravity  Gravitron amusement park ride – normal force  Child spins toy on end of rope – tension  Car making a turn on flat road - friction

14. Centripetal Force – Example 1 A car is going in a circle at 30 m/s and radius 150 m Your body’s mass is 80 kg What centripetal force does you body feel  F c = mv 2 /r = 80(30 2 )/150 = 480 N

15. Centripetal Force – Example 2 A car of mass 1.5 x 10 3 kg rounds a circular turn of radius 20 m The road is flat and the coefficient of static friction between the tires and the road is 0.50 How fast can the car travel without skidding?  Centripetal force is provided by static friction between the tires and the road  Car can make the turn without skidding if the centripetal force equals the maximum static friction  F c = mv 2 /r   s N =  s (mg) = mv 2 /r  v = sqrt(  s gr) = sqrt(0.50 x 9.8 x 20) = 9.9 m/s

16. Centripetal Force – Example 3 A car goes over the top of a hill with a radius of 150 m Mass of the driver is 80 kg What normal force does driver feel?  F c = mv 2 /r  N - mg = -mv 2 /r  N = mg - mv 2 /r  N = 80(9.8 - 30 2 /180)=384 N At what speed is N = 0 0 = m(g-v 2 /r) v = sqrt(rg) = sqrt(150x9.8) v = 38.3 m/s At this speed the car leaves the road

17. Centripetal Force – Example 4 Fighter plane dives towards ground and then pulls up (Position 1 in diagram) Velocity is 400 m/s Turning radius is 3600 m Mass of the pilot is 60 kg What is normal force on pilot?  F c = mv 2 /r  N – mg = mv 2 /r  N = m(g + v 2 /r)  N = 60(9.8 + 400 2 /3600) = 3254 N  Pilots weight is mg = 60(9.8) = 588 N  N = 5.53 times weight of pilot

18. Centripetal Force – Example 5  What is the slowest velocity if bike is to stay on the track at position 3?  F c = mv 2 /r  -N – mg = -mv 2 /r  N + mg = mv 2 /r  v = sqrt[r(N/m + g)]  Just leave track when N = 0  v = sqrt(rg)

19. Centripetal Force - Example 6 Plane turns Main force is the lift of the wing Inclining plane in turn causes lift to act as normal force Concorde flies at 600 m/s Passenger can’t take normal force greater than 1.5 mg What is sharpest turn it can make?

20. Centripetal Force - Example 6 Find  L = 1.5 mg Lcos  = mg 1.5(mg)cos  = mg cos  = 1/1.5 = 2/3  = 48.19 0 Find r F c = mv 2 /r Lsin  = mv 2 /r 1.5(mg)sin  = mv 2 /r r = v 2 /1.5gsin  r=360000/[1.5(9.8)sin48.19]=32.9 km

21. Centripetal Force – Example 7 Coin sits on long playing record spinning at 33.33 rpm Coefficient of friction between penny and record is 0.3 What is maximum r for penny to remain on the record v = 33.33 rpm = 33.33 x 2  r/60 m/s =3.49r m/s F c = mv 2 /r  s N =  s mg = mv 2 /r  s g = 3.49 2 r 2 /r r =  s g/3.49 2 =0.3(9.8)/12.18 = 0.241 m

22. Centripetal Force – Example 8 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

23. Centripetal Force – Example 8 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

24. PHY 101: Lecture 4 Dynamics of Uniform Circular Motion 5.4 Banked Road

25. Centripetal Force Banked Road F c = F N sin  = mv 2 /r F N cos  = mg Divide first equation by second equation tan  = v 2 /gr

26. Centripetal Force Banked Road - Example A car with a velocity of 30 m/s makes a turn of radius 150 m on a banked road of angle  Calculate   tan  = v 2 /gr = (30) 2 /(9.8 x 150) = 0.6122   = 31.47 0

27. PHY 101: Lecture 4 Dynamics of Uniform Circular Motion 5.5 Satellites in Circular Orbits

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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

35. 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

36. 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

37. 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

38. Stars Orbiting Edge of Galaxy - 2

39. Graph

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

41. PHY 101: Lecture 4 Dynamics of Uniform Circular Motion 5.6 Apparent Weightlessness and Artificial Gravity

42. Apparent Weightlessness Person is on a scale in a freely falling elevator and in a satellite in a circular orbit. When the person is standing stationary on the earth, his weight is 800 N. In each case, the scale records 0 N. N is apparent weight N = ma + mg a = -9.8, g = -9.8 N = m(-9.8 – 9.8) = 0

43. Artificial Gravity A space station rotating about an axis. Because of the rotational motion, any object located at point P on the interior surface of the station experiences a centripetal force directed toward the axis. The surface of the station provides this force by pushing on the feet of an astronaut. The centripetal force can be adjusted to match the astronaut’s earth-weight by selecting the rotational speed of the space station.

44. Artificial Gravity - Example At what speed must the interior surface of the space station (r = 1700 m) move, so that the astronaut experiences a push on his feet that equals his weight on earth (800 N)? F c = mv 2 /r F c = mg mg = mv 2 /r v 2 = rg v = sqrt(rg) = sqrt (1700 x 9.8) = 130 m/s

45. PHY 101: Lecture 4 Dynamics of Uniform Circular Motion 5.7 Vertical Circular Motion

46. Vertical Circular Motion - 1

47. Vertical Circular Motion - 2 m = mass of motorcycle and rider At point 3 when F N3 = 0, the motorcycle will leave the track Then v 3 = sqrt(rg)