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 (a R )

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

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 (360 o ) revolution – Measured in seconds Frequency – Number of revolutions per second – Measured in rev/s or Hertz (Hz) T = 1 f

7. Formulas v = 2  rv =  r T a R = v 2 a =   r 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/s 2 )

9. The moon has a radius with the earth of about 384,000 km and a period of 27.3 days. A.Calculate the acceleration of the moon toward the earth. (2.72 X 10 -3 m/s 2 ) B.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

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

16. Centripetal Motion  F = ma R = mv 2 r A 0.150 kg yo-yo is attached to a 0.600 m string and twirled at 2 revolutions per minute. a.Calculate the velocity in m/s () b.Calculate the centripetal force in the string (14.2 N)

17. 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)

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

19. What angle will the hammer take with the horizontal? FRFR mg  Let’s resolve the F R vector into it’s components: F Rx = F R sin  F Ry = F R cos   F y = 0 (the hammer is not rising or falling)  F y = 0 = F R cos  – mg F R cos  = mg cos  = mg/F R  = 57o How about if he swings faster?

20. 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 (  )? (14 rpm)

21. 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  F = F T + mg mgFTFT

22.  F = F T + mg F R = F T + mg mv 2 = F T + mg r (tricky part: assume F T = 0, just as the cord goes slack, but before the ball falls) mv 2 = mg r v 2 = gr v = 3.28 m/s

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

24. 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) mg FTFT At the bottom of the circle, the weight opposes the centripetal force.  F = F T – mg mv 2 = F T - mg r F T = mv2 + mg r F T = 7.34 N

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

26. 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  s =0.60? F fr = F R mg FNFN Let’s first solve for the Normal Force F N = mg = (1000 kg)(9.8 m/s2) F N = 9800 N

27.  F x = F fr F R = F fr mv 2 =  sF N 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.

28. Will the car make it if it is icy and the  s = 0.25  F x = F fr F R = F fr mv 2 =  sF N 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.

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

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

31. F R = F fr + F N sin  (ideally, we bank the road so that no friction is required: F fr = 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? mg  FNFN  F N = mgcos 

33. Now we will simply work with the Normal Force to find the component that points to the center of the circle mg  FNFN  First consider the y forces.  F y = F N cos  - mg Since the car does not move up or down:  F y = 0 0 = F N cos  – mg F N cos  = mg F N = mg/cos  F N cos  F N sin  

34. mv 2 = F N sin  r mv 2 = mgsin  r cos  v 2 = gtan  r r v 2 = tan  gr

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

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

37. Circular Orbits Orbits are freefall (not true weightlessness) Orbital velocity must match the weight mg = mv 2 r g = v 2 v = √ gr r

38. A satellite wishes to orbits at a height of 200 miles above the earth’s surface. a.Calculate the height above the center of the earth if R earth = 6.37 X 10 6 m. (6.69 X 10 6 m) b.Calculate the orbital velocity. (8098 m/s) c.Calculate the period in minutes. (86.5 min)

39. 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. a.Draw a free body diagram of the block. b.Calculate the tangential acceleration of the block (due to friction). (-5.88 m/s 2 ) c.Calculate the angular acceleration. (-7.35 rad/s2) d.Calculate the time until the block comes to a rest. (2.8 s) e.Calculate the number of revolutions. (4.7 rev)