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Circular Motion and Gravitation

Circular Motion and Gravitation. What happens in these situations. Your wet dog shakes his body You swing a bucket of water over your head You ride on the scrambler with another person. The water from the dog, the water in the Bucket, and Bubba are pulled out, right?

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Circular Motion and Gravitation

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  1. Circular Motion and Gravitation

  2. What happens in these situations • Your wet dog shakes his body • You swing a bucket of water over your head • You ride on the scrambler with another person The water from the dog, the water in the Bucket, and Bubba are pulled out, right? CENTRIFUGAL FORCE X

  3. Imagine if you will. . . • A car calmly drives down a straight street at 30 mph (13m/s) • Suddenly a deer jumps into the road and the driver turns the steering wheel suddenly to the right. But the car does not turn, what went wrong? • Someone cut the power steering • A mysterious, ghostly force is present • The coefficient of friction was not great enough to provide a force so the moving car maintained it’s straight line course.

  4. Nothing can turn unless • A force causes the object to change it’s motion. • The force must be in the direction of the turn

  5. Turning is always caused by an inward force Which is the true force? • Centripetal – inward force • Centrifugal – outward force Why is centrifugal a word? • Wrong people still use it • Faith Hill – This Kiss • Just because everybody’s doing it doesn’t make it right! No F words! don't use centriFugal!

  6. Inertia • Imagine being in a merry go round! • If you didn’t hold on, you would be thrown off! • This is due to inertia, not centrifugal force!

  7. Circular Motion Section 1

  8. Tangential Speed • The tangential speed (vt) is the object’s speed along an imaginary line drawn tangent to the circular path. • Tangential speed depends on the distance from the object to the center of the circular path. • When the tangential speed is constant, the motion is described as uniform circular motion.

  9. Tangential Speed • Picture this . . . • A pair of horses side by side on a carousel. Each completes one full circle in the same time period, but the horse on the outside covers more distance than the inside horse does, so the outside horse has a greater tangential speed.

  10. Centripetal Acceleration • The acceleration of an object moving in a circular path and at constant speed is due to a change in direction. • An acceleration of this nature is called a centripetal acceleration.

  11. Centripetal Acceleration • (a) As the particle moves from A to B, the direction of the particle’s velocity vector changes. • (b) For short time intervals, ∆v is directed toward the center of the circle. • Centripetal acceleration is always directed toward the center of a circle.

  12. What Did He Just Say? • At point A the object has a tangential velocity, vi. At point B the object has a tangential velocity, vf. • Assume they have the same magnitude, but differ in direction. • ∆v = vf - vi

  13. Example • A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2, what is the car’s tangential speed?

  14. Solution • Vt = 19.7 m/s

  15. Your Turn I • A rope attaches a tire to an overhanging tree limb. A girl swinging on the tire has a centripetal acceleration of 3.0 m/s2. If the length of the rope is 2.1 m, what is the girl’s tangential speed? • A kid swings a yo-yo parallel to the ground and above his head, the yo-yo has a tangential velocity of 11.2 m/s. If the yo-yo’s string is 0.50 m long, what is the yo-yo’s centripetal acceleration? • A race car moving along a circular tack has a centripetal acceleration of 15.4 m/s2. If the car has a tangential speed of 30.0 m/s, what is the distance between the car and the center of the track.

  16. Centripetal Force • Consider a ball of mass m that is being whirled in a horizontal circular path of radius r with constant speed. • The force exerted by the string has horizontal and vertical components. The vertical component is equal and opposite to the gravitational force. Thus, the horizontal component is the net force. • This net force, which is directed toward the center of the circle, is a centripetal force.

  17. Centripetal Force • Newton’s second law can be combined with the equation for centripetal acceleration to derive an equation for centripetal force:

  18. Centripetal Force • Centripetal force is simply the name given to the net force on an object in uniform circular motion. • Any type of force or combination of forces can provide this net force. • For example, friction between a race car’s tires and a circular track is a centripetal force that keeps the car in a circular path. • As another example, gravitational force is a centripetal force that keeps the moon in its orbit.

