Chapter 7


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Chapter 7

  1. 1. +Chapter 7Pg. 232-261
  2. 2. +7.1 Circular MotionPg. 234- 239
  3. 3. + Circular Motion  Any object that revolves about a single axis  The line about which the rotation occurs is called the axis of rotation
  4. 4. + Tangential Speed (vt) Speed of an object in circular motion Uniform circular motion: vt has a constant value  Onlythe direction changes  Example shown to the right How would the tangential speed of a horse near the center of a carousel compare to one near the edge? Why?
  5. 5. + Centripetal Acceleration (ac) Acceleration directed toward the center of a circular path Acceleration is a change in velocity (size or direction). Direction of velocity changes continuously for uniform circular motion.
  6. 6. + Centripetal Acceleration (magnitude)  How do you think the magnitude of the acceleration depends on the speed?  How do you think the magnitude of the acceleration depends on the radius of the circle?
  7. 7. + Example  A test car moves at a constant speed around a circular track, If the car is 48.2m from the center and has the centripetal acceleration of 8.05m/s2, what is the car’s tangential speed?  ac = vt2 r
  8. 8. + Example  ac = vt2 r  acr = vt2  vt =√ac r  vt= √(8.05) (48.2)  vt= 19.69 m/s
  9. 9. + Tangential Acceleration Occurs if the speed increases Directed tangent to the circle Example: a car traveling in a circle  Centripetal acceleration maintains the circular motion.  directed toward center of circle  Tangential acceleration produces an increase or decrease in the speed of the car.  directed tangent to the circle
  10. 10. + Centripetal Acceleration Click below to watch the Visual Concept. Visual Concept
  11. 11. + Centripetal Force Maintains motion in a circle Can be produced in different ways, such as  Gravity  A string  Friction Which way will an object move if the centripetal force is removed?  Ina straight line, as shown on the right
  12. 12. + Centripetal Force (Fc)Fc mac vt 2 and ac r mvt 2 so Fc r
  13. 13. + Example  A pilot is flying a small plane at 56.6 m/s in a circular path with a radius of 188.5m. The centripetal force needed to maintain the plane’s circular motion is 1.89 X 104 N. What is the plane’s mass?  Given:  vt= 56.6 m/sr= 188.5 m  Fc= 1.89X 104 N m= ??
  14. 14. + Example  Fc = mvt2m = Fcr r vt2 m = (1.89 X 104)(188.5m) (56.6)2 m = 1112.09 kg
  15. 15. + Describing a Rotating System  Imagine yourself as a passenger in a car turning quickly to the left, and assume you are free to move without the constraint of a seat belt.  How does it “feel” to you during the turn?  How would you describe the forces acting on you during this turn?  Thereis not a force “away from the center” or “throwing you toward the door.”  Sometimes called “centrifugal force”  Instead, your inertia causes you to continue in a straight line until the door, which is turning left, hits you.
  16. 16. +7.2 Newton’s Law of UniversalGravitation Pg. 240-247
  17. 17. + Gravitational Force  The mutual force of attraction between particles of matter.  Gravitational force depends on the masses and the distance of an object.
  18. 18. + Simpson’s Video  s/ScienceOnSimpsons/Clips_files/3D- Homer.m4v
  19. 19. + Family Guy Video
  20. 20. + Newton’s Thought Experiment What happens if you fire a cannonball horizontally at greater and greater speeds? Conclusion: If the speed is just right, the cannonball will go into orbit like the moon, because it falls at the same rate as Earth’s surface curves. Therefore, Earth’s gravitational pull extends to the moon.
  21. 21. + Law of Universal Gravitation  Fg is proportional to the product of the masses (m1m2).  Fgis inversely proportional to the distance squared (r2).  Distance is measured center to center. G converts units on the right (kg2/m2) into force units (N).  G = 6.673 x 10-11 N•m2/kg2
  22. 22. + Law of Universal Gravitation
  23. 23. + The Cavendish Experiment  Cavendish found the value for G.  He used an apparatus similar to that shown above.  He measured the masses of the spheres (m1 and m2), the distance between the spheres (r), and the force of attraction (Fg).  Hesolved Newton’s equation for G and substituted his experimental values.
