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Circular Motion

Circular Motion. Chapter 7.3. Motion & Forces. What you already know: Velocity – a measure of the change in displacement (distance with direction. Mass – A measure of the amount of matter an object contains. Acceleration – A measure of the change in velocity over change in time.

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Circular Motion

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  1. Circular Motion Chapter 7.3

  2. Motion & Forces • What you already know: • Velocity – a measure of the change in displacement (distance with direction. • Mass – A measure of the amount of matter an object contains. • Acceleration – A measure of the change in velocity over change in time. • Force – A push or pull that is equal to the mass of the object multiplied by its acceleration (F = ma).

  3. Uniform Circular Motion • Uniform circular motion is defined by any object that is moving at constant speed in a circular path. • Determining Speed: • The distance an object moving in a circular path is equal to the circumference (C = 2r). • The time it takes an object to complete one revolution is called the period (T). • It then follows that the speed of an object moving in a circular path can be determined by: v = d/t = C/T = 2r/T

  4. Uniform Circular Motion • If an object is moving at constant speed in a circular path, can it be accelerating? • Yes • Although the speed may be constant, the direction is changing. • If direction is changing over time, then the velocity must be changing. • Acceleration is the change in velocity over time (a = v/t). • If the velocity is changing over time, then the object must be accelerating.

  5. r2  r1 r Instantaneous v (arrived at by letting  get smaller and smaller) is tangent to the circle. v2 r2 v1  r1 Circular Motion – Instantaneous Velocity • Note that the velocity vector is at right angles to the position vector and tangent to the circle at any given point along the circle. v = r/t

  6. v v2  v1 Circular Motion – Centripetal Acceleration (ac) • The acceleration of an object moving in a circular path always points towards the center of the circle, and is perpendicular to the velocity vector. v2 v a v1 a = v/t r

  7. Centripetal Acceleration • The angle between r1 and r2 is the same as the angle between v1 and v2. • Therefore, the triangles these vectors make are similar such that: r/r = v/v • If you divide both sides by t: r/(t r) = v/(t·v) • Since : • r/t = v and v/t = a • Hence: v/r = a/v and ac = v2/r

  8. Centripetal Acceleration • An alternative representation for centripetal acceleration can be derived using the circumference and period of revolution. • d = 2πr • v = d/T = 2πr/T • Substituting into ac = v2/r • ac = (2πr/T)2/r • ac = 4π2r/T2

  9. Circular Motion – Centripetal Force • To make an object move in a circular path, an external force must act perpendicular or at right angles to its direction of motion. • This force is called centripetal force. Instantaneous direction of velocity Direction of force required to make object move in a circular path(towards the center)

  10. Centripetal Force • Centripetal force is affected by: • The mass of the object (m). • The speed of the object around the circle (v). • The radius of the circle (r). • Using Newton’s 2nd Law of Motion (F = ma), centripetal force is mathematically represented as follows: F = mv2 r Note: Centripetal force is an unbalanced “net” force

  11. Force Force Force How the Factors Affect Centripetal Motion • Which graph shows the proper relationship with respect to force: • Force vs. Mass. • Force vs. Speed. • Force vs. Radius. Speed Radius Mass

  12. Objects that travel in circular paths.What is the cause of the force? • The Earth – Sun System: • Gravity. • A racecar traveling around a turn on the racetrack: • Friction. • An athlete throwing the hammer: • Tension in the cable attached to the hammer.

  13. The path of objects. • If the centripetal force were suddenly removed from an object moving in a circular path, what trajectory (or path) would it follow?

  14. (a) (b) (c) Which Path? • Why? • Because of Inertia – An object in motion wants to remain in motion at constant speed in a straight line. • If the unbalanced centripetal force is removed, the object will continue in a straight path.

  15. Example #1: • A 1.5 kg cart moves in a circular path of 1.3 meter radius at a constant speed of 2.0 m/s. • Determine the magnitude of the centripetal acceleration. (ac = v2/r ) • Determine the magnitude of the centripetal force. (Fc = mac ) • Determine the period. (T = d/v = c/v) • Get out your calculators!

  16. Example #1: (cont.) • Centripetal Acceleration: ac = v2/r = (2.0 m/s)2/(1.3 m) = 3.1 m/s2 • Centripetal Force: Fc = mac = (1.5 kg)(3.1 m/s2) = 4.6 N • Period: T = c/v = 2r/v = 2(1.3 m)/(2.0 m/s) = 4.08 s

  17. Example #2: Roller Coaster The centripetal force is the force keeping the object in a circle. It is either going to be Fg, FN, or some combination. m = 1kg r = 10m v = 9.9m/s ac = ??m/s^2 Fc = ??N Here Fg = Fc. Fg IS Fc. There is no Normal force. You feel weightless. Fg = mg = Fc Fc = N Fg = mg Fg = mg Here Fg = Fc but point in opposite directions. The Normal force has to oppose Fg and provide Fc. You feel 2 “g”s. Here Fc = FN and points to the center. Fg points straight down but if you are accelerating with it you only feel Fc. (And feel confused if you have time.) Fg = mg

  18. Example #2: Roller Coaster The track of the roller coaster provides the centripetal force via the Normal force, FN

  19. Example #3: Ball on a rope A tetherball of mass 0.28kg is moving at 8m/sec around a pole. (For this example, we will ignore gravity.) The length of the rope is 2.4m. What is the period of motion? What is the centripetal force? Ignoring gravity, what is the tension in the rope? (For this example, we will ignore gravity.)

  20. Example #3: Ball on a rope Period T = d/v = 2Pr/v = 2P(2.4m)/(8m/s) = 1.9s (This is not actually needed to solve the problem.) What is the centripetal force? Fc = mac = mv2/r = 0.28kg * (4m/s) 2/2.4m What is the tension in the rope? The tension = Fc

  21. Example #4: Car on a curve A 1000kg car is traveling at 22m/s on a curve of radius 100m. What provides the centripetal force required to keep the car from sliding off the road? What is the magnitude of the centripetal force required to keep the car on the road? What is the coefficient of friction between the car and the road? Fc = mv2/r = 1000kg *(22m/s)2/100m = 4840N. What is the coefficient of friction? U = Ff/Fn

  22. Example #4: Car on a curve What provides the centripetal force required to keep the car from sliding off the road? Friction What is the magnitude of the centripetal force required to keep the car on the road? Fc = mv2/r = 1000kg *(22m/s)2/100m = 4840N. What is the coefficient of friction between the car and the road? m = Ff/FN = Fc/FN = Fc/(mg) = 4840N/(1000kg * 9.81m/s2) = 0.5 What is the coefficient of friction? U = Ff/Fn

  23. Example #5: Moon around the Earth • The moon has a mass of 7.35 × 1022 kg and orbits at an average distance of 3.84 × 108 m in 27.32 days (2.36 × 106sec). • What centripetal acceleration does the moon experience and what is the required centripetal force? • ac = 0.0027m/s2 • Fc = mac = (7.35 × 1022 kg)(0.0027m/s2) = 1.98 × 1020N

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