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Circular Motion

Circular Motion. Notes (Ref p234-239 HRW). Uniform Circular Motion. Definition Motion exhibited by a body following a circular path at constant speed. Frequency (f) # of cycles, rotations or revolutions per unit time. SI units are cycles/sec or hertz (Hz) Period (T)

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Circular Motion

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  1. Circular Motion Notes (Ref p234-239 HRW)

  2. Uniform Circular Motion • Definition • Motion exhibited by a body following a circular path at constant speed. • Frequency (f) • # of cycles, rotations or revolutions per unit time. • SI units are cycles/sec or hertz (Hz) • Period (T) • Time taken for 1 complete rotational cycle or revolution. • SI units are seconds. • Frequency-period relationship • f = 1/T

  3. Example • Q: A merry-go-round takes 5 seconds to make 1 complete revolution. • What is the frequency (f) of rotation? • A: Given T, find f = 1/T • T = 5 seconds • f = 1/T = 1/5 • f = 0.2 rev/sec (or cycles/sec)

  4. Example 2 • Q: A CD rotates at 500 rpm near the center. • What is the period (T) of the rotation? • A: Given f, find T = 1/f • f = 500 rpm • T = 1/f = 1/500 = 0.002 minutes • T = 0.12 sec (= 0.002 x 60) • T = 120 milliseconds (ms)

  5. Linear (tangential/orbital) speed • v = distance traveled around a circle per unit time • v = circumference/period • v = 2(radius)/period • v = 2r/T • Directionof velocity vector is tangential (perpendicular) to radius • Units are m/s

  6. Example • What is the orbital speed (v) of a tennis ball that is being swung on the end of a 1.5 m (r) string and takes 1.25 seconds (T) for a complete revolution? • Given: r = 1.5m, T = 1.25 s • Solve v = 2(radius)/period • v = 2 (1.5)/1.25 • v = 7.54 m/s

  7. Centripetal Acceleration (ac) • ac = linear velocity2 radius of rotation • ac = v2/r • Directionof acceleration vector is inward along radius (center seeking). • Units are m/s2

  8. Example • What is the centripetal acceleration (ac) of the tennis ball in the previous example? • Solve ac = v2/r (v = 7.54 m/s, r=1.5) • ac = 7.542/1.5 • ac = 37.89 m/s2 • Per Newton, if there is acceleration there is…

  9. Centripetal force (Fc) • Fc = mass x ac (Newton’s 2nd Law) • Fc = mac = mv2/r • Direction of force vector is inward along radius (center-seeking). • Units are Newtons (N)

  10. Example – Merry Go Round • A merry-go-round has a diameter of 10m and a period of 10 seconds. There are 2 circles of horses, at 3 m radius and at the outer edge. If a rider has a mass of 50 kg, calculate: • Speed of the horses at the edge and 3 m • Centripetal acceleration of each horse • Centripetal force on each horse

  11. Solution – Merry Go Round • v(edge) = 2π r/T = 2π(5)/10 = 3.14 m/s • v(3 m) = 2π r/T = 2π(3)/10 = 1.88 m/s • ac (edge) = v2/r = 3.142/5 = 1.972 m/s2 • ac (3m) = v^2/r = 1.882/3 = 1.178 m/s2 • Fc(edge) = mac = 50(1.972) = 98.6N (inward) • Fc (3m) = mac = 50(1.178) = 58.9N (inward)

  12. Circular Motion Summary

  13. Torque () • Turning force or “moment” of a force • Product of the perpendicular component of a force applied at a distance from the axis of rotation (pivot or fulcrum) of an object. •  = Force┴ x lever arm •  = F┴ x L • Units are N-m. • Convention direction • Counterclockwise (ccw) = +ve • Clockwise (cw) = -ve pivot

  14. Torque – Change angle and lever arm

  15. Models • 2 models for analysis • Single pivot (see-saw) • Double pivot (plank, stretcher, bridge)

  16. d2 d1 W2 W1 Single Pivot (Seesaw) Model • To achieve balance • total ccw torque = total cw torque • total means all objects contributing to the respective torques. • ccw = cw

  17. Center of Gravity • The point on an object through which all the mass is deemed to be acting. • The object is assumed to be uniform in structure • Important concept when considering double pivot models

  18. Double Pivot (Bridge/Plank) Model • For balance, • the sum of the CW torques = the sum of the CCW torques mass FL FR Mass of box Mass of plank CCW CW

  19. Moment of Inertia or Rotational Inertia (I) Symbol • Defn: The tendency of a body rotating about a fixed axis to resist change in rotational motion. • Rotational equivalent of mass. • Dependent on the distribution of the mass of the object (where is the mass located?) • Formula • I = “shape” constant x mass x linear dimension (squared) • Where shape is a multiplier dependent upon the object shape • I = kmr2– see text for models • Units are kg-m2

  20. Rotational Inertia Models

  21. Example • What is the moment of inertia of of a disk-shaped wheel that has a mass of 15 kg and a diameter of 2.8 m? • Solve: Refer to formulas on p 103 • Model = cylinder > I = ½ mr2 • I = ½ (15)(2.8/2)2 • I = 14.7 kg.m2

  22. Angular Momentum (L) Symbol • A measure of the how much a rotating body resists stopping. • Rotational equivalent of linear momentum • L = mass x linear velocity x radius • L = mvr

  23. Example of Angular Momentum • Jupiter orbits the sun with a speed of 2079 m/s and orbit of 71,398,000 m from the sun. If Jupiter’s mass is 1.9 x 1027 kg, what is its angular momentum (L)? • Solve: L = mvr • L = 1.9 x 1027 x 2,079 x 71,398,000 • L = 2.82 x 1038kg.m2/s

  24. Conservation of Angular Momentum • Linear momentum is conserved, so… • Angular momentum is conserved across an event • Vizualize and ice skater doing pirouettes… • Skater has arms out (larger radius) – smaller spinning velocity – arms in (smaller radius) – greater spinning velocity. • L before = Lafter • (mvr)before = (mvr)after • v1r1 = v2r2 (mass (m) cancels)

  25. Skater Example of Cons of Angular Momentum Before After v * R V * r v*r product does not change across an event

  26. Example of Conservation of Angular Momentum • A physics student is spinning around in a chair @ 1.5 m/s with his arms stretched out 0.6 m from the center of his body, holding in each hand a 2 kg mass. If he pulls in his arms to 0.2 m from the center of his body, how fast does he now spin? • Solve: Conservation of Angular momentum • Angular momentum (before) = angular momentum (after) • v1 x r1 = v2 x r2 • 1.5 x 0.6 = v2 x 0.2 • v2 = 0.9/0.2 • v2 = 4.5 m/s

  27. Linear vs Rotational Model

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