1. General Physics 1, Lec 8 By/ T.A. Eleyan 1 Lecture 8 Circular Motion & Relative Velocity

2. General Physics 1, Lec 8 By/ T.A. Eleyan 2 Circular Motion  Consider an object moving at constant speed in a circle. The direction of motion is changing, so the velocity is changing (even though speed is constant). Therefore, the object is accelerating.  The direction of the acceleration is toward the center of the circle and so we call it centripetal acceleration.  The magnitude of the acceleration is

3. General Physics 1, Lec 8 By/ T.A. Eleyan 3 Centripetal Acceleration The best estimate of the acceleration at P is found by calculating the average acceleration for the symmetric interval 1  2.

4. General Physics 1, Lec 8 By/ T.A. Eleyan 4 if Then,

5. General Physics 1, Lec 8 By/ T.A. Eleyan 5 Example: What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun? Solution: But Then

6. General Physics 1, Lec 8 By/ T.A. Eleyan 6 Tangential acceleration The tangential acceleration component causes the change in the speed of the particle. This component is parallel to the instantaneous velocity, and is given by Tangential and Radial acceleration Note: If the speed is constant then the tangential acceleration is zero (uniform Circular Motion)

7. General Physics 1, Lec 8 By/ T.A. Eleyan 7 The radial acceleration component arises from the change in direction of the velocity vector and is given by Radial acceleration

8. General Physics 1, Lec 8 By/ T.A. Eleyan 8 Total acceleration The total acceleration vector a can be written as the vector sum of the component vectors: Since the component perpendicular to other

9. General Physics 1, Lec 8 By/ T.A. Eleyan 9 Example: A car exhibits a constant acceleration of 0.300 m/s 2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like a circle of radius 500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What is the direction of the total acceleration vector for the car at this instant?

10. General Physics 1, Lec 8 By/ T.A. Eleyan 10 If the angle between

11. General Physics 1, Lec 8 By/ T.A. Eleyan 11 Problem: A train slows down as it rounds a sharp horizontal turn, slowing from 90km/h to 50km/h in the 15s that it takes to round the bend. The radius of the curve is 150m. Compute the acceleration at the train.

12. General Physics 1, Lec 8 By/ T.A. Eleyan 12 Example: A particle moves in a circular path 0.4m in radius with constant speed. If the particle makes five revolution in each second of its motion, find: (a)The speed of the particle. (b) Its acceleration. (a) Since r =0.4m, the particle travels a distance 0f 2 r = 2.51m in each revolution. Therefore, it travels a distance of 12.57m in each second (since it makes 5 rev. in the second). v = 12.57m/1sec = 12.6 m/s (b)

13. General Physics 1, Lec 8 By/ T.A. Eleyan 13 Centripetal Force A string cannot push sideways or lengthwise. A string in tension only pulls. The string pulls the ball inward toward the center of the circle

14. General Physics 1, Lec 8 By/ T.A. Eleyan 14 What if we cut the sting? The ball should move off with constant velocity This means the ball will continue along the tangent to the circle.

15. General Physics 1, Lec 8 By/ T.A. Eleyan 15 Centripetal Force If there is a centripetal acceleration, then the net force must also be a centripetal force:

16. General Physics 1, Lec 8 By/ T.A. Eleyan 16 The Conical Pendulum As the ball revolves faster, the angle increases What’s the speed for a given angle? Example:

17. General Physics 1, Lec 8 By/ T.A. Eleyan 17

18. General Physics 1, Lec 8 By/ T.A. Eleyan 18 Problem: I rotate a ball at an angle of 30 o. What is the centripetal acceleration? If the string is 1 meter long, how fast is it rotating?

19. General Physics 1, Lec 8 By/ T.A. Eleyan 19 Problem Driving in your car with a constant speed of 12 m/s, you encounter a bump in the road that has a circular cross section, as indicated in the Figure. If the radius of curvature of the bump is 35 m, find the apparent weight of a 67-kg person in your car as you pass over the top of the bump. N mg a=v 2 /r

20. General Physics 1, Lec 8 By/ T.A. Eleyan 20 Relative Velocity Two observers moving relative to each other generally do not agree on the outcome of an experiment For example, observers A and B below see different paths for the ball

21. General Physics 1, Lec 8 By/ T.A. Eleyan 21 Relative Velocity equations The positions as seen from the two reference frames are related through the velocity The derivative of the position equation will give the velocity equation These are called the Galilean transformation equations

22. General Physics 1, Lec 8 By/ T.A. Eleyan 22 Central concept for problem solving: “x” and “y” components of motion treated independently. Again: man on the cart tosses a ball straight up in the air. You can view the trajectory from two reference frames: Reference frame on the ground. Reference frame on the moving train. y(t) motion governed by 1) a = -g y 2) v y = v 0y – g t 3) y = y 0 + v 0y – g t 2 /2 x motion: x = v x t Net motion : R = x(t) i + y(t) j (vector)

23. General Physics 1, Lec 8 By/ T.A. Eleyan 23 Acceleration in Different Frames of Reference The derivative of the velocity equation will give the acceleration equation  v’ = v – v o  a’ = a The acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer moving at a constant velocity relative to the first frame.

24. General Physics 1, Lec 8 By/ T.A. Eleyan 24 Questions [1]You are on a train traveling 40 mph North. If you walk 5 mph toward the front of the train, what is your speed relative to the ground? A) 45 mphB) 40 mphC) 35 mph [2]You are on a train traveling 40 mph North. If you walk 5 mph toward the rear of the train, what is your speed relative to the ground? A) 45 mphB) 40 mphC) 35 mph [3]You are on a train traveling 40 mph North. If you walk 5 mph sideways across the car, what is your speed relative to the ground? A) 40 mph