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CHAPTER 6: Circular motion (3 Hours)

CHAPTER 6: Circular motion (3 Hours). LEARNING OUTCOME:. 6.1 Uniform circular motion (1 hour). At the end of this chapter, students should be able to: Describe graphically the uniform circular motion in terms of the change in direction of velocity. Circular motion. Uniform.

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CHAPTER 6: Circular motion (3 Hours)

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  1. CHAPTER 6: Circular motion(3 Hours)

  2. LEARNING OUTCOME: 6.1 Uniform circular motion (1 hour) At the end of this chapter, students should be able to: • Describe graphically the uniform circular motion in terms of the change in direction of velocity.

  3. Circular motion Uniform Non - Uniform 6.1 UNIFORM CIRCULAR MOTION Circular motion– motion which occurs when bodies rotate in circular path. (Vertical plane) (Horizontal plane) • Examples: • a ball is swung in horizontal circle. • a ca /motorcycle turning a corner. • cone pendulum. • merry go round. • Examples: • a bucket of water is swung in vertical circle. • roller coaster cars

  4. 6.1 Uniform Circular Motion Uniform circular motionis the motion of an object in a circle with a …………………………speed. • …………. of its velocity remains constant. • …………. of its velocity changes continually. Comparison Of Linear And Circular Motion

  5. B r A 6.1 Uniform Circular Motion MOTION CHARACTERICTICS FOR CIRCULAR MOTION Linear Distance ( ) The arc length between A and B Angular Displacement ( ) The angle subtended by the arc length. Unit : radian ( rad ) Relation between , r,

  6. 6.1 Uniform Circular Motion Motion Characterictics For Circular Motion ANGULAR VELOCITY ( ω ) • Rate of change of angular displacement. angular displacement (rad) time taken (s) - Unit ω: ……………… - Other units:……………. - ……………………and its direction is perpendicular to the plane of motion (right hand rule)

  7. Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion THE RELATIONSHIP BETWEEN LINEAR VELOCITY, V AND ANGULAR VELOCITY, ω Divide both sides by , - The directionof linear velocity at every point along the circular path is ………………..to the point. • The direction of the angular velocity, ωdepends ………………… • of the object (clockwise or counterclockwise ).

  8. Period, T • is defined as the …………………………………..revolution (cycle/rotation). • The unit of the period is …………………... • Frequency, f • is defined as the …………………………(cycles/rotations) completed in one second. • The unit of the frequency is ……………….or …….. • Equation : • Let the object makes one complete revolution in circular motion, thus • the distance travelled is (circumference of the circle), • the time interval is one period, T.

  9. Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion LINEAR VELOCITYcan be written in terms of period, T and frequency, f : ANGULAR VELOCITYcan be written in terms of period, T and frequency, f :

  10. Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion EXAMPLE 6.1.1 An object undergoes circular motion with uniform angular speed 100 rpm. Calculate : (a) the period, T (b) the frequency of revolution, f. SOLUTION 6.1.1 Given : ω = 100 rpm Convert to rad s-1 : (a) (b)

  11. Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion EXAMPLE 6.1.2 An object travels around the circumference of a circle of radius 6 m at a rate of 30 rev/min. Calculate (a) its angular speed in rad/s. (b) its linear speed around the circle. SOLUTION 6.1.2 r = 6m ω = 30 rev/min (a) (b)

  12. EXAMPLE 6.1.3 The diameter of a tire is 64.8 cm. A tack is embedded in the tread of the right rear tire. What is the magnitude and direction of the tack's angular velocity vector if the vehicle is traveling at 10.0 km/h? Solution 6.1.3:

  13. Learning Outcome: 6.2 Centripetal force (2 hours) At the end of this chapter, students should be able to: • Define and use centripetal acceleration: • Define and solve problems on centripetal force:

