Chapter 5: Circular Motion; Gravitation

1. Chapter 5: Circular Motion; Gravitation

2. Curves • Forces act in straight lines • An object will move in a straight line if there is NO force acting on it or if the net force acts in the direction of the motion. • BUT, what if the net force acts at an angle to the motion?

3. Come Full Circle • From the last example, what would happen if the magnitude of the force and its angle to the motion remained constant?

4. Uniform Circular Motion • An object that moves in a circle at constant speed is said to experience uniform circular motion.

5. Kinematics of UCM • aR = v2/ r • Remember:frequency (f) is the number of times the object completes a circle per second.period (T) is the amount of time it takes to complete a circle, or T = 1/f • Also remember v = d/t. In a circle, distance means circumference(2πr) and time means period. So v =

6. Example • A 150g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600m. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

7. Answer • v = (2πr)/T = (2*3.14*0.600m)/(0.500s) = 7.54m/s • aR = v2/r = (7.54m/s)2 / (0.600m) = 94.8m/s2

8. Dynamics of UCM • From chapter 4 we learned that acceleration requires a force. • So what causes the centripetal acceleration that causes UCM? • That would be the centripetal force, ΣFR • We start with ΣF = ma and through the wonders of substitution we get ΣFR=maR or • ΣFR = mv2/ r

9. Beware • Centripetal force is NOT a new kind of force. • The term “centripetal” simply tells you that the force is “center seeking”. • Example: Picture swinging a child around by the arms. In order to keep the child moving in a circle you have to pull in with your arms. You pulling with your arms is no different from you pulling a box.

10. Example • Estimate the force a person must exert on a string attached to a 0.150kg ball to make the ball revolve in a horizontal circle of radius 0.600m. Also note the ball makes 2.00 revolutions per second.

11. Solution • ΣFR = mv2 / r = m(2πr)2/ r = (0.150kg)(7.54m/s)2 / (0.600m) ≈ 14N

12. Example • A 0.150kg ball on the end of a 1.10m long cord(of negligible mass) is swung in a vertical circle. • A.) determine the minimum speed the ball must have at the top of its arc so that it continues moving in a circle. • B.) Calculate the tension in the cord at the bottom of the arc assuming the ball is moving at twice the speed from part A.

13. You Going Into a Curve • When you are in a car that enters a curve you get pulled toward the door. • This is because of Newton’s First Law; you want to keep going straight. • So what makes you move toward the center of the curve?

14. The Car Going Into the Curve • The car also wants to go straight when it enters a curve. • What makes the car move in a curve? • What happens if this force is not enough?

15. Example • A 1000kg car rounds a curve on a flat road of radius 50m at a speed of 14m/s. Will the car make the turn or will it skid off if…a. the pavement is dry and the coefficient of static friction is μs = 0.6b. the pavement is icy and μs = 0.25

16. Solution • The only force acting in the horizontal direction is friction. • We need to see if this force provides enough centripetal force to maintain the curve. • FR = maR = mv2/r = (1000kg * (14m/s)2)/50m = 3900N this is how much force is needed. • Ffr = μsFN= μsmg = (0.60)(1000kg)(9.8m/s2)=5900N this is the max friction by the car, so we are fine.

17. Solution • On the icy road the Ffr becomes,Ffr = μsFN= μsmg = (0.25)(1000kg)(9.8m/s2) = 2500N • This is too low. The car would skid off the road.

18. Slamming on the Brakes • The situation gets even worse if your wheels lock up. (granted this doesn’t happen much anymore thanks to the wonders of anti-lock brakes) • When the wheels are rolling there is a spot on the tire that is stationary to the ground, so the friction is static. • When the tire locks and slides as a whole unit, the friction is kinetic, which is lower.

19. Handling the Curves • There are a few ways we can make curves safer. • Reduce v (slow down when entering a curve) • Increase r (make the curve more gradual) • Increase friction (new asphalt technology) • Make the curve an incline (called a banked curve)

20. Banked Curves • By banking the curve we create a normal force that is at an angle, which creates 2 components. • The x component of FN (FN sinθ) adds to the FR. • We can set the horizontal normal force, Fnsinθ equal FR, mv2/r to find the correct angle.

21. Example • What banking angle is need for an express way off-ramp curve of radius 50m at a design speed of 50km/hr (or 14m/s)

22. Solution • FN sinθ = mv2/r • We need to find FN • We know the vertical forces are balanced (because the car is not floating) So FNcosθ = mgor FN = mg/cosθ • We can plug this into FN sinθ = mv2/r

23. Solution • sinθ = mv2/r • mg tanθ = mv2/r • tanθ = v2 /rg • tanθ = (14m/s)2 / (50m)(9.8m/s2) = 0.40 • Using tan-1we get θ = 22o

24. Falling Bodies • At this point Newton had already constructed his 3 laws when he was pondering falling objects. • He already concluded that falling objects must have a force exerted on them, but what was doing the exerting?

