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Circular Motion

Circular Motion. Chapter 7 (already). We commonly use Cartesian or rectangular coordinate system where (x, y) identifies a point in two dimensions. We could equivalently use polar coordinates to identify a point (r, θ ). Polar Coordinates. Given r and θ , X = r·cos θ Y = r·sin θ

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Circular Motion

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  1. Circular Motion Chapter 7 (already)

  2. We commonly use Cartesian or rectangular coordinate system where (x, y) identifies a point in two dimensions. We could equivalently use polar coordinates to identify a point (r, θ) Polar Coordinates

  3. Given r and θ, X = r·cosθ Y = r·sinθ This converts from a polar coordinate to a rectangular coordinate Given x and y, r = (x2 + y2) ½ θ = tan -1 (y/x) This converts from a rectangular coordinate to a polar coordinate Converting Between Coordinate Systems

  4. Imagine a point moving along the circle at constant speed. Which variables change and which stay the same? Circular Motion

  5. The angle, θ, changes but the radius does not change. The coordinates x- and y- both change. The speed does not change but the velocity does change (why?) Circular Motion

  6. Angles may be measured in degrees or in radians. 360° = 2π The length of an arc subtending a circle of radius r is given by S = r·θ Angle Measure

  7. Examples • A spectator is standing at the center of a circular running track and observe a runner start practice 256 m due east of her position. The runner runs on the track in a counter clockwise direction to the finish line, located due north of the observer. What was the distance of the run? • A sailor sights a distant ship and finds that it subtends an angle of 1.15°. He knows from the shipping charts that the tanker is 150 m in length. How far away is the tanker?

  8. Angular Speed / Velocity • Angular speed, ω, is defined as the rate of change of θ. ω = Δθ/ Δt [radians/sec] and θ = ωt Then s = rθ becomes s = r (ωt) • And v = r ω

  9. Speeds in Circular Motion • Angular velocity is denoted with ω, omega, and the direction is at a right angle to the plane of the circular motion (use right hand rule) • Tangential velocity is denoted with v, velocity.

  10. Radians~! • Note that ω is measured in radians/sec. • The formulas involving ω are true only if θ (angular displacement) is measured in radians since only then is s = rθ a true statement.

  11. Period and Frequency • The Period, T, is the time necessary for one complete cycle. Period, T, is measured in [seconds]. • The frequency, f, is the number of cycles that happen in one second. Frequency is measured in [cycles/second] or [Hertz] • T = 1/f • Angular speed, ω = 2π/T = 2πf

  12. Examples • A merry go round at its constant operational speed makes one complete rotation in 45 seconds. Two children are on horses, on at 3.0 m from the center and one 6.0 m from the center. Find the angular speed and the tangential speed for each child. • A CD rotates in the player at a constant speed of 200 rpm. What are the CD’s a) frequency and b) Period of revolution?

  13. Centripetal Acceleration • An object in Uniform Circular Motion (constant speed) experiences a Centripetal Force ( and therefore a centripetal acceleration) that is pointed towards the center. • ac = v2/r = (rω)2/r = rω2 • Fc = m(v2/r) = mrω2

  14. Examples • A lab centrifuge operates at a rotational speed of 12,000 rpm. A) What is the magnitude of the centripetal acceleration of a red blood cell at a radial distance of 8.0 cm from the centrifuge’s axis of rotation? Compare this with g. • A ball attached to a string is swung with uniform motion in a horizontal circle over a person’s head. If the string breaks, what will be the trajectory of the ball?

  15. More Examples… • A car approaches a level circular curve with radius of 45.0 m. If the concrete pavement is dry, what is the maximum (constant) speed at which the car can negotiate the curve? • Two masses, m1 = 2.5 kg and m2 = 3.5 kg are connected by light strings and are in uniform circular motion on a horizontal frictionless surface, with r1 = 1.0m and r2 = 1.3m. The forces acting an the masses are T1 and T2 respectively. Find the centripetal acceleration and the tangential speed of mass 2 and mass 1.

  16. Angular Acceleration • When circular motion isn’t uniform, angular velocity changes. • α , alpha, is used to denote angular acceleration. We consider only the case of constant angular acceleration so • αav = α = Δω/t • Then Δω = αt and ωf = ωi + αt

  17. Tangential Acceleration • If the rate of angular velocity changes, then the object undergoes angular acceleration and tangential acceleration: • at = Δv/t = Δ(rω)/t = rα • at = rα

  18. X = vav t Vav = (vf + vi)/2 Vf = vi + at Xf = xi + vit + ½ at2 Vf2 = vi2 + 2aΔx θ = ωavt ωav = (ωf+ωi)/2 ωf = ωi + αt θf =θi + ωit + ½ αt2 ωf2 = ωi2 + 2αΔθ Linear and Rotational Kinematics

  19. Examples • A CD accelerates uniformly from rest to its operational speed of 500 rpm in 3.0 sec. What is the angular acceleration of the CD during this time? If the CD comes to a stop in 4.0 sec, what is the angular acceleration during that part of the motion? • A microwave oven has a 30 cm rotating plate. The plate accelerates from rest to a uniform rate of 0.87 rad/s2 for 0.50 sec before reaching its constant operational speed. How many revolutions does the plate make before reaching its operational speed? What are the operational angular speed and tangential speed at its rim?

  20. Newton’s Law of Gravitation • Newton wondered whether the force pulling apples to the ground was the same force pulling the moon towards the Earth. • He recognized that, although the moon orbits the Earth in a circular path, that the moon really is ‘falling’ towards the Earth

  21. Newton’s Cannon Thought Experiment

  22. Gravitational Force

  23. Gravitational Force • Is always attractive • Is proportional to the product of the masses • Is inversely proportional to the square of the distance between the masses • G = 6.67 X 10 -11 Nm2/kg2

  24. Examples • Compute the gravitational force between Earth and the Moon. Mm = 7.4 x 1022 kg, ME = 6.0 X 1024 kg, rEM = 3.8 X 108 m. • Find the acceleration due to gravity acting on an object with mass Mo near the Earth’s surface. rE = 6.4 X 106m

  25. Satellite Example • Some communication and weather satellites are launched into circular orbits above the Earth’s equator so they are synchronous with Earth’s rotation. At what altitude are these satellites? RE = 6.4 X 106 km; ME = 6.0 X 1024 kg, T = 365 days

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