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Uniform Circular Motion

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Presentation on theme: "Uniform Circular Motion"— Presentation transcript:

1 Uniform Circular Motion
Horizontal Circle Vertical Circle Banked Curve Unbanked Curve Newton’s Law of Universal Gravitation Kepler’s 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
ac = V2 R Acceleration is occurring because the direction of the velocity is changing. Acceleration is always toward the center of the circular path. ac Finkster 2007

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

8 Centrifugal Force Fictitious force. Doesn’t exist
Force and acceleration acting outward. Finkster 2007

9 Derived Equations for ac and Fc
a = V2 R a = (2πR/T)² a = 4π²R F = ma F = m V2 R F = m 4π²R 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 = 2πr R = 2850 m T T = ? T = 2πr V = 2π(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 = 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 = 2πr T = 0.30 s T = 2π(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 = V2 v = 7.5 m/s R r = 25 m = (7.5 m/s)² a = ? 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: ac = 4π² r R = 2.6 x 10³ m T² T = 360 s = 4 π² (2.6 x 10³ m ) (360s)² ac = 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 Fc = m V² R = 0.25 m r V = 0.68 m/s = (0.015 kg)(0.68 m/s)² Fc = ? m Fc = 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 Fc = m 4π² r R = m T² T = s = ( kg)4 π² (0.100 m) (0.500 s)² Fc = 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: Fc = mac m = kg = (0.350 kg)(0.75 m/s²) a = 0.75 m/s² Fc = 0.26 N Finkster 2007

18 Vertical Circle Finkster 2007

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

20 F c equations for vertical circle
Fc = Fnet Object at Top T + W = mV2 R Object at Bottom T – W = mV2 R 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, (fs)
The static friction acts toward the center of circular motion. fs = Fc µsFN = m V²/R µsmg = m V²/R µsg = 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: µs g = V²/R V = 32 m/s µs =(32 m/s)² /(56 m)(9.80 m/s²) r = 56 m µs = 1.87 Finkster 2007

26 Banked Curve Finkster 2007

27 Banked Curve Fc = Wx The x-component of the weight is acting towards the center of the circular path Wx = Fc mg sin Ө = Fc 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 Fc = Tx
The x-component of T provides the Fc to keep the object moving in a circular path. T sinӨ = Fc T sinӨ = m V²/R Finkster 2007

33 Using two equations to find V
T cos Ө = mg T sin Ө = mV2/r T = mg T = mV2/r cos Ө sin Ө T sin Ө = mg T cos Ө mV2/r V2 = 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 Newton’s Law of Universal Gravitation
Finkster 2007

36 Newton’s 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 Fg = gravitational force of attraction
G = gravitational constant, 6.67 x Nm²/kg² (Henry Cavendish) Finkster 2007

38 Fg is directly proportional to m1m2 indirectly proportional to R²
Mass of one object doubled, Fg is doubled. Mass of one of the objects is halved, Fg is halved. Distance between the centers of masses is doubled, Fg is decreased to ¼. Finkster 2007

39 Three Laws of Planetary Motion
Kepler’s 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 Finkster 2007

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. Ta² = ra³ Tb² rb³ 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 = Gm1m2 ma = Gm1m2 a = Gm Finkster 2007

50 Example 2. F = F W = mg Fg = Gm1m2 mg = Gm1m2 g = Gm Finkster 2007

51 Example 3. F = F Fc= mV² Fg = Gm1m2 r r² mV² = Gm1m2 r r² V² = Gm r
Finkster 2007

52 Example 4. F = F Fc = m4π²r Fg = Gm1m2 T² r² m4π²r = Gm1m2 T² r²
GT² Finkster 2007

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