Centripetal Acceleration – acceleration towards the center of a circle. - PowerPoint PPT Presentation

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Centripetal Acceleration – acceleration towards the center of a circle.
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Centripetal Acceleration – acceleration towards the center of a circle.

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  1. Circular Motion and Gravitation Centripetal Acceleration – acceleration towards the center of a circle. • a.k.a. Radial Acceleration (aR)

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

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

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

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

  6. Formulas aR = v2v = 2pr r T T = 1 v = 2prf f

  7. Centripetal Acceleration: Ex. 1 A 150-g 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)

  8. Centripetal Acceleration: Ex. 2 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. • Convert the answer to g’s. (Ans: 2.72 X 10-3 m/s2, 2.78 X 10-4 g )

  9. Jupiter is about 778 X 106 km from the sun. It takes 4332.6 days to orbit the sun. • Calculate the circumference of Jupiter’s . orbit. (4.89 X 1012 m) • Calculate Jupiter’s period in seconds. (3.74 X 108 s) • Calculate Jupiter’s orbital speed. (1.30 X 104 m/s) • Calculate Jupiter’s centripetal acceleration towards the sun. (2.18 X 10-4 m/s2)

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

  11. “Centrifugal Force?” • Doesn’t exist • “apparent outward force” • When you let the string go, the ball will continue in a straight line path. No new acceleration involved. • Water in swinging cup example

  12. Direction water wants to go Centripetal Force of string

  13. Circular 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 second. • Calculate the circumference of the circle (3.77 m) • Calculate the linear speed (7.54 m/s) • Calculate the centripetal force in the string (14.2 N)

  14. An electron orbits the nucleus with a radius of 0.5 X 10-10 m. The electron has a mass of 9 X 10-31 kg and a speed of 2.3 X 106 m/s. • Calculate the centripetal force on the electron. (9.52 X 10-8 N) • Calculate the frequency of an electron. (7.32 X 1015 Hz) • Convert the velocity to miles/s. (1429 miles/s) • What provides the centripetal force?

  15. Circular Motion: Example 2 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)

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

  17. Mass = 10 kg r = 50 cm v = 3 m/s What angle will the hammer take with the horizontal? FH q mg

  18. Let’s resolve the FR vector into it’s components: FHx = Fhcosq FHx = mv2/r FHcosq = mv2/r SFy = 0 (the hammer is not rising or falling) 0 = FHsinq – mg Two Equations: Two Unknowns What angle will the hammer take with the horizontal? FR q mg

  19. FHcosq = mv2/rFHsinq = mg FH = mv2 rcosq mv2 sinq= mg r cosq sinq= gr tan q = gr q = 28.6o cosq v2 v2

  20. A 0.0050 kg walnut is swung in a radius of 50.0 cm. The walnut makes 2.50 revolutions per second. • Calculate the linear speed of the walnut. (7.85 m/s) • Draw a free-body diagram of the walnut. • Calculate the centripetal force needed to keep it in a circle. (0.616 N) • Calculate the force of tension and the angle the string makes with the horizontal. (0.618 N, 4.5o)

  21. Circular Motion: Example 3 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 v = 3.28 m/s mg FT

  22. What is the tension in the cord at the bottom of the arc if the ball moves at the minimum speed? (v = 3.28 m/s) FT = 2.94 N FT mg

  23. A rollercoaster vertical loop has a radius of 20.0 m. Assume the coaster train has a mass of 3,000 kg. • Calculate the minimum speed the coaster needs to have to make the loop. (14.0 m/s) • Calculate the normal force the tracks provide to the train at the bottom of the curve if the train is travelling at 25.0 m/s. (123,150 N) • Calculate the normal force the tracks provide at the top of the curve if the train is travelling at 25.0 m/s. (64, 350 N)

  24. The ferris wheel at Knoebels has a radius of 16.8 m and travels at a speed of 3.52 m/s. • Calculate the frequency and period (0.033 Hz, 30 s) • Calculate the normal force that the seat provides to a 56.0 kg rider at the top. (507 N) • Calculate the normal force that the seat provides to a 56.0 kg rider at the bottom. (590 N)

  25. A car travels over a round hill (radius = 50.0 m). • Calculate the maximum speed at which the car can take the hill. (22.0 m/s) • Calculate the normal force on a 1000.0 kg car if it is travelling over the hill at 10.0 m/s. (7.80 X 103 N)

  26. Car Rounding a Turn • Friction provides centripetal force • Use (ms). 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.