  14. 6.2 Centripetal Acceleration, ac • When an object moving in a circle of radius, r at a ……………………., v, the direction of the object ………….... Thus it has an acceleration called the centripetal acceleration. • acis defined as the acceleration of an object moving in circular path and it directed ……………………….of the circle. • Direction of ac – graphically. FIGURE 6.2.1

  15. 6.2 Centripetal Acceleration, ac speed of the object Magnitude of ac : radius v : linear tangential velocity ω: angular velocity (angular frequency) r : radius of circular path since v = r ω,

  16. Uniform Circular Motion Problem Solving

  17. 6.2 Centripetal Acceleration, ac Example 6.2.1 Calculate the centripetal acceleration of a car traveling on a circular racetrack of 1000 m radius at a speed of 180 km h-1. Solution Given : r = 1000 m v = 180 km h-1

  18. Centripetal Force, Fc Fc is defined as the …………………..required to keep an object of mass, m moving at a speed v on a circular path of radius, r. Examples : As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion. FIGURE 6.2.2 As a car makes a turn, the force of friction acting upon the turned wheels of the car provides the centripetal force required for circular motion. FIGURE 6.2.3

  19. Centripetal Force, Fc Magnitude of Fc : since v = r ω , thus Direction of Fc- …………………………of the circle and …………………..of the centripetal acceleration. Fcis perpendicular to the direction v, so it does no work on the object. FIGURE 6.2.4

  20. Circular motion in horizontal plane (uniform circular motion) Case 1 : object moves in a horizontal circle with steady speed. T mg FIGURE 6.2.5 • Two forces acting on the object : • The force of ………………….( weight i.e. mg ) • The ……………….in the string - is the only component in the radial direction that provided thecentripetal force. Applying Newton 2nd Law :

  21. Circular motion in horizontal plane (uniform circular motion) Example 6.2.2 A 0.25 kg rock attached to a string is whirled in a horizontal circle at a constant speed of 10.0 ms-1. The length of the string is 1.0m. Neglecting the effects of gravity, find the tension in the string. Solutio 6.2.2

  22. Case 2 :Motion of car round a curve : Flat curve road R fs mg If the coefficient of static friction between the tires & the road is μ then, f s= µR Vertical comp. : Horizontal comp.: …( 2 ) … ( 1 ) (1) into (2): ∴ The maximum velocity without slipping on the road is :

  23. R fs mg Motion of car round a curve Example 6.2.3 A car travels around a flat curve of radius r = 50m. The coefficient of the static friction between the tires & the road is μs = 0.75. Calculate the maximum speed at which the car can travel without skidding. Solution 6.2.3 Given: μs = 0.75 , r = 50m fs supplies the centripetal force

  24. R fs mg Motion of car round a curve Example 6.2.4 A 1200 kg car with a velocity of 8.0 m/s travels around a flat curve of radius r = 9.0m. a) Calculate the horizontal force must the pavement exert on the tires to hold the car in the circular path ? b) What coefficient of friction must exist for the car not to slip ? Solution 6.2.4 a) b)

  25. Case 3 :Conical Pendulum A conical pendulum moving in uniform circular motion with speed v: T sin θ supplies centripetal force. r = L sin θ Component – x : r FIGURE 6.2.7 Component – y : … (1) … (2)

  26. Case 3 :Conical Pendulum r = L sin θ … (1) … (2) r FIGURE 6.2.8

  27. r = L sin 30˚ r Conical Pendulum Example 6.2.6 A 0.15 kg ball attached to a string which is 1.2 m in length moves in a horizontal circle. The string makes an angle of 30° with the vertical. Find the tension in the string & the speed of the ball. Solution 6.2.6 Component - x : Component - y :

  28. CIRCULAR MOTION IN VERTICAL PLANE (NON-UNIFORM CIRCULAR MOTION) Case 1 : A ball is attached to a string & moves in a vertical circle. FIGURE 6.2.9 FIGURE 6.2.10 At the top of the circle ( point A ) : both T & mg are directed downwards At the bottom of the circle ( point B ) : T & mg point in opposite direction (T is minimum) (T is maximum)

