1. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company What do you think? Consider the following objects moving in circles at constant speeds`: A ball tied to a string being swung in a circle The moon as it travels around Earth A child riding rapidly on a playground merry-go-round A car traveling around a circular ramp on the highway For each example above, answer the following: What is keeping the object in the circular path? Are the objects accelerating?

2. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company Ok, what did we learn? An object moving in a circle is accelerating. So, there must be a force. The force is always pointed towards the center! centripetal forceThis “center- seeking” force is called a centripetal force (LEARN THIS!!!) The feeling that you are being pulled outward is your INERTIA and is called centrifugal (center fleeing) and is a FALSE FORCE!!!

3. Section 1Circular Motion and Gravitation © Houghton Mifflin Harcourt Publishing Company Tangential Speed (v t ) Speed in a direction tangent to the circle - AKA linear speed Uniform circular motion: v t has a constant value –Only the direction changes Angular speed 2  /T  =  /  t = 2  /TAngular speed is 2  /T where T is the period.  =  /  t = 2  /T How would the angular speed of a horse near the center of a carousel compare to one near the edge? Tangential? Why?

4. Section 1Circular Motion and Gravitation © Houghton Mifflin Harcourt Publishing Company Tangential Speed (v t ) (2  r)/ T(2  r)/ T Tangential speed = the circumference divided by the Period (T) The Period is the time of one revolution.

5. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company Centripetal Acceleration (a c ) Acceleration is a change in velocity (speed and/or direction). Direction of velocity changes continuously for uniform circular motion. What direction is the acceleration? –the same direction as  v –toward the center of the circle Centripetal means “center seeking”

6. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company 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. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company Tangential Acceleration Occurs if the speed CHANGES 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

8. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company Click below to watch the Visual Concept. Visual Concept Centripetal Acceleration

9. Section 1Circular Motion and Gravitation © Houghton Mifflin Harcourt Publishing Company Centripetal Force (F c )

10. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company 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? –In a straight line, as shown on the right

11. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company 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? There is 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.

12. Circular Motion and GravitationSection 1 © Houghton Mifflin Harcourt Publishing Company Classroom Practice Problems A 35.0 kg child travels in a circular path with a radius of 2.50 m as she spins around on a playground merry-go-round. She makes one complete revolution every 2.25 s. –What is her speed or tangential velocity? (Hint: Find the circumference to get the distance traveled.) –What is her centripetal acceleration? –What centripetal force is required? `Answers: 6.98 m/s, 19.5 m/s 2, 682 N

13. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company What do you think? Imagine an object hanging from a spring scale. The scale measures the force acting on the object. What is the source of this force? What is pulling or pushing the object downward? Could this force be diminished? If so, how? Would the force change in any way if the object was placed in a vacuum? Would the force change in any way if Earth stopped rotating?

14. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company 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.

15. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Law of Universal Gravitation F g is proportional to the product of the masses (m 1 m 2 ). F g is inversely proportional to the distance squared (r 2 ). –Distance is measured center to center. G converts units on the right (kg 2 /m 2 ) into force units (N). –G = 6.673 x 10 -11 Nm 2 /kg 2

16. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Law of Universal Gravitation

17. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Gravitational Force If gravity 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.

18. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company 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?

19. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company 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.

20. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company 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.

21. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company 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 a g (9.81 m/s 2 ) The value for g (on Earth) can be calculated as shown below.

22. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Classroom Practice Problems Find the gravitational force that Earth (m E = 5.97  10 24 kg) exerts on the moon (m m = 7.35  10 22 kg) when the distance between them is 3.84 x 10 8 m. –Answer: 1.99 x 10 20 N Find the strength of the gravitational field at a point 3.84 x 10 8 m from the center of Earth. –Answer: 0.00270 N/m or 0.00270 m/s 2

23. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company RPM = rotations per minute  T If 33 1/3 RPM Then 1 minute = 33 1/3 rotations So 60 seconds = 33 1/3 rotations And 60/33.333 = the period = T

24. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company What do you think? Electric forces and gravitational forces are both field forces. Two charged particles would feel the effects of both fields. Imagine two electrons attracting each other due to the gravitational force and repelling each other due to the electrostatic force. Which force is greater? Is one slightly greater or much greater than the other, or are they about the same? What evidence exists to support your answer?

25. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Coulomb’s Law The force between two charged particles depends on the amount of charge and on the distance between them. –Force has a direct relationship with both charges. –Force has an inverse square relationship with distance.

26. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Coulomb’s Law Use the known units for q, r, and F to determine the units of k c. –k c = 9  10 9 Nm 2 /C 2 The distance (r) is measured from center to center for spherical charge distributions.

27. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Classroom Practice Problem The electron and proton in a hydrogen atom are separated, on the average, a distance of about 5.3  10 -11 m. Find the magnitude of both the gravitational force and the electric force acting between them. –Answer: F e = 8.2  10 -8 N, F g = 3.6  10 -47 N The electric force is more than 10 39 times greater than the gravitational force. –Atoms and molecules are held together by electric forces. Gravity has little effect.

28. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Compare and contrast

29. Circular Motion and GravitationSection 2 © Houghton Mifflin Harcourt Publishing Company Classroom Practice Problem A balloon is rubbed against a small piece of wool and receives a charge of -0.60  C while the wool receives an equal positive charge. Assume the charges are located at a single point on each object and they are 3.0 cm apart. What is the force between the balloon and wool? Answer: 3.6 N attractive

30. Circular Motion and GravitationSection 4 © Houghton Mifflin Harcourt Publishing Company Torque Where should the cat push on the cat-flap door in order to open it most easily? –The bottom, as far away from the hinges as possible Torque depends on the force (F) and the length of the lever arm (d).

31. Circular Motion and GravitationSection 4 © Houghton Mifflin Harcourt Publishing Company Torque Torque also depends on the angle between the force (F) and the distance (d). Which situation shown above will produce the most torque on the cat-flap door? Why? –Figure (a), because the force is perpendicular to the distance

32. Circular Motion and GravitationSection 4 © Houghton Mifflin Harcourt Publishing Company Torque SI units: Nm –Not joules because torque is not energy The quantity “d ” is the perpendicular distance from the axis to the direction of the force.

33. Circular Motion and GravitationSection 4 © Houghton Mifflin Harcourt Publishing Company Torque as a Vector Torque has direction. –Torque is positive if it causes a counterclockwise rotation. –Torque is negative if it causes a clockwise rotation. Are the torques shown to the right positive or negative? –The wrench produces a positive torque. –The cat produces a negative torque. Net torque is the sum of the torques.

34. Circular Motion and GravitationSection 4 © Houghton Mifflin Harcourt Publishing Company Classroom Practice Problems Suppose the force on the wrench is 65.0 N and the lever arm is 20.0 cm. Calculate the torque. If that was not enough torque to do the job, what could you do? Using a cheat pipe, you increase the lever arm to 60.0 cm. What is the torque now?