  27. Car Rounding a Turn: Example 1 A 1000-kg car rounds a curve (r=50 m) at a speed of 14 m/s. • Calculate the centripetal force needed to keep the car on the road • Calculate if the car will skid if the road is dry and ms = 0.60 • Calculate if the car will skid if the road is icy and ms = 0.25 FN Ffr = Fc mg

  28. Car Rounding a Turn: Example 2 • Calculate the centripetal force needed to keep the car on the road (40,000 N ) • Calculate the coefficient of static friction (0.27) • Calculate the maximum speed a 1000 kg Cube car can take the turn. (20 m/s) A 15,000-kg truck can safely round a 150 m curve at a speed of 20 m/s.

  29. The Rotor The Rotor at an amusement park has a radius of 7.0 m and makes 30 rev/min. • Calculate the speed of the rotor. ( 22.0 m/s) • Draw a free body diagram of a person in the rotor. What causes the FN? • Calculate the coefficient of static friction between the person and the wall. (0.14)

  30. Banked Curves • Banked to reduce the reliance on friction • Part of the Normal Force now contributes to the centripetal force

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

  32. 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 q mg q

  33. 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

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

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

  36. A 2,000-kg Nascar car rounds a 300 m radius turn at 200 miles/hr. • Convert the speed to m/s. (89.4 m/s) • What angle should the road be banked so that no friction is required? (70o) • Suppose a track is only banked at 35.0o, calculate the maximum speed that a car can take the turn. (45.3 m/s, 101 mph) • Looking at the formula for banking angle, how could a track designer decrease that angle?

  37. Weightlessness • True weightlessness exists only very far from planets • “Apparent weightlessness” can be achieved on earth

  38. Elevator at Constant Velocity a= 0 SF = FN – mg 0 = FN – mg FN = mg Suppose Chewbacca has a mass of 102 kg: FN = mg = (102kg)(9.8m/s2) FN = 1000 N FN mg a is zero

  39. Elevator Accelerating Upward a = 4.9 m/s2 SF = FN – mg ma = FN – mg FN = ma + mg FN = m(a + g) FN=(102kg)(4.9m/s2+9.8 m/s2) FN = 1500 N FN mg a is upward

  40. Elevator Accelerating Downward a = 4.9 m/s2 SF = mg - FN ma = mg - FN FN = mg - ma FN = m(g - a) FN=(102kg)(9.8m/s2 – 4.9 m/s2) FN = 500 N FN mg a is down

  41. At what acceleration will he feel weightless? FN = 0 SF = mg - FN ma = mg -FN ma = mg - 0 ma = mg a = 9.8 m/s2 Apparent weightlessness occurs if a > g FN mg

  42. Calculate the apparent weight of a 56.0 kg man in an elevator if the elevator is: • Accelerating upwards at 2.00 m/s2. (661 N) • Accelerating downwards at 2.00 m/s2 (436 N) • Accelerating downwards at 9.80 m/s2 (0 N) • Accelerating sideways at 9.80 m/s2. (549 N)

  43. Other examples of apparent weightlessness Even when you are running, you fell weightless between strides.

  44. Why don’t satellites fall back onto the earth? • Speed • They are “falling” due to the pull of gravity • Can feel “weightless” (just like in the elevator)

  45. Gravitation Is gravity caused by the earth’s rotation? Will a man down here fall off if the earth stops rotating?

  46. Gravitation Newton’s Law of Universal Gravitation • Every object in the universe is attracted to every other object. (based on mass) • The force drops off with the distance squared. (As distance increases, the force of gravity drops very quickly)