  19. Example • A pilot is flying a small plane at 56.6 m/s in a circular path with a radius of 188.5 m. The centripetal force needed to maintain the plane’s circular motion is 1.89 x 104 N. What is the plane’s mass?

  20. Solution • m = 1.11 x 103 kg

  21. Your Turn II • A 2.10 m rope attaches a tire to an overhanging tree limb. A girl swinging on the tire has a tangential speed of 2.50 m/s. If the magnitude of the centripetal force is 88.0 N, what is the girls mass? • A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track. The magnitude of the centripetal force is 377 N, and the combined mass of the bike and the rider is 86.5 kg. What is the track’s radius? • A 905 kg car travels around a circular track with a circumference of 3250 m. If the magnitude of the centripetal force is 2140 N, what is the car’s tangential speed? (circumference = 2πr)

  22. Centripetal Force • If the centripetal force vanishes, the object stops moving in a circular path. • A ball that is on the end of a string is whirled in a vertical circular path. • If the string breaks at the position shown in (a), the ball will move vertically upward in free fall. • If the string breaks at the top of the ball’s path, as in (b), the ball will move along a parabolic path.

  23. Describing a Rotating System • To better understand the motion of a rotating system, consider a car traveling at high speed and approaching an exit ramp that curves to the left. • As the driver makes the sharp left turn, the passenger slides to the right and hits the door. • What causes the passenger to move toward the door?

  24. Describing a Rotating System • As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to move along the original straight path. • If a sufficiently large centripetal force acts on the passenger, the person will move along the same curved path that the car does. The origin of the centripetal force is the force of friction between the passenger and the car seat. • If this frictional force is not sufficient, the passenger slides across the seat as the car turns underneath.

  25. Newton’s Law of Universal Gravitation Section 2

  26. Gravitational Force • Gravitational force is the mutual force of attraction between all particles of matter. • Gravitational force depends on the mass of both objects and the distance between them. • Near earths surface, Fgravity = weight

  27. Gravitational Force • Newton realized that the centripetal force that holds the planets in orbit is the very same force that pulls an apple to the ground – Gravitational Force. • So how does gravity create an orbit?

  28. Gravitational Force • To see how this happens, we can use a thought experiment that Newton developed. Consider a cannon sitting on a high mountaintop. Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing.

  29. Gravitational Force • Satellites stay in orbit for the same reason. • The force that pulls the apple toward the Earth is the same force that keeps the moon in orbit around the Earth.

  30. Gravitational Force The constant G, called the constant of universal gravitation, equals 6.67  10–11 N•m2/kg.

  31. Gravitational Force • The gravitational forces that two masses exert on each other are always equal in magnitude and opposite in direction. (Newton’s third law of motion) • So this means that even if the earth exerts a Fg, we also exert THE SAME Fg on earth! • This is easily explained using F = ma

  32. Gravitational Force • Gravitational forces exist between any two masses, regardless of size. • For example, two desks in a classroom have a mutual attraction because of gravitational force. • Why don’t your desks go flying together? • Use Newton’s Law of Universal Gravitation to see why.

  33. Flying desks? • Earth = 5.97 x 1024 kg • Desk = 5.0 kg • Distance between desks = 1.5 m • Distance between desk and Earth = 6.38 x 106 m • Use this equation for the desks then the desk and the Earth

  34. Gravitational Force • If gravitational force acts between all masses then why doesn't the Earth accelerate up toward a falling apple? • IT DOES!!!!!!!!!!!!!! • Earth acceleration is so small you cannot detect it. • Earth’s acceleration = 2.5 x 10-25 m/s2

