POP QUIZ: 1. List the SI Units for the following: (a) Acceleration: (b) Displacement: (c) Velocity. (d) Time. (e) Speed.

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1 POP QUIZ: 1. List the SI Units for the following: (a) Acceleration: (b) Displacement: (c) Velocity (d) Time (e) Speed (f) Distance

2 NOTES 3.3 2D Motion: Uniform Circular Motion Physics Honors I

3 OBJECTIVES: Explain why an object moving in a circle at a constant speed is accelerated. Describe how centripetal acceleration depends upon the object s speed and the radius of the circle. Identify the force that causes centripetal acceleration. Define Uniform Circular Motion, Centripetal Acceleration, Centripetal Force. Briefly Discuss Relative Motion.

4 Active Physics Reference: Chapter 2, Section 7: Centripetal Force Page , 426 Chapter 4, Section 10: Safety is Required but Thrills are Desired Page Physics: Principles and Problems (RED BOOK in classroom) Chapter 6, Section 2: Circular Motion Page Chapter 6, Section 3: Relative Velocity Page

5 Further Learning: Khan Academy Centripetal Acceleration: Physics Classroom Uniform Circular Motion: Crash Course Physics: Uniform Circular Motion

6 Crash Course Physics:

7 Swing and a Miss: Imagine swinging a ball on a string around and around in a circular motion. Let s imagine you are swinging it with a constant velocity. Is there any acceleration? YES!!! There is an acceleration.

8 Swing and a Miss: Normally, when we discuss constant velocity, it means there is no acceleration. Right? However, in the case of uniform circular motion, there is an acceleration even though the ball is being swung at a constant velocity. How can this be?

9 Change in Velocity, doesn t mean a change in speed: Remember that we have been learning about motion in twodimensions thus velocity has two components where one is on the x- axis, v x, and one on the y-axis, v y.

10 Change in Velocity, doesn t mean a change in speed: If we look at the ball being swung a string from an overhead view, we will see that it make a circle. Now let s take this circle that the ball is making and analyze it on an x-y plane.

11 Change in Velocity, doesn t mean a change in speed: So, what we are seeing is that the ball has a constant speed; however, it is shifting between the x-axis and y-axis. What this means is that the velocity on the x-axis, v x, is changing while the velocity on the y-axis, v y, is also changing. Because there is a change in velocity on the x-axis and a change in velocity on the y-axis, we have an acceleration. Remember, there is an acceleration anytime there is a change in velocity.

12 Change in Velocity, doesn t mean a change in speed: Recall that: Also, recall that: a x = v x t and a y = v y t Ԧa = a x 2 + a y 2

13 Uniform Circular Motion: So, this is what uniform circular motion is: Something going around in a circle at a constant velocity. That is, a particle is said to have uniform circular motion if it travels around a circle or a circular arc at constant (uniform) speed. Although the speed does not vary, the particle is accelerating because the velocity changes in direction. The velocity is always directed tangent to the circle in the direction of motion. The acceleration is directed radially inward.

14 Uniform Circular Motion: The acceleration is directed radially inward. (Velocity and acceleration vectors for uniform circular motion).

15 Uniform Circular Motion: The acceleration is directed radially inward. IMPORTANT: The acceleration associated with uniform circular motion is called a centripetal acceleration.

16 Centripetal Acceleration: Centripetal means center-seeking. The magnitude of centripetal acceleration Ԧa is: Where, a = acceleration v = velocity r = radius a = v2 r

17 Example 1: The moon orbits about the Earth with an average speed of just over 1000 m/s. If the distance from the Earth to the moon is 384,400 km, what is the acceleration of the moon around the Earth? 1. Convert 384,400 km to meters. 2. Use equation a = Τ v 2 r. 3. Plug in and solve for mτ s 2.

18 Check Point 1: 1. What is the magnitude of the acceleration of a sprinter running 10 m/s when rounding a turn on a radius 25 m. 2. A 500-kg roller coaster car is at the top of the loop on the Shockwave. The radius of the loop is 4.0 m and the speed is 8.0 m/s. What is the acceleration? 3. A rock tied to a string is moving at a constant speed of in a circle of radius Calculate the approximate magnitude of the centripetal acceleration of the rock.

