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MAIN IDEA

Circular Motion. SECTION 6.2. MAIN IDEA An object in circular motion has an acceleration toward the circle’s center due to an unbalanced force toward the circle’s center. Why is an object moving in a circle at a constant speed accelerating?

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MAIN IDEA

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  1. Circular Motion SECTION6.2 MAIN IDEA An object in circular motion has an acceleration toward the circle’s center due to an unbalanced force toward the circle’s center. • Why is an object moving in a circle at a constant speed accelerating? • How does centripetal acceleration depend upon the object’s speed and the radius of the circle? • What causes centripetal acceleration? Essential Questions

  2. New Vocabulary Uniform circular motion Centripetal acceleration Centripetal force Circular Motion SECTION6.2 • Review Vocabulary • Average velocity the change in position divided by the time during which the change occurred; the slope of an object’s position-time graph.

  3. Circular Motion SECTION6.2 Describing Circular Motion Click image to view movie.

  4. Circular Motion SECTION6.2 Centripetal Acceleration (cont.) • Solve the equation for acceleration and give it the special symbol ac, for centripetal acceleration. • Centripetal acceleration always points to the center of the circle. Its magnitude is equal to the square of the speed, divided by the radius of motion.

  5. Circular Motion SECTION6.2 Centripetal Acceleration (cont.) • One way of measuring the speed of an object moving in a circle is to measure its period, T, the time needed for the object to make one complete revolution. • During this time, the object travels a distance equal to the circumference of the circle, 2πr. The object’s speed, then, is represented by v = 2πr/T. • Velocity is always tangent to the circular motion

  6. Circular Motion SECTION6.2 Centripetal Acceleration (cont.) • The acceleration of an object moving in a circle is always in the direction of the net force acting on it, there must be a net force toward the center of the circle. This force can be provided by any number of agents.

  7. Circular Motion SECTION6.2 Centripetal Acceleration (cont.) • When an Olympic hammer thrower swings the hammer, the force is the tension in the chain attached to the massive ball. • However, when the thrower releases the hammer, it will go in the direction of its velocity.

  8. Circular Motion SECTION6.2 Centripetal Acceleration (cont.) • When an object moves in a circle, the net force toward the center of the circle is called the centripetal force. • To analyze centripetal acceleration situations accurately, you must identify the agent of the force that causes the acceleration. Then you can apply Newton’s second law for the component in the direction of the acceleration in the following way.

  9. Circular Motion • Newton’s Second Law for Circular Motion SECTION6.2 Centripetal Acceleration (cont.) • The net centripetal force on an object moving in a circle is equal to the object’s mass times the centripetal acceleration.

  10. Circular Motion SECTION6.2 Centripetal Acceleration (cont.) • When solving problems, it is useful to choose a coordinate system with one axis in the direction of the acceleration. • For circular motion, the direction of the acceleration is always toward the center of the circle.

  11. Circular Motion SECTION6.2 Centrifugal “Force” • According to Newton’s first law, you will continue moving with the same velocity unless there is a net force acting on you. • The passenger in the car would continue to move straight ahead if it were not for the force of the car acting in the direction of the acceleration.

  12. Circular Motion SECTION6.2 Centrifugal “Force” (cont.) • The so-called centrifugal, or outward force, is a fictitious, nonexistent force. • You feel as if you are being pushed only because you are accelerating relative to your surroundings. There is no real force because there is no agent exerting a force.

  13. Section Check SECTION6.2 Explain why an object moving in a circle at a constant speed is accelerating.

  14. Section Check SECTION6.2 Answer Acceleration is the rate of change of velocity, the object is accelerating due to its constant change in the direction of its motion.

  15. Section Check SECTION6.2 What is the direction of the velocity vector of an accelerating object? A. toward the center of the circle B. away from the center of the circle C. along the circular path D. tangent to the circular path

  16. Section Check SECTION6.2 Answer Reason:While constantly changing, the velocity vector for an object in uniform circular motion is always tangent to the circle. Vectors are never curved and therefore cannot be along a circular path.

  17. Relative Velocity SECTION6.3 MAIN IDEA An object’s velocity depends on the reference frame chosen. • What is relative velocity? • How do you find the velocities of an object in different reference frames? Essential Questions

  18. New Vocabulary Reference frame – a coordinate system from which motion is viewed is a frame of reference Relative Velocity SECTION6.3 • Review Vocabulary • resultant a vector that results from the sum of two other vectors.

  19. Relative Velocity SECTION6.3 Relative Motion in One Dimension • Suppose you are in a school bus that is traveling at a velocity of 8 m/s in a positive direction. You walk with a velocity of 1 m/s toward the front of the bus. • If a friend of yours is standing on the side of the road watching the bus go by, how fast would your friend say that you are moving?

  20. Relative Velocity SECTION6.3 Relative Motion in One Dimension (cont.) • If the bus is traveling at 8 m/s, this means that the velocity of the bus is 8 m/s, as measured by your friend in a coordinate system fixed to the road. • When you are standing still, your velocity relative to the road is also 8 m/s, but your velocity relative to the bus is zero.

  21. In the previous example, your motion is viewed from different coordinate systems. A coordinate system from which motion is viewed is a reference frame. Relative Velocity SECTION6.3 Relative Motion in One Dimension (cont.)

  22. Walking at 1m/s towards the front of the bus means your velocity is measured in the reference from of the bus. Your velocity in the road’s reference frame is different You can rephrase the problem as follows: given the velocity of the bus relative to the road and your velocity relative to the bus, what is your velocity relative to the road? Relative Velocity SECTION6.3 Relative Motion in One Dimension (cont.)