  35. Example • Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force between them is 8.92 x 10 -11 N.

  36. Solution • r = 3.00 x 10-1 m

  37. Universal Gravitation & Centripetal Force • Since the path the satellite follows is circular, centripetal force must be present. • The force that provides Cf in satellite motion is Gravity! Leads us to Fc = Fg

  38. Formula Relation • Derive on board: (provides us with an equation to find tangential speed of an orbiting object) • Final Equation: vt2 = Gmi/r • Mi = mass of larger object

  39. Remember • When an object is orbiting earth, you have to add radius of earth (6.38 X 106m) to distance satellite is above earth: • Ex: Satellite 1000 m above earth; • r = 6.38 X 106m + 1000m • Radius is measured in m, not km! • If asked to find acceleration of orbiting object, F=ma (F is Fg)

  40. Your Turn III • What must the distance be between two 0.800 kg blocks if the magnitude of the gravitational force between them is 8.92 x 10-11 N? • Find the gravitational force a 66.5 kg person would experience while standing on the surface of the Earth: (Earth – mass = 5.98 x 1024 kg – r = 6.38 x 106 m) • From problem #2, What is the acceleration of the person? Of Earth? • A 1200 kg satellite orbits earth at an altitude of 25,500 m. What is the Vt of the satellite?

  41. Applying the Law of Gravitation • Newton’s law of gravitation accounts for ocean tides. • High and low tides are partly due to the gravitational force exerted on Earth by its moon. • The tides result from the difference between the gravitational force at Earth’s surface and at Earth’s center.

  42. Applying the Law of Gravitation • On the side of the Earth that is nearest to the moon, the gravitational force is greater. Water is pulled to toward the moon, causing high tide. • On the opposite side, gravitational force is less, all the mass is pulled toward the moon, but water is pulled the least, causing high tide. • So, high tide is occurring in two places opposite each other. Each coast goes through two high tides a day, since the Earth will rotate past each one during its rotation.

  43. Applying the Law of Gravitation • Henry Cavendish applied Newton’s law of universal gravitation to find the value of G and Earth’s mass. (Remember both were previously unknown) • When two masses, the distance between them, and the gravitational force are known, Newton’s law of universal gravitation can be used to find G. • Once the value of G is known, the law can be used again to find Earth’s mass.

  44. Applying the Law of Gravitation • Newton was not able to describe how objects can exert a forces on one another without coming in contact with each other. • His theory described gravity, but he didn’t explain how it worked. • Scientists later developed the theory of fields. • Masses create a gravitational field in space. A gravitational force is an interaction between a mass and the gravitational field created by other masses.

  45. Applying the Law of Gravitation • When you raise a ball to a certain height above the Earth, the ball gains potential energy. • Where is this potential energy stored? • The physical properties of the ball and the Earth have not change. • The gravitational field between the ball and the Earth has changed, since the ball changed position relative to the Earth. • So gravitational energy is stored in the gravitational field itself.

  46. Applying the Law of Gravitation • Gravity is a field force. • Gravitational field strength, g, equals Fg/m. • The gravitational field, g, is a vector with magnitude g that points in the direction of Fg. • Gravitational field strength equals free-fall acceleration. The gravitational field vectors represent Earth’s gravitational field at each point. Does it make sense that the vectors get smaller the further away you go?

  47. Applying the Law of Gravitation • weight = mass  gravitational field strength • Because it depends on gravitational field strength, weight changes with location: • On the surface of any planet, the value of g, as well as your weight, will depend on the planet’s mass and radius.

  48. PNBW • Page 247 • Physics – 1-4 • Honors – 1-5

  49. Motion in Space Section 3

  50. Kepler’s Laws • Kepler’s laws describe the motion of the planets. • First Law: Each planet travels in an elliptical orbit around the sun, and the sun is at one of the focal points. • Second Law: An imaginary line drawn from the sun to any planet sweeps out equal areas in equal time intervals. • Third Law: The square of a planet’s orbital period (T2) is proportional to the cube of the average distance (r3) between the planet and the sun.

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