1. Arc Length Formula Pre-Calculus Unit #4, Day 5

2. Arc Length and Central Angles

3. Example Find the measure of a rotation in radians when a point 2 m from the center of rotation travels 4 m.

4. Example Find the length of an arc of a circle of radius 5 cm associated with an angle of  /3 radians.

5. Linear & Angular Velocity Things that turn have both a linear velocity and an angular velocity. www.vcsc.k12.in.us/lincoln/math/CaseNotes/.../LinearAngVel.ppt

6. Linear Velocity Linear Velocity is distance/time: Definition: Linear Speed Distance Time

7. Angular Velocity Angular Velocity is turn/time: Definition: Angular Speed (omega) Rotation in radians Time

8. Linear & Angular Velocity Definition of Linear Velocity: Recall Arc Length Formula  Linear Velocity in terms of Angular Velocity:

9. Let us take 2 pendulums hung on a slim rotating rod for analysis. If the 2 pendulums (A and B) rotate one full cycle, the time taken by them is the same. They covered the same amount of angular distance (360 degree) within the same amount of time. This showed that they have exactly the SAME angular speed. But is the Linear speed the same? http://mathsisinteresting.blogspot.com/2008/08/linear-angular-speed.html

10. Let us take 2 pendulums hung on a slim rotating rod for analysis. The length of the 2 circumferences travelled by the individual pendulums are not the same. The linear length or distance is therefore NOT the same. Length = 2 x (pi) x radius. They took the same time to complete one full cycle, though. The linear speed is thus DIFFERENT, having travelled different length for the same amount of time. http://mathsisinteresting.blogspot.com/2008/08/linear-angular-speed.html

11. Example: A satellite traveling in a circular orbit approximately 1800 km. above the surface of Earth takes 2.5 hrs. to make an orbit. The radius of the earth is approximately 6400 km. a) Approximate the linear speed of the satellite in kilometers per hour. b) Approximate the distance the satellite travels in 3.5 hrs.

12. Example: r = 6400 + 1800 = 8200 t = 2.5 hrs. a) Approximate the linear speed of the satellite in kilometers per hour.

13. Example: r = 6400 + 1800 = 8200 t = 2.5 hrs. a) Approximate the linear speed of the satellite in kilometers per hour.

14. Example: r = 6400 + 1800 = 8200 t = 2.5 hrs. a) Approximate the linear speed of the satellite in kilometers per hour.

15. Example: r = 6400 + 1800 = 8200 t = 2.5 hrs. b) Approximate the distance the satellite travels in 3.5 hrs.

16. Example: r = 6400 + 1800 = 8200 t = 2.5 hrs. b) Approximate the distance the satellite travels in 3.5 hrs.

17. Example: r = 6400 + 1800 = 8200 t = 2.5 hrs. b) Approximate the distance the satellite travels in 3.5 hrs.

18. A small pulley 6 cm in diameter is connected by a belt to a larger pulley 15cm in diameter. The small pulley is turning at 120 rpm. a) Find the angular velocity of the small pulley in radians per second. b) Find the linear velocity of the rim of the small pulley.

19. A small pulley 6 cm in diameter is connected by a belt to a larger pulley 15cm in diameter. The small pulley is turning at 120 rpm. a) Find the angular velocity of the small pulley in radians per second.

20. A small pulley 6 cm in diameter is connected by a belt to a larger pulley 15cm in diameter. The small pulley is turning at 120 rpm. a) Find the angular velocity of the small pulley in radians per second.

21. A small pulley 6 cm in diameter is connected by a belt to a larger pulley 15cm in diameter. The small pulley is turning at 120 rpm. b) Find the linear velocity of the rim of the small pulley.

22. A small pulley 6 cm in diameter is connected by a belt to a larger pulley 15cm in diameter. The small pulley is turning at 120 rpm. b) Find the linear velocity of the rim of the small pulley.

23. A small pulley 6 cm in diameter is connected by a belt to a larger pulley 15cm in diameter. The small pulley is turning at 120 rpm. b) Find the linear velocity of the rim of the small pulley.