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Circular Motion

Circular Motion. Circular motion equations:. v = 2 π r f. Example 1 : A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its period and linear speed. m = 2 kg r = 2 m

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Circular Motion

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  1. Circular Motion

  2. Circular motion equations: v = 2π r f

  3. Example 1: A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its period and linear speed m = 2 kg r = 2 m f = 3 rev/s = 0.33 s = 37.8 m/s

  4. Remember these?

  5. v Constant velocity tangent to path. Constant force toward center. Uniform Circular Motion Uniform circular motionis motion along a circular path in which there is no change in speed, only a change in direction. Fc Question: Is there an outwardforce on the ball?

  6. v Uniform Circular Motion (Cont.) The question of an outwardforce can be resolved by asking what happens when the string breaks! Ball moves tangent to path, NOT outward as might be expected. When central force is removed, ball continues in straight line. Centripetal forceis needed to change direction.

  7. CENTRIPETAL ACCELERATION An object experiencing uniform circular motion is continuallyaccelerating.The direction and velocity of a particle moving in a circular path of radius r are shown at two instants in the figure. The vectors are the same size because the velocity is constant but the changing direction means acceleration is occurring.

  8. To calculate the centripetal acceleration, we will use the linear velocity and the radius of the circle Or substituting for v We get ac

  9. Example 2: A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration. = 5.92 m/s2

  10. CENTRIPETAL FORCE Theinward forcenecessary to maintain uniform circular motion is defined ascentripetal force.From Newton's Second Law, the centripetal force is given by: Fc = mac Or

  11. Example 3: A 1000-kg car rounds a turn of radius 30 m at a velocity of 9 m/s a. How much centripetal force is required? m = 1000 kg r = 30 m v = 9 m/s = 2700 N b. Where does this force come from? Force of friction between tires and road.

  12. Like a ball on a string, Satellites are an example of uniform circular motion. There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius. The gravitational pull of the Earth provides the centripetal force and acts like an invisible guideline for a satellite.

  13. Fc = mac

  14. Example 4:A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Determine: a. The height above the Earth’s surface such a satellite must orbit and FUG = FC mE = 5.98x1024 kg RE = 6.38x106 m T = 1 day = 86400 s GMET2 = 4 π2r3 r = 4.23x107 m from the Earth’s center

  15. height = r - rE = 4.23x107 - 6.38x106 m = 3.592x107 m b. The satellite’s speed = 3070 m/s

  16. T L q h T mg R The Conical Pendulum A conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L. T cos q q T sin q Note: The inward component of tension T sin q gives the needed central force.

  17. T cos q q T T sin q L q h mv2 R Solve two equations to find angle q T sin q = T v 2 gR mg R tan q = Angle qand velocity v: T cos q = mg

  18. L q h T R Example 5:A 2-kg mass swings in a horizontal circle at the end of a cord of length 10 m. What is the constant speed of the mass if the rope makes an angle of 300 with the vertical? 1. Draw & label sketch. q = 300 2. Recall formula for pendulum. Find: v = ? 3. To use this formula, we need to find R = ? R = L sin 300 = (10 m)(0.5) R = 5 m

  19. R = 5 m q = 300 L q h T R Example 5(Cont.): Find vfor q = 300 4. Use given info to find the velocity at 300. R = 5 m g = 9.8 m/s2 Solve for v = ? v = 5.32 m/s

  20. T cos q q T T sin q L q h T mg R mg cos q (2 kg)(9.8 m/s2) cos 300 T = = Example 6:Now find the tension T in the cord if m= 2 kg, q= 300, and L= 10 m. 2 kg SFy = 0: T cos q - mg = 0; T cos q= mg T = 22.6 N

  21. T cos q q T T sin q mv2 R Fc = L q h T mg R Example 7:Find the centripetal force Fc for the previous example. q = 300 Fc 2 kg m= 2kg; v = 5.32m/s;R = 5 m; T = 22.6 N Fc = 11.3 N or Fc = T sin 300

  22. Circular motion equations: Fc = mac v = 2π r f

  23. HOMEWORK: • 1-10

  24. Circular Motion-2

  25. Let’s imagine that you are riding in a car going around a curve.  Sitting on your dashboard is a cassette tape.  As you go around the curve, the tape moves to outside edge of the car. 

  26. n Fc Fc = fs R fs m R v mv2 R Fc = mg Finding the maximum speed for negotiating a turn without slipping. The car is on the verge of slipping when FC is equal to the maximum force of static friction fs. fs = msmg Fc = fs

  27. n fs R Fc R m mg v mv2 R = msmg v = msgR Maximum speed without slipping (Cont.) Fc = fs Velocity v is maximum speed for no slipping.

