Example using equation 1

1. Example using equation 1 • An animal jumps vertically. The animal’s takeoff or initial velocity is 3 m/s. How long does it take to reach the highest point of the jump? (Air resistance negligible). • - 0.3 s • 0.03 s • - 0.03 s • 0.3 s

2. Example using equation 1 • An animal jumps vertically. The animal’s takeoff or initial velocity is 3 m/s. How long does it take to reach the highest point of the jump? (Air resistance negligible). • vi = 3 m / s; a = -9.81 m / s2 • At point when animal is highest: vf = 0 • vf= vi + at ---> t = (vf - vi) / a • t = (0 - 3) / (-9.81) = 0.3 seconds

3. Example using equation 2 What was the net jump height of the animal in example #1. In other words, how high did it jump once its feet left the ground? • 46 m • 46 cm • 1.34 m • -1.34 m

4. Example using equation 2 • What was the net jump height of the animal in example #1 Knowns: vi = 3 m/s; t = 0.3 sec, a = -9.81 m/s2 Want to solve for rf rf - ri = vit + 1/2 * at2 ri = 0 when legs are extended rf = 3t + (0.5) (-9.81) (t2) rf = 3 *(0.3) + (0.5) (-9.81) (0.3)2 rf = 0.9 + (0.5) (-9.81) (0.09) rf = 0.9 + (-0.44) = 0.46 m about 18 inches

5. Example using Equation 3 A runner starts at a standstill and accelerates uniformly at 1 m/s2. How far has she travelled at the point when she reaches her maximum speed of 10 m/s? • 5 m • 10 m • 50 m • 100 m

6. Example using Equation 3 • A runner starts at a standstill and accelerates uniformly at 1 m/s2. How far has she travelled at the point when she reaches her maximum speed of 10 m/s? • Given: vmax= 10 m/s; a=1 m/s2; Dr=? vf2 = vi2 + 2a (rf - ri) ∆r = (vf2 - vi2)/ 2a ∆r = (102 - 02) / (2 * 1) ∆r = 100 / 2 = 50 meters

7. A person sits in a chair and does a knee extension. The knee angle changes by 0.5 radians in 0.5 seconds. What is the magnitude of the angular velocity of the knee, the linear velocity of the foot and the linear displacement of the foot? (Shank length is 0.5 meters) • 1 rad/s, 0.5 m/s, 0.25rad • 0.5 rad/s, 0.5 rad/s, 0.25rad • 1 rad/s, 0.5 m/s, 0.25 m • 0.25 rad/s, 0.5 m/s, 0.25m

8. Given: ∆=0.5rad ∆t = 0.5s r=0.5m • Angular velocity of the knee •  = ∆ / ∆t = 0.5 rad / 0.5 s = 1 rad / s • Linear velocity of the foot • v = r = (1 rad/s) * (0.5 m) = 0.5 m/s • Linear displacement of foot • s = r = (0.5 meters)(0.5 rad) = 0.25 m

9. A 100 kg person is riding a bike around a corner (radius = 2 m) with a constant angular velocity of 1 rad /sec. What were the magnitudes of the tangential & radial components of the acceleration? • at = 981 m/s2; ar= 2 m/s2 • at = 0 m/s2; ar= 4 m/s2 • at = 0 m/s2; ar= 2 m/s2 • at = 0 m/s2; ar= 2 rad/s2 • E) at = 981 m/s2; ar= 4 rad/s2

10. A 100 kg person is riding a bike around a corner (radius = 2 m) with a constant angular velocity of 1 rad /sec. What were the magnitudes of the tangential & radial components of the acceleration? Given:  = 0; r = 2 m;  = 1 rad/sec at = r at = r= (2 m)(0 rad/s2) = 0 m/s2 ar = r2 ar= (2 meters) (1 rad/s)2 = 2 m/s2

11. Angular Kinematics Example # 1 A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed for the duration of the race. a. How long does it take for her to run the 200 m?   b. What is her speed when she crosses the finish line? c. What is her angular velocity when she finishes? d. What was her angular acceleration during the first two seconds? e. What is her angular acceleration when she finishes? f. Determine her radial and tangential accelerations 1 second after the start of the race.

12. Angular Kinematics Example # 1 A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed for the duration of the race. • How long does it take for her to run the 200 m? Given: Solve for total time of race Dr = 200m m=60kg From t = 0s to t= 2s at=5 m/s2 From t = 2s to end at=0 m/s2 To solve for the time, you need to break the race into two time periods: one with a constant tangential acceleration (at) of 5 m/s2 (the first two seconds) and one with a constant tangential acceleration of 0 m/s2 (after two seconds have gone by). Find the distance traveled in the first 2 seconds: Dr = vit + 0.5 at2 = (0)(2) + 0.5(5)(2)2 = 10 m After the first two seconds, the runner has (200 m - 10 m) = 190 m left to go. Find the tangential velocity after the 2 seconds of acceleration: v2sec = v0sec + at = 0 + (5)(2) = 10 m/s Find the time to complete the remaining distance at this velocity: Dr = vit + 0.5 at2 = (10)t + 0 = 190 m t = 19 s The total time is 19 s plus the first 2 s which equals 21s.

13. Angular Kinematics Example # 1 A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed for the duration of the race. Given: Dr = 200m m=60kg From t = 0s to t= 2s at=5 m/s2 From t = 2s to end at=0 m/s2 b. What is her speed when she crosses the finish line? Since there is no tangential acceleration after the first two seconds, her speed is constant. Therefore her speed when she crosses the finish line is 10 m/s. c. What is her angular velocity when she finishes? Since the circumference of the track is 200 m = 2pr , then r = 31.83 m. Her angular velocity when she finishes is w = v / r = (10m/s) / (31.83m) = 0.31 rad/s

14. Angular Kinematics Example # 1 A 60 kg sprinter is running a race on an unbanked circular track with a circumference of 200 m. At the start, she increases speed at a constant rate of 5 m/s2 for 2 seconds. After two seconds, she maintains a constant speed for the duration of the race. d. What was her angular acceleration during the first two seconds? Her angular acceleration during the first two seconds is a = at / r = (5 m/s2) / (31.83 m) = 0.16 rad/s2 e. What is her angular acceleration when she finishes? a = at / r = (0 m/s2) / (31.83 m) = 0 rad/s2 f. Determine her radial and tangential accelerations 1 second after the start of the race. To find the radial acceleration after one second, you need to find the angular velocity after one second. Remember the angular acceleration a = 0.16 rad/s2 (from part d) and the initial angular velocity w0s = 0 a = Dw/t Dw= w1s – w0s = at w1s = at +w0s = 0.16(1) + 0 = 0.16 rad/s Therefore her radial acceleration after one second is ar = rw1s2 = (31.83)(.16)2 = 0.81 m/s2 The tangential acceleration after one second is given, at = 5 m/s2

15. Usain Bolt • http://www.youtube.com/watch?v=o-urnlaJpOA

16. How fast could Usain Bolt have run?A Dynamical Study • http://www.youtube.com/watch?v=o-urnlaJpOA • Eriksen et al. 2008 – could 9.69s have been 9.52s?

17. Could Usain Bolt have run a faster 100m? • http://www.youtube.com/watch?v=o-urnlaJpOA • Eriksen et al. 2008 – could 9.62s have been 9.52s?

21. Could Usain Bolt have run a faster 100m? • http://www.youtube.com/watch?v=o-urnlaJpOA • Eriksen et al. 2008 – could 9.69s have been 9.52s? • Close – but more like 9.55s