1. Day 17 – June 17 – WBL 8.4-8.5 8.4 Rotational Work and Kinetic Energy PC141 Intersession 2013Slide 1

2. Day 17 – June 17 – WBL 8.4-8.5 8.4 Rotational Work and Kinetic Energy PC141 Intersession 2013Slide 2

3. Day 17 – June 17 – WBL 8.4-8.5 8.4 Rotational Work and Kinetic Energy PC141 Intersession 2013Slide 3

4. Day 17 – June 17 – WBL 8.4-8.5 Problem #1: Bowling Ball PC141 Intersession 2013Slide 4 WBL LP 8.15 A bowling ball rolls without slipping on a flat surface. The ball has… A …rotational kinetic energy B …translational kinetic energy C …both translational and rotational kinetic energy D …neither translational nor rotational kinetic energy

5. Day 17 – June 17 – WBL 8.4-8.5 Problem #2: Tipped Pencil PC141 Intersession 2013Slide 5 A pencil 18 cm long stands vertically on its point end on a horizontal table. If it falls over without slipping, with what tangential speed does the eraser end strike the table? Solution: In class WBL EX 8.55

6. Day 17 – June 17 – WBL 8.4-8.5 Problem #3: Loop-the-Loop Revisited PC141 Intersession 2013Slide 6 WBL EX 8.63 had to be released from a height h = 2.5R in order to achieve this minimum speed. For the rolling ball, the required minimum speed is the same, but the required height of release is not. Calculate the new h. Solution: In class

7. Day 17 – June 17 – WBL 8.4-8.5 8.5 Angular Momentum PC141 Intersession 2013Slide 7

8. Day 17 – June 17 – WBL 8.4-8.5 8.5 Angular Momentum PC141 Intersession 2013Slide 8

9. Day 17 – June 17 – WBL 8.4-8.5 8.5 Angular Momentum PC141 Intersession 2013Slide 9 in a chair while holding weights in his outstretched arms. When he brings the weights in toward his body (lowering the moment of inertia), his angular velocity increases.

10. Day 17 – June 17 – WBL 8.4-8.5 Figure skaters also exploit the conservation of angular momentum. They start to spin with their arms (and one leg) stretched outward. Bringing these limbs together along the rotation axis leads to a substantial increase in the rotational 8.5 Angular Momentum PC141 Intersession 2013Slide 10 Videos are not embedded into the PPT file. You need an internet connection to view them. speed (up to 7 revolutions per second).

11. Day 17 – June 17 – WBL 8.4-8.5 8.5 Angular Momentum PC141 Intersession 2013Slide 11

12. Day 17 – June 17 – WBL 8.4-8.5 As a final example of angular momentum, consider the helicopter shown below. Before it leaves the ground, there is no angular momentum (nothing is moving at all). If there was only one rotor, conservation of L dictates that as it rotated, the body of the helicopter would have to rotate in the opposite direction (the torque required to turn the rotor is internal; there are no external torques). By using a second rotor with an oppositely-directed rotation, the angular momentum can be kept at zero without requiring the body to rotate. 8.5 Angular Momentum Slide 12PC141 Intersession 2013

13. Day 17 – June 17 – WBL 8.4-8.5 8.5 Angular Momentum PC141 Intersession 2013Slide 13 The table shown here illustrates the many similarities between the translational and rotational dynamical equations.

14. Day 17 – June 17 – WBL 8.4-8.5 Problem #4: Angular Momentum Increase PC141 Intersession 2013Slide 14 WBL LP 8.19 (corrected) The angular momentum may be increased by… A …decreasing the moment of inertia B …decreasing the angular velocity C …increasing the product of the angular speed and moment of inertia D None of these

15. Day 17 – June 17 – WBL 8.4-8.5 Problem #5: Rotating Disk PC141 Intersession 2013Slide 15 A 10-kg rotating disk of radius 0.25 m has an angular momentum of 0.45 kg·m 2 /s. What is the angular speed of the disk? Solution: In class WBL EX 8.65

16. Day 17 – June 17 – WBL 8.4-8.5 Problem #6: Figure Skater PC141 Intersession 2013Slide 16 An ice skater spinning with outstretched arms has an angular speed of 4.0 rad/s. She tucks in her arms, decreasing her moment of inertia by 7.5%. a)What is the resulting angular speed? b)By what factor does the skater’s kinetic energy change? c)Where does the extra kinetic energy come from? Solution: In class WBL EX 8.71

17. Day 17 – June 17 – WBL 8.4-8.5 Problem #7: Angular Momentum of an Airplane PC141 Intersession 2013Slide 17 A 1200 kg airplane is flying in a straight line at 80 m/s, 1.3 km above the ground. What is the magnitude of its angular momentum with respect to a point on the ground directly under the path of the plane? Note: The purpose of this problem is to illustrate that the presence of a non-zero angular momentum does NOT imply that an object is rotating or moving in a curved path. Solution: In class