  28. m Fc R mv2 R Fc = ms = 0.7 v v = msgR Example 8:A car negotiates a turn of radius 70 m when the coefficient of static friction is 0.7. What is the maximum speed to avoid slipping? fs = msmg From which: g = 9.8 m/s2; R = 70 m v = 21.9 m/s

  29. Fc R m v fs = 0 fs n n n fs w w w q q q slow speed Optimum Banking Angle By banking a curve at the optimum angle, the weight Wcan provide the necessary centripetal force without the need for a friction force. fast speed optimum

  30. n mg mv2 R Apply Newton’s 2nd Law to x and y axes. FN sin q = q Optimum Banking Angle (Cont.) q FN Fw SFx = mac FNcos q = mg SFy = 0

  31. n mg mv2 R FN sin q = Optimum Banking Angle q q Optimum Banking Angle (Cont.) FNcos q= mg

  32. n mg v 2 gR (12 m/s)2 (9.8 m/s2)(80 m) tan q = = q= 10.40 q Example 9:A car negotiates a turn of radius 80 m. What is the optimum banking angle for this curve if the speed is to be equal to 12 m/s? tan q = 0.184 How might you find the centripetal force on the car, knowing its mass?

  33. Think about the sensations you feel as you ride the loop… Where are the forces greatest? Where are the forces least?

  34. v v v + v Bottom mg T T T mg T + T T v v Top of Path Left Side + Top Right Top Right Tension is minimum as weight helps Fc force Weight has no effect on T Maximum tension T, W opposes Fc Weight causes small decrease in tension T Weight has no effect on T Bottom + + + mg mg mg mg Motion in a Vertical Circle Consider the forces on a ball attached to a string as it moves in a vertical loop. Note also that the positivedirection is always along acceleration, i.e., toward the centerof the circle. Note changes as you click the mouse to show new positions.

  35. + 10 N v T T R + 10 N v As an exercise, assume that a central force of Fc = 40 Nis required to maintain circular motion of a ball and W = 10 N. The tension T must adjust so that central resultant is 40 N. T = _?_ At top: 10 N + T = 40 N T = 30 N Bottom: T – 10 N = 40 N T = 50 N T = __?___

  36. v T R mv2 R Fc = mv2 R v mg + T = mv2 R T = - mg + mg mg T Motion in a Vertical Circle Resultant force toward center Consider TOP of circle: AT TOP:

  37. v T R mv2 R Fc = mv2 R mg v T - mg = mv2 R T T = + mg + mg Vertical Circle; Mass at bottom Resultant force toward center Consider bottom of circle: AT Bottom:

  38. v R mg T mv2 R v mg + T = mv2 R T = - mg Example 10:A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its highest point is 10 m/s. What is tension Tin rope? At Top: - T = 5.40N T = 25N - 19.6 N The force are least here

  39. v mv2 R T - mg = R v T mg mv2 R T = + mg Example 10 cont :A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes itslowest point is 10 m/s. What is tension T in rope? At Bottom: T = 44.6N T = 25N + 19.6 N The force are greatest here

  40. v 0 At Top: R mg T mv2 R mv2 R v mg + T = mg = vc = gR v = gR = (9.8 m/s2)(8 m) Example 11:What is the critical speed vc at the top, if the 2-kg mass is to continue in a circle of radius 8 m? vc occurs when T = 0 vc = 8.85 m/s

  41. v mv 2 R n = - mg R v AT BOTTOM: mv 2 R n= + mg n + + mg mg n The Loop-the-LoopSame as cord, n(force of track)replaces T AT TOP:

  42. v mv 2 R mg - n = AT TOP: n R mv 2 R n = mg - v AT BOTTOM: mv 2 R n n = + mg + + mg mg n is in a direction opposite to the centripetal force at the top of the circle. The Ferris Wheel:

  43. n v mv 2 R mg - n = R v mv 2 R + n = mg - mg Example 12:What is the apparent weight of a 60-kg person as she moves through the highest point when R = 45 m and the speed at that point is 6 m/s? Apparent weight will be the normal force at the top: n= 540 N

  44. v mv2 R T = - mg R v AT BOTTOM: mv2 R T T = + mg + + mg mg T For Motion in Circle AT TOP: The force on you is the least The force on you is the greatest

  45. v mv 2 R mg - n = AT TOP: n R mv 2 R n = mg - v AT BOTTOM: mv 2 R n n = + mg + + mg mg Summary: Ferris Wheel

  46. Conical pendulum: v 2 gR v = gR tanq tan q= Summary Centripetal acceleration: Homework: 11-17

  47. Rotational Kinematics – its just like linear kinematics except stuff spins instead of going in a straight line…

  48. A Rotational Displacement,  Consider a disk that rotates from A to B: B Angular displacement q:  Measured in revolutions, degrees, or radians. 1 rev=360 0= 2rad The best measure for rotation of rigid bodies is the radian.

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