25. High Minded • The story goes that Newton watched an apple fall to the ground and was struck by an inspiration: if gravity can reach tree tops and even mountain tops can it reach the moon? • It was this simple question (along with a lot of math) that set in motion a chain of thinking that connects Isaac Newton to Neil Armstrong to Curiosity and beyond.

26. Haters gonna hate • Newton ran into a great deal of resistance to his theory of gravity. • What do you think was/were the issue(s)?

27. Pro-active • Newton did more than just bring up good questions; he set out to find the answers. • We already know that the acceleration due to gravity on the surface of the Earth is 9.8m/s2. • Let’s do an example that will cast some light on the nature of gravity.

28. That’s no space station • The Moon is 3.84x108 m from the Earth and travels in a near perfect circle. The period of the moon is 27.3days = 2.36x106 s. Find the centripetal acceleration, aR, of the Moon.

29. Analysis • The distance from the center of the Earth to the Moon is 60 times the distance from the center of the Earth to its surface. • 602 = 3600 • Divide 9.8m/s2 by 3600 and compare it to your answer from the last problem. • What do you conclude?

30. Mass Effect • We already stated in chapter 4 that objects with more mass need more force to reach the same acceleration as a less massive object. • We also know from Newton’s 3rd that if the Earth is pulling on the Moon, then Moon is pulling on the Earth. • Therefore, Newton concluded that the force of gravity is proportional to both the masses.

31. Put it all together • So far we got the force is related to 1/r2 and also related to m1 x m2 • Together they make NEWTON’S LAW OF UNIVERSAL GRAVITATION:

32. Near Earth Gravity • When we use on the surface of the earth, m1 becomes the mass of the Earth (5.98E24kg), m2 becomes the mass of the object on the surface, and r becomes the radius of the Earth (6.38E6m).

33. Weighty Subject Matter • As we have already stated in chapter 4, we call the force of gravity the weight of the object, mg. • So we can rewrite as • The mass cancels out and we get

34. Example • Estimate g on the top of Mt. Everest, 8848m above the Earth’s surface.

35. Solution • We need to add the distance from the surface to the summit to the radius of the Earth. • 6.38E6m + 8.8E3m = 6.389E6m • We plug this value in as “r” in

36. Geosynchronous Orbit • An object in geosynchronous orbit is an object that hovers over the same spot above the Earth. It revolves at the same rate that the Earth rotates. • Why is this important? • How is this possible?

37. Ready for Launch • What is the necessary height above the Earth a satellite must reach to obtain geosynchronous orbit? • What is that satellite’s speed? (That is how fast does the Earth rotate?)

38. Solution • Start at FR = maR, • Substituting (GmsatmE)/r2 in for FRand v2/r for aR gives us: • (GmsatmE)/r2 = msatv2/r • Substituting (2πr)/T for v gives us • (GmE)/r2 = (2πr)2/rT2 (note that T = 1day = 86,400s) • We need to solve this for r.

39. Solution Cont. • Algebra gives us • Which equals 7.54 x 1022 m3 • Taking the cube root gives us r = 4.23 x 107m. • Finally we are ready to find v

40. As The World Turns • We can solve (GmmE)/r2 = mv2/r for v and get,

41. Meet Johannes Kepler • Over 50 years before Newton published his 3 laws of motion and the law of gravitation, German astronomer Johannes Kepler(1571-1630) published many works including his 3 laws of planetary motion. (Kepler btw based his work off the earlier work of Tycho Brahe(1546-1601)).

42. Kepler’s 3 Laws • 1: The path of each planet about the Sun is an ellipse with the Sun at one focus. • 2: Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal periods of time. • 3: The ratio of the squares of the periods of any two planets revolving about the Sun is equal to the ratio of the cubes of their mean distances from the Sun.

43. Our Focus is on 3 • The last law can be represented mathematically as follows: • We can rearrange them to get

44. Newton over Kepler • Where Newton really shined was in his ability to show that all 3 of Kepler’s Laws can be derived mathematically using the Law of Universal Gravitation.

45. Imperfections Reveal the True Beauty • Since Kepler’s time we have gotten somewhat better at measuring the orbits the planets and we have seen that they do not exactly follow Kepler’s perfect ellipses. • Newton fixes this as well. He stated that everything pulls on everything else. Other planets tug on a planet and cause disturbances in its orbit.

46. Something is Wrong with Uranus • These disturbances actually lead to the discovery of 2 planets! • Astronomers could not account for the weirdness in the orbit of Uranus. • The only thing that could fix the math would be if there was more mass beyond Uranus. • This lead us to look for and find Neptune and Pluto.