Ch10-1 Angular Position, Displacement, Velocity and Acceleration

1. Chapter 10: Rotational Kinematics and Energy Ch10-1 Angular Position, Displacement, Velocity and Acceleration Rigid body: every point on the body moves through the same displacement and rotates through the same angle.  CCW +

2. CT1:A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular speed is A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine.

3. Angular Position Counterclockwise is positive

4. Radians  = s/r (a dimensionless ratio)

5. Angular