  29. Circular motion in vertical plane (non-uniform circular motion) Circular motion is possible as long as the cord remain taut, thus there is a critical (minimum) speed to be maintained. If the rope is sagging, T = 0 , thus : T=0

  30. Circular motion in vertical plane (non-uniform circular motion) Example 6.2.7 constant speed A 1.2 kg rock is tied to the end of a 90 cm length of string. The rock is then is whirled in a vertical circle at a constant speed of 8 m/s. What are the tensions in the string at the top and bottom of the circle ? Solution 6.2.7 m = 1.2 kg , r = 90 cm , v = 8 m/s Top Bottom

  31. Circular motion in vertical plane (non-uniform circular motion) Example 6.2.8 not constant speed A 2 kg ball is tied to the end of a 80 cm length of string. The ball is then is whirled in a vertical circle and has a velocity of 5 m/s at the top of the circle. a) What is the tension in the string at that instant ? b) What is the minimum speed at the top necessary to maintain circular motion ? Solution 6.2.8 m = 2 kg , r = 80 cm , v = 5 m/s at the top. a) b) Top

  32. Circular motion in vertical plane (non-uniform circular motion) Example 6.2.9 A rope is attached to a bucket of water and the bucket is then rotated in a vertical circle of 0.70 m radius. Calculate the minimum speed of the bucket of water such that the water will not spill out. Solution 6.2.9 The water will not spill out if the T=0, thus :

  33. Circular motion in vertical plane (non-uniform circular motion) Case 2 : Roller coaster on a circular track / Ferris wheel Top of the circle : Bottom of the circle : R R mg mg A minimum velocity (when R = 0) is required in order to keep a roller coaster car on a circular track.

  34. Circular motion in vertical plane (non-uniform circular motion) Example 6.2.10 What minimum speed must a roller coaster be traveling when upside down at the top of a circle (refer to the figure) if the passengers are not to fall out ? Assume R = 8.0 m. Solution 6.2.10 r = 8.0 m Top For vmin , R = 0

  35. v v Example 6.2.11 : A rider on a Ferris wheel moves in a vertical circle of radius, r = 8 m at constant speed, v as shown in Figure 6.13. If the time taken to makes one rotation is 10 s and the mass of the rider is 60 kg, Calculate the normal force exerted on the rider a. at the top of the circle, b. at the bottom of the circle. (Given g = 9.81 m s-2) Figure 6.13

  36. Solution 6.2.11: a. The constant speed of the rider is The free body diagram of the rider at the top of the circle :

  37. Solution 6.2.11: b. The free body diagram of the rider at the bottom of the circle :

  38. Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion EXERCISES 6.1 • A particle is moving on a circular path of radius 0.5 m at a constant speed of 10 m/s. Calculate the time taken to complete 20 revolutions. (t = 6.28 s) 2. Two wheels of a machine are connected by a transmission belt. The radius of the first wheel r1 = 0.5 m, the radius of the second wheel r2 = 0.125 m. The frequency of the bigger wheel equals 3.5 Hz. What is the frequency of the smaller wheel ? (f2 = 14 Hz)

  39. Motion Characterictics For Circular Motion 6.1 Uniform Circular Motion 3. The astronaut orbiting the Earth is preparing to dock with Westar VI satellite. The satellite is in a circular orbit 600 km above the Earth’s surface, where the free fall acceleration is 8.21 m s2. Take the radius of the Earth as 6400 km. Determine : a. the speed of the satellite, b. the time interval required to complete one orbit around the Earth. ANS. : 7581 m s1; 5802 s 4. A pendulum bob of mass 1 kg is attached to a string 1 m long and made to revolve in a horizontal circle of the radius 60 cm. Calculate the period of the motion and the tension of the string. Tension , T = 12.25 N

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