2. CIRCULAR MOTION

3. REMEMBER : The curved path of a projectile was due to a force (gravity) acting on a body in a direction NOT parallel to its line of motion.

4. CIRCULAR MOTION is due to a net force acting perpendicular to its line of motion (radially inward) V F

5. Uniform Circular Motion (UCM) Path is circular and the speed is constant Such as the motion of the moon Variable Circular Motion Path is circular and the speed changes Such as vertical circular motion

6. NOTE: For a body traveling in a circular path, the velocity is not constant because the direction is always changing. Thus it must be accelerating. NOTE: For a body traveling in a circular path, the velocity is not constant because the direction is always changing. Thus it must be accelerating.

7. DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCM R R V0V0 V1V1 A B O

8. R R V0V0 V1V1 A B O

9. V0V0 V1V1 -V0-V0 ΔVΔV R R A B O NOTE : If θ is very small, ΔV points radially inward (centripetally)

10. DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCM R R A B O V V ΔVΔV X Y Z Since in UCM, |V 0 | = |V 1 | = V, ΔXYZ is isosceles --- as is ΔOAB. Isosceles triangles with congruent vertex angles are similar.

11. DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCM R R A B O V V ΔVΔV X Y Z ΔOAB ~ ΔXYZ So corresponding parts are in proportion

12. DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCM As θ → 0 0, chord AB → arc AB ≡ Δd From kinematics, Δd = VΔt

13. DERIVING CENTRIPETAL ACCELERATION FOR A BODY IN UCM ( )

14. By Newton’s Law of Acceleration (a=F/m) Centripetal force

15. Horizontal Circles - UCM A 2.5 kg ball is spun in a horizontal circle at 5.0 m/s at the end of a rope 0.75 m long. Find (a) the centripetal acceleration and (b) the tension in the rope. A 2.5 kg ball is spun in a horizontal circle at 5.0 m/s at the end of a rope 0.75 m long. Find (a) the centripetal acceleration and (b) the tension in the rope. m = 2.5 kg R = 0.75 m V = 5.0 m/s T V R

16. Horizontal Circles - UCM a) = 33 m/s 2 b) T = 83 N

17. VERTICAL CIRCLES – VARIABLE CIRCULAR MOTION V min V max R FwFw FwFw T top T bot At the top F net =F c =F w +T top At the bottom F net =F c =T bot -F w

18. VERTICAL CIRCLES – VARIABLE CIRCULAR MOTION Critical velocity, V crit The lowest possible speed for a body at the top of a vertical circle to maintain that circular path.

19. V min V max R FwFw FwFw T top T bot At the top : if V min = V crit then T top = 0 F c = F w = mg NOTE THAT THE GREATER THE RADIUS AT THE TOP, THE GREATER THE V CRIT.

20. VERTICAL CIRCLES – VARIABLE CIRCULAR MOTION V min V max R FwFw FwFw T top T bot At the bottom F c = T bot - F w T bot = F c + F w Note that the greater the radius at the bottom, the smaller the tension in the string. ( )

21. VERTICAL CIRCLES – VARIABLE CIRCULAR MOTION You may have noticed that modern looping roller coasters are never designed with circular loops. They are designed with “clothoid” loops. These have small radii at the top and large radii at the bottom. WHY? r R

22. CENTRIFUGAL FORCE ? The apparent outward force acting on a body rounding a curve. The inertial effect of a centripetal force acting on a body. NOT A TRUE FORCE !!! The body is in a non-inertial (accelerating) frame of reference.

23. PROBLEM The moon has a mass of 7.3 x 10 22 kg and orbits the Earth at a radius of 3.8 x 10 8 m once every 27.4 days. Find a) The orbital speed of the moon, b) The acceleration of the moon towards the Earth, and c) The gravitational force the Earth exerts on the moon.

24. PROBLEM m = 7.3 x 10 22 kg R = 3.8 x 10 8 m T = 27.4 days= 2.37 x 10 6 s V V R R m m a) For 1 revolution, Δd = C = 2πR and Δt = T =1.0x10 3 m/s F grav

25. PROBLEM b) a c = ? = 2.7 x 10 -3 m/s 2 [ ]

26. PROBLEM c) F grav = ? F grav = F c F grav = 2.0 x 10 20 N [ ]

27. PROBLEM The track at Talladega Superspeedway has curves that are banked at 33 0 with a radius of 450 m. At what speed must a 750 kg car make these turns if there is no friction due to oil on the track?

28. FcFc R FwFw N R V The only two forces acting on the car are gravity and the normal force. Of these the only one having a component that runs radially inward is the normal force.

29. FwFw N NyNy NxNx Since N x acts radially inward N x = F c N sin θ = mv 2 /R Given : θ = 33 o R = 450 m m = 750 kg Vertically, the car is at rest At equilibrium, Σ F up = Σ F down N y = F w N cos θ = mg

30. PROBLEM N cos θ = mgN sin θ = mv 2 /R V = 54 m/s (120 MPH) VERT HORIZ

31. SATELLITES The gravitational force acting on a satellite is the centripetal force responsible for its “circular” path. F grav

32. SATELLITES F c = F grav Where m is the mass of the satellite and M is the mass of the “planet” Remember this !!!

33. SATELLITES So, for satellites orbiting the same body : Orbital radius is inversely proportional to the speed squared

34. SATELLITES

35. SATELLITES REMEMBER THIS !!!

37. 1 st LAW Click here for Next slide Click here to view Movie

38. 2 nd LAW a Click here for Next slide Click here to view Movie

39. 2 nd LAW b Click here for Next slide SOUND FAMILIAR ? Click here to view Movie

40. 3 rd LAW Click here for Next slide SOUND FAMILIAR ? Click here to view Movie

41. END OF CIRCULAR MOTION