Download presentation

Presentation is loading. Please wait.

Published byVanessa Scovel Modified over 2 years ago

1
Uniform Circular Motion Horizontal Circle Vertical Circle Banked Curve Unbanked Curve Newtons Law of Universal Gravitation Keplers Laws Finkster 2007

2
Uniform Circular Motion Motion of an object traveling at a constant speed in a circular path. Finkster 2007

3
Uniform Circular Motion r Finkster 2007

4
Horizontal Circle Finkster 2007

5
Velocity Velocity is constant in magnitude only. The direction is changing with time. V Finkster 2007

6
Centripetal Acceleration a c = V 2 R Acceleration is occurring because the direction of the velocity is changing. Acceleration is always toward the center of the circular path. acac Finkster 2007

7
Centripetal Force F c = ma c Net force (F c ) and acceleration always act in the same direction. The centripetal force keeps the object moving in a circle FcFc Finkster 2007

8
Centrifugal Force Fictitious force. Doesnt exist Force and acceleration acting outward. Finkster 2007

9
Derived Equations for a c and F c a = V 2 R a = (2 πR/T)² R a = 4π²R T² F = ma F = m V 2 R F = m 4π²R T² Finkster 2007

10
Example 1. Period How long does it take a plane traveling at a speed of 110 m/s, to fly once around a circle whose radius is 2850 m? Given: V = 110 m/s V = 2 πr R = 2850 m T T = ? T = 2 πr V = 2 π(2850 m) 110 m/s T = 160 s Finkster 2007

11
Example 2. Frequency It takes 2 min to twirl a ball in a circle 50 times. What is the frequency? f = 50 2 min. X 60 s min f = 0.4/s Finkster 2007

12
Example 3. Velocity A stopper is twirled in a horizontal circle whose radius is 0.75 m. The stopper travels once around the circle in 0.30 s. What is its velocity? Given: R = 0.75 m V = 2 πr T = 0.30 s T = 2 π(0.75 m) 0.30 s V = 16 m/s Finkster 2007

13
Example 4. Acceleration A runner moving at a speed of 7.5 m/s rounds a bend with a radius of 25 m. Determine the centripetal acceleration of the runner. Givens: a = V 2 v = 7.5 m/s R r = 25 m = (7.5 m/s)² a = ? 25 m a = 2.3 m/s² Finkster 2007

14
Example 5. Acceleration A race car travels at a constant speed around a circular track whose radius is 2.6 km. If the car goes around the track in 360 s, what is the magnitude of the centripetal acceleration of the car? Given: a c = 4 π ² r R = 2.6 x 10³ m T² T = 360 s = 4 π ² (2.6 x 10³ m ) (360s)² a c = 0.79 m/s² Finkster 2007

15
Example 6. Centripetal Force A kg ball is shot from the plunger of a pinball machine. Because of a centripetal force, the ball follows a circular arc whose radius is 0.25 m at a speed of 0.68 m/s. Calculate the centripetal force. Given: M = kg F c = m V ² R = 0.25 m r V = 0.68 m/s = (0.015 kg)(0.68 m/s)² F c = ? 0.25 m F c = N Finkster 2007

16
Example 7. Centripetal Force A child is twirling a kg ball on a string in a horizontal circle whose radius is 0.l00 m. The ball travels once around the circle in s. Determine the centripetal force on the ball. Given: M = kg F c = m 4 π ² r R = m T² T = s = ( kg)4 π ² (0.100 m) (0.500 s)² F c = N Finkster 2007

17
Example 8. Centripetal Force A kg ball is twirled in a horizontal circle. The centripetal acceleration is 0.75 m/s². Calculate the centripetal force. Given: F c = ma c m = kg = (0.350 kg)(0.75 m/s²) a = 0.75 m/s² F c = 0.26 N Finkster 2007

18
Vertical Circle Finkster 2007

19
Direction of forces when object is at top and bottom of vertical circle Top F net = T + W Bottom Fnet = T - W F c = F net Finkster 2007

20
F c equations for vertical circle Object at Top T + W = mV 2 R Object at Bottom T – W = mV 2 R F c = F net Finkster 2007

21
Example 9. Top of Vertical Circle A 2.0 kg object is attached to a 1.5 m long string and swung in a vertical circle at a constant speed of 12 m/s. What is the tension in the string when the object is at the top of its path? Given: T + W = mV² m = 2.0 kg R r = 1.5 m T = mV² - W V = 12 m/s R T = ? = (2.0 kg)(12 m/s)² - (2.0 kg)(9.80 m/s²) 1.5 m T = 1700 N Finkster 2007

22
Example 10. Vertical Circle What is the tension in the string when the object is at the bottom of its path? Given: T – W = m V²/r m= 2.0 kg T = m V² + W V = 12 m/s R r = 1.5 m = (2.0 kg)(12m/s)² N 1.5 m T = 2100 N Finkster 2007

23
Unbanked Curve Finkster 2007

24
F c is the Static Friction, (f s ) The static friction acts toward the center of circular motion. f s = F c µ s F N = m V²/R µ s mg = m V²/R µ s g = V²/R Finkster 2007

