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Chapter 7
- 1.
- 2. + 7.1 Circular Motion Pg.234- 239
- 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. + 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. + 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. + 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. + 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. + Example ac = vt2 r acr = vt2 vt =√ac r vt= √(8.05) (48.2) vt= 19.69 m/s
- 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. + Centripetal Acceleration Click below to watch the Visual Concept. Visual Concept
- 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. + Centripetal Force (Fc) Fc mac vt 2 and ac r mvt 2 so Fc r
- 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. + Example Fc = mvt2m = Fcr r vt2 m = (1.89 X 104)(188.5m) (56.6)2 m = 1112.09 kg
- 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. + 7.2 Newton’s Lawof Universal Gravitation Pg. 240-247
- 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. + Simpson’s Video http://www.lghs.net/ourpages/users/dburn s/ScienceOnSimpsons/Clips_files/3D- Homer.m4v
- 19. + Family Guy Video
- 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. + Law ofUniversal 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. + Law of Universal Gravitation
- 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. + 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. + 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. + 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. + 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. + 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. + 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. + 7.3 Motion inSpace Pg. 248- 253
- 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. + 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. + 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. + Equations for Planetary Motion UsingSI units, prove that the units are consistent for each equation shown below.
- 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. + 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. + 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. + 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.