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Motion in a Plane

Motion in a Plane. Chapter 8. Centripetal Acceleration. Centripetal Acceleration – acceleration that points towards the center of a circle. Also called Radial Acceleration (a R ). v. Ball rolling in a straight line (inertia). v. Same ball, hooked to a string. a R = v 2

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Motion in a Plane

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  1. Motion in a Plane Chapter 8

  2. Centripetal Acceleration • Centripetal Acceleration – acceleration that points towards the center of a circle. • Also called Radial Acceleration (aR)

  3. v Ball rolling in a straight line (inertia) v Same ball, hooked to a string aR = v2 r aR aR v

  4. If you are on a carousel at constant speed, are you experiencing acceleration?

  5. If you twirl a yo-yo and let go of the string, what way will it fly?

  6. Period and Frequency Period (T) • Time required for one complete (360o) revolution • Measured in seconds Frequency • Number of revolutions per second • Measured in rev/s or Hertz (Hz) T = 1 f

  7. Formulas v = 2pr v = wr T aR = v2 a = w2r r

  8. A 150-kg ball is twirled at the end of a 0.600 m string. It makes 2.00 revolutions per second. Find the period, velocity, and acceleration. (0.500 s, 7.54 m/s, 94.8 m/s2)

  9. The moon has a radius with the earth of about 384,000 km and a period of 27.3 days. • Calculate the acceleration of the moon toward the earth. (2.72 X 10-3 m/s2) • Calculate the previous answer in “g’s” (2.78 X 10-4 g)

  10. Centripetal Force – the “center seeking” force that pulls an object in a circular path. • Yo-yo • Planets • Merry-go-round • Car rounding a curve

  11. Centrifugal Force A word about Centrifugal Force • Doesn’t really exist. • “apparent outward force” • Water in swinging cup example Direction water wants to go Centripetal Force of string

  12. Centripetal Motion SF = maR = mv2 r A 0.150 kg yo-yo is attached to a 0.600 m string and twirled at 2 revolutions per minute. • Calculate the velocity in m/s () • Calculate the centripetal force in the string (14.2 N)

  13. Thor’s Hammer (mjolnir) has a mass of 10 kg and the handle and loop have a length of 50 cm. If he can swing the hammer at a speed of 3 m/s, what force is exerted on Thor’s hands? (Ans: 180 N)

  14. Can Thor swing his hammer so that it is perfectly parallel to the ground? FR

  15. What angle will the hammer take with the horizontal? • Let’s resolve the FR vector into it’s components: • FRx = FRsinq • FRy = FRcosq • SFy = 0 (the hammer is not rising or falling) • SFy = 0 = FRcosq – mg • FRcosq = mg • cosq = mg/FR • = 57o How about if he swings faster? q FR mg

  16. A father places a 20.0 kg child on a 5.00 kg wagon and twirls her in a circle with a 2.00 m rope of tension 100 N. How many rpms does the wagon make (w)? (14 rpm)

  17. A 0.150 kg ball is swung on a 1.10-m string in a vertical circle. What minimum speed must it have at the top of the circle to keep moving in a circle? At the top of the circle, both the weight and the tension in the string contribute to the centripetal force SF = FT + mg mg FT

  18. SF = FT + mg FR = FT + mg mv2 = FT + mg r (tricky part: assume FT = 0, just as the cord goes slack, but before the ball falls) mv2 = mg r v2 = gr v = 3.28 m/s

  19. Note: this equation is also the minumum velocity for orbit of a satellite v = \/rg

  20. What is the tension in the cord at the bottom of the arc if the ball moves at twice the minimum speed? (v = 6.56 m/s) At the bottom of the circle, the weight opposes the centripetal force. SF = FT – mg mv2 = FT - mg r FT = mv2 + mg r FT = 7.34 N FT mg

  21. Car Rounding a Turn • Friction provides the centripetal force • Use the coefficient of static friction (ms). The wheels are turning, not sliding, across the surface • Wheel lock = kinetic friction takes over. mk is always less than ms, so the car is much more likely to skid.

  22. A 1000-kg car rounds a curve (r=50 m) at a speed of 14 m/s. Will the car skid if the road is dry and ms=0.60? Let’s first solve for the Normal Force FN = mg = (1000 kg)(9.8 m/s2) FN = 9800 N FN Ffr = FR mg

  23. SFx = Ffr FR = Ffr mv2 = msFN r (1000 kg)(14m/s)2 = (0.60)(9800 N) (50 m) 3920 N < 5800 N The car will make it. 3920 N are required, and the frcition provides 5800 N.

  24. Will the car make it if it is icy and the ms = 0.25 SFx = Ffr FR = Ffr mv2 = msFN r (1000 kg)(14m/s)2 = (0.25)(9800 N) (50 m) 3920 N > 2450 N The car will not make it. 3920 N are required, and the friction only provides 2450 N.

  25. What is the maximum speed a 1500 kg car can take a flat curve with a radius of 50 m (ms = 0.80)

  26. BANKED CURVES • Banked to reduce the reliance on friction • Part of the Normal Force now contributes to the centripetal force

  27. FR = Ffr + FNsinq (ideally, we bank the road so that no friction is required: Ffr = 0)

  28. Banked Curves: Example 1 A 1000-kg car rounds a 50 m radius turn at 14 m/s. What angle should the road be banked so that no friction is required? FN FN = mgcosq q mg q

  29. Now we will simply work with the Normal Force to find the component that points to the center of the circle First consider the y forces. SFy = FNcosq - mg Since the car does not move up or down: SFy = 0 0 = FNcosq – mg FNcosq = mg FN = mg/cosq FNcosq FN q FNsinq q mg q

  30. mv2= FNsinq r mv2= mgsinq r cosq v2= gtanq r v2 = gtanq r v2 = tanq gr

  31. tan q = (14 m/s)2 = 0.40 (50 m)(9.8m/s2) q = 22o

  32. Fred Flintstone places a 1.00 kg rock in a 1.00 m long sling. The vine breaks at a tension of 200 N. • Calculate the angle below the horizontal plane that the rock will take. (2.81o) • Calculate the maximum linear velocity the rock can twirl. (14.1 m/s) • Calculate the angular velocity in rpm’s. (135 rpm)

  33. Circular Orbits • Orbits are freefall (not true weightlessness) • Orbital velocity must match the weight mg = mv2 r g = v2 v = √ gr r

  34. A satellite wishes to orbits at a height of 200 miles above the earth’s surface. • Calculate the height above the center of the earth if Rearth = 6.37 X 106 m. (6.69 X 106 m) • Calculate the orbital velocity. (8098 m/s) • Calculate the period in minutes. (86.5 min)

  35. Review of Angular Kinematics A motor spins a 2.0 kg block on an 80.0 cm arm at 200 rpm. The coefficient of kinetic friction is 0.60. • Draw a free body diagram of the block. • Calculate the tangential acceleration of the block (due to friction). (-5.88 m/s2) • Calculate the angular acceleration. (-7.35 rad/s2) • Calculate the time until the block comes to a rest. (2.8 s) • Calculate the number of revolutions. (4.7 rev)

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