25
Unbanked Curve Racing on a flat track, a car going 32 m/s rounds a curve 56 m in radius. What would be the minimum coefficient of static friction between tires and road that would be needed for the car to round the curve without skidding? Given: µ s g = V²/R V = 32 m/s µ s =(32 m/s)² /(56 m)(9.80 m/s²) r = 56 m µ s = 1.87 Finkster 2007

26
Banked Curve Finkster 2007

27
Banked Curve F c = W x The x-component of the weight is acting towards the center of the circular path W x = F c mg sin Ө = F c mg sin Ө = m V²/R g sin Ө = V²/R Finkster 2007

28
Example 11. Banked Curve A 60.0 kg speed skater comes into a curve of 20.0 m radius and banked at 12º. Calculate the speed needed to negotiate the curve? Given: g sin Ө = V²/R m = 60.0 kg V² = Rg sin Ө R = 20.0m =(20.0 m)(9.80 m/s²)(sin 12º) V = ? V = 6.38 m/s Finkster 2007

29
Motion and Forces on a Banked Curve These problems deal with an object (a car, for example) navigating a circular curve at constant speed. The radius of the curve is fixed at 10 m, but the speed and mass of the object, the angle at which the curve is banked, the coefficient of static friction, and the gravitational field strength can all be varied. Two views of the object are shown: i) overhead showing the circular path with acceleration and velocity vectors, ii) ground level showing the forces acting on the object. As the values of input parameters are changed, the vectors change in response. A "fric-o-meter" gauges the ratio of the actual static friction to the maximum available static friction. This allows the user to determine whether the object will skid off the curve or slide down to the center Finkster 2007

30
Motion and Forces on a Banked Curve Finkster 2007

31
Conical Circular Path Finkster 2007

32
Conical Circular Path F c = T x The x-component of T provides the F c to keep the object moving in a circular path. T sinӨ = F c T sinӨ = m V²/R Finkster 2007

33
Using two equations to find V T cos Ө = mg T sin Ө = mV 2 /r T = mg T = mV 2 /r cos Ө sin Ө T sin Ө = mg T cos Ө mV 2 /r V 2 = rg tan Ө Finkster 2007

34
Example 12. Conical Path The Swing Ride at an amusement park travels at 15.0 m/s and the 12.0 m swings make an angle of 25.0º with the vertical as they move in a circle. What is the tension in the rope of a swing carrying a 50.0 kg person? Given: T sinӨ = m V²/R V = 15.0 m/s T = m V² Ө = 25.0º r sinӨ M = 50.0 kg = (50.0 kg)(15.0 m/s)² R = 12.0 m (12.0 m)(sin 25.0º) T = 2218 N Finkster 2007

35
Newtons Law of Universal Gravitation Finkster 2007

36
Newtons Law of Universal Gravitation If the path of a planet were an ellipse, then the net force on the planet varies inversely with the square of the distance between the planet and the sun. Finkster 2007

37
Symbols represent F g = gravitational force of attraction G = gravitational constant, 6.67 x Nm²/kg² (Henry Cavendish) Finkster 2007

38
F g is directly proportional to m 1 m 2 indirectly proportional to R² Mass of one object doubled, F g is doubled. Mass of one of the objects is halved, F g is halved. Distance between the centers of masses is doubled, F g is decreased to ¼. Finkster 2007

39
Keplers Laws Three Laws of Planetary Motion Finkster 2007

40
Law 1 Path of the planets are ellipses with the sun at one focus. Finkster 2007

41

42
Law 2 An imaginary line extending from the sun to a planet will sweep out equal areas in equal amounts of time. Planets move fastest when closest to the sun. Planets move slowest when farthest from the sun. Finkster 2007

43
Shaded segments = 30 days Finkster 2007

44
Law 3 The ratio of the squares of the periods of any two planets in orbit around the Sun is equal to the ratio of the cubes of their distances from the sun. T a ² = r a ³ T b ² r b ³ Finkster 2007

45
Motion of Planets and Satellites Finkster 2007

46
Satellite in Orbit Uniform circular motion if always at the same height above Earth Path has both vertical and horizontal components Velocity of horizontal component great enough… path will follow a curve that matches the curve of Earth. Projectile is in orbit. Finkster 2007

47
Weightlessness Occurs when the downward force of gravity is unbalanced and there is no upward force acting on a satellite in orbit. Finkster 2007

48
Deriving Equations F = F Finkster 2007

49
Example 1. F = F F = ma F = Gm 1 m 2 r² ma = Gm 1 m 2 r² a = Gm r² Finkster 2007

50
Example 2. F = F W = mg F g = Gm 1 m 2 r² mg = Gm 1 m 2 r² g = Gm r² Finkster 2007

51
Example 3. F = F F c = mV² F g = Gm 1 m 2 r r² mV² = Gm 1 m 2 r r² V² = Gm r Finkster 2007

52
Example 4. F = F F c = m4 π ²r F g = Gm 1 m 2 T² r² m4 π ²r = Gm 1 m 2 T² r² m = 4 π ²r³ GT² Finkster 2007

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google