  24. 24. + Gravitational Force  Ifgravity is universal and exists between all masses, why isn’t this force easily observed in everyday life? For example, why don’t we feel a force pulling us toward large buildings?  The value for G is so small that, unless at least one of the masses is very large, the force of gravity is negligible.
  25. 25. + Ocean Tides  What causes the tides?  How often do they occur?  Why do they occur at certain times?  Are they at the same time each day?
  26. 26. + Ocean Tides  Newton’s law of universal gravitation is used to explain the tides.  Since the water directly below the moon is closer than Earth as a whole, it accelerates more rapidly toward the moon than Earth, and the water rises.  Similarly, Earth accelerates more rapidly toward the moon than the water on the far side. Earth moves away from the water, leaving a bulge there as well.  As Earth rotates, each location on Earth passes through the two bulges each day.
  27. 27. + Gravity is a Field Force  Earth,or any other mass, creates a force field.  Forces are caused by an interaction between the field and the mass of the object in the field.  The gravitational field (g) points in the direction of the force, as shown.
  28. 28. + Calculating the value of g Since g is the force acting on a 1 kg object, it has a value of 9.81 N/m (on Earth).  The same value as ag (9.81 m/s2) The value for g (on Earth) can be calculated as shown below. Fg GmmE GmE g 2 2 m mr r
  29. 29. + Classroom Practice Problems  Find the gravitational force that Earth (mE= 5.97 1024 kg) exerts on the moon (mm= 7.35 1022 kg) when the distance between them is 3.84 x 108 m.  Answer: 1.99 x 1020 N  Findthe strength of the gravitational field at a point 3.84 x 108 m from the center of Earth.  Answer: 0.00270 N/m or 0.00270 m/s2
  30. 30. +7.3 Motion in SpacePg. 248- 253
  31. 31. + Kepler’s Laws  Johannes Kepler built his ideas on planetary motion using the work of others before him.  Nicolaus Copernicus and Tycho Brahe
  32. 32. + Kepler’s Laws  Kepler’s first law  Orbits are elliptical, not circular.  Some orbits are only slightly elliptical.  Kepler’s second law  Equal areas are swept out in equal time intervals.  Basically things travel faster when closer to the sun
  33. 33. + Kepler’s Laws Kepler’s third law  Relatesorbital period (T) to distance from the sun (r)  Period is the time required for one revolution.  Asdistance increases, the period increases.  Not a direct proportion  T2/r3 has the same value for any object orbiting the sun
  34. 34. + Equations for Planetary Motion  UsingSI units, prove that the units are consistent for each equation shown below.
  35. 35. + Classroom Practice Problems  A large planet orbiting a distant star is discovered. The planet’s orbit is nearly circular and close to the star. The orbital distance is 7.50 1010 m and its period is 105.5 days. Calculate the mass of the star.  Answer: 3.00 1030 kg  What is the velocity of this planet as it orbits the star?  Answer: 5.17 104 m/s
  36. 36. + Weight and Weightlessness Bathroom scale  A scale measures the downward force exerted on it.  Readings change if someone pushes down or lifts up on you.  Your scale reads the normal force acting on you.
  37. 37. + Apparent Weightlessness  Elevator at rest: the scale reads the weight (600 N).  Elevatoraccelerates downward: the scale reads less.  Elevator in free fall: the scale reads zero because it no longer needs to support the weight.
  38. 38. + Apparent Weightlessness  Youare falling at the same rate as your surroundings.  No support force from the floor is needed.  Astronautsare in orbit, so they fall at the same rate as their capsule.  Trueweightlessness only occurs at great distances from any masses.  Even then, there is a weak gravitational force.