  23. Relative Velocity SECTION6.3 Relative Motion in One Dimension (cont.) • When a coordinate system is moving, two velocities are added if both motions are in the same direction, and one is subtracted from the other if the motions are in opposite directions. • In the given figure, you will find that your velocity relative to the street is 9 m/s, the sum of 8 m/s and 1 m/s.

  24. Relative Velocity SECTION6.3 Relative Motion in One Dimension (cont.) • You can see that when the velocities are along the same line, simple addition or subtraction can be used to determine the relative velocity.

  25. Relative Velocity SECTION6.3 Relative Motion in One Dimension (cont.) • Mathematically, relative velocity is represented asvy/b+vb/r = vy/r. • The more general form of this equation is: Relative Velocityva/b + vb/c + va/c • The relative velocity of object a to object c is the vector sum of object a’s velocity relative to object b and object b’s velocity relative to object c.

  26. Relative Velocity SECTION6.3 Relative Motion in One Dimension (cont.) • Can a moving car be a frame of reference? • Yes

  27. Relative Velocity SECTION6.3 Relative Motion in Two Dimensions • The method for adding relative velocities also applies to motion in two dimensions. • As with one-dimensional motion, you first draw a vector diagram to describe the motion and then you solve the problem mathematically by resolving vectors into x and y components and finding the resultant magnitude and direction

  28. Relative Velocity SECTION6.3 Relative Motion in Two Dimensions (cont.) • For example, airline pilots must take into account the plane’s speed relative to the air, and their direction of flight relative to the air. They also must consider the velocity of the wind at the altitude they are flying relative to the ground.

  29. Relative Velocity SECTION6.3 Relative Motion in Two Dimensions (cont.)

  30. Relative Velocity SECTION6.3 Relative Motion in Two Dimensions (cont.) How are vectors used to describe relative motion in two dimensions? Resolve the velocity vectors into their x and y components. Each component gives the speed in the corresponding direction relative to a given reference frame.

  31. Relative Velocity SECTION6.3 Relative Velocity of a Marble Ana and Sandra are riding on a ferry boat that is traveling east at a speed of 4.0 m/s. Sandra rolls a marble with a velocity of 0.75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water?

  32. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Step 1: Analyze and Sketch the Problem • Establish a coordinate system.

  33. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) • Draw vectors to represent the velocities of the boat relative to the water and the marble relative to the boat.

  34. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Identify known and unknown variables. Known: vb/w = 4.0 m/s vm/b = 0.75 m/s Unknown: vm/w = ?

  35. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Step 2: Solve for the Unknown

  36. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Because the two velocities are at right angles, use the Pythagorean theorem.

  37. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s

  38. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Find the angle of the marble’s motion.

  39. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s = 11° north of east The marble is traveling 4.1 m/s at 11° north of east.

  40. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Step 3: Evaluate the Answer

  41. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Are the units correct? Dimensional analysis verifies that the velocity is in m/s. Do the signs make sense? The signs should all be positive. Are the magnitudes realistic? The resulting velocity is of the same order of magnitude as the velocities given in the problem.

  42. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Step 1: Analyze and Sketch the Problem Establish a coordinate system. Draw vectors to represent the velocities of the boat relative to the water and the marble relative to the boat. The steps covered were:

  43. Relative Velocity SECTION6.3 Relative Velocity of a Marble (cont.) Step 2: Solve for the Unknown Use the Pythagorean theorem. Step 3: Evaluate the Answer The steps covered were:

  44. Section Check SECTION6.3 Steven is walking on the top level of a double-decker bus with a velocity of 2 m/s toward the rear end of the bus. The bus is moving with a velocity of 10 m/s. What is the velocity of Steven with respect to Anudja, who is sitting on the top level of the bus and to Mark, who is standing on the street? A. The velocity of Steven with respect to Anudja is 2 m/s and is 12 m/s with respect to Mark. B. The velocity of Steven with respect to Anudja is 2 m/s and is 8 m/s with respect to Mark. C. The velocity of Steven with respect to Anudja is 10 m/s and is 12 m/s with respect to Mark. D. The velocity of Steven with respect to Anudja is 10 m/s and is 8 m/s with respect to Mark.

  45. Section Check SECTION6.3 Answer Reason:The velocity of Steven with respect to Anudja is 2 m/s since Steven is moving with a velocity of 2 m/s with respect to the bus, and Anudja is at rest with respect to the bus. The velocity of Steven with respect to Mark can be understood with the help of the following vector representation.

  46. Section Check SECTION6.3 Which of the following formulas correctly relates the relative velocities of objects a, b, and c to each other? A.va/b + va/c = vb/c B.va/bvb/c = va/c C.va/b + vb/c = va/c D.va/bva/c = vb/c

  47. Section Check SECTION6.3 Answer Reason:The relative velocity equation is va/b + vb/c = va/c. The relative velocity of object a to object c is the vector sum of object a’s velocity relative to object b and object b’s velocity relative to object c.

  48. Section Check C. D. SECTION6.3 An airplane flies due south at 100 km/hr relative to the air. Wind is blowing at 20 km/hr to the west relative to the ground. What is the plane’s speed with respect to the ground? A. (100 + 20) km/hr B. (100 − 20) km/hr

  49. Section Check SECTION6.3 Answer Reason:Since the two velocities are at right angles, we can apply the Pythagorean theorem. By using relative velocity law, we can write: vp/a2 + va/g2 = vp/g2

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