19 Periodic Motion: So, we know that the ball goes in a circle thus it goes around and around in a circle and crosses the same point over and over again. This is called periodic motion.

20 Periodic Motion: Previously when dealing with constant velocity, we used the formula: v = x t We can still use this formula but with some adjustments. What is the distance traveled in a circle?

21 Periodic Motion: Well the distance the ball travels is the circumference of a circle. C = 2πr, where r is the radius of the circle. In our example, r is essentially just the length of the string that the ball is attached to. Thus the x can be replaced by the circumference of the circle so that we get: v = 2πr t

22 Periodic Motion: We know the velocity is constant and we don t need to change it. But let s solve the equation for time by multiplying both sides by time, then dividing both sides by the velocity. This gives us the equation: t = 2πr v

23 Periodic Motion: In regards to uniform circular motion, we call the time it takes for an object to make a loop around the circle to be its period of revolution, and we write it as an uppercase T instead of a lowercase t so that we can distinguish time t from a period T. So our equation for determining the period is: T = 2πr v t = total time that the ball may be traveling. T = the time for it to make one complete circle. The period, T, is the time it takes for a particle to go around a closed path exactly once.

24 Periodic Motion: Also, mathematically, we can combine the two equations of a = v2 and T = 2πr v by solving the periodic equation for r, and substituting it into the acceleration equation to get: a = 2πv T One other equation that we can attain from the combination of these two equations is: a = 4π2 r T 2 r

25 Periodic Motion: Frequency the number of rotations per second. f = 1 T Or T = 1 f NOTE: SI UNITS FOR FREQUENCY IS HERTZ or 1/s.

26 Example 2: A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what distance does the tip move in one revolution? (b) What is the blade tip s speed? (c) What is the magnitude of the blades acceleration? (d) What is the period of the motion?

27 Check Point 2: 4. An Earth satellite moves in a circular orbit 640 km above Earth s surface with a period of 98.0 min. (a) What is the speed of the satellite? (b) What is the magnitude of the centripetal acceleration of the satellite?

28 Relative Velocity: Suppose that 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 3 m/s toward the back of the bus. Your friend is on the street and sees this all taking place. From your friends point of view, how fast are you moving? 8 m/s 3 m/s = 5 m/s

29 Relative Velocity: What is going on here? Well we have different frames of reference going on. There is the motion of the bus relative to the earth at 8 m/s There is the motion of you, the passenger, relative to the bus. There is the motion of you, the passenger, relative to your friend on the side of the street.

30 Relative Velocity: What is going on here? Well we have different frames of reference going on. There is the motion of the bus relative to the earth at 8 m/s There is the motion of you, the passenger, relative to the bus. There is the motion of you, the passenger, relative to your friend on the side of the street.

31 Relative Velocity: So the frame of reference is different for different observers. Relative motion is the calculation of the motion of an object with regard to some other moving object. Thus, the motion is not calculated with reference to the earth, but is the velocity of the object in reference to the other moving object as if it were in a static state. NOTE: Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving particle.

32 Relative Velocity: Suppose that Alex (at the origin frame A) is parked by the side of a highway, watching car P (the particle ) speed past. Barbara (at the origin of the frame B) is driving along the highway at constant speed and is also watching car P. Suppose that they both measure the velocity of the car at a given moment. We could quantify this as: v PA = v PB + v BA The velocity v PA of P as measured by observer A is equal to the velocity of v PB of P as measured by observer B plus the velocity of v BA of B as measured by observer A.

33 Check Point 3: 5. Suppose that Barbara s velocity relative to Alex is constant v BA = 52 km h and car P is moving in the negative direction of the x axis. If Alex measures a constant v PA = 78 km h v PB will Barbara measure? for car P, what velocity 78 km h = v PB = 52 km h v PB = 130 km h

34 Exit Ticket: 6. Accelerating objects are. Choose the one most inclusive answer. a. going fast b. speeding up (only) c. speeding up or slowing downd. changing their velocity 7. (True or False) If an object moves in a circle at a constant speed, its velocity vector will be constant. 8. You are at the airport. The moving walkway is moving at 3 m/s. You are in a rush and running at 7 m/s. To an observer waiting at their airport terminal gate, how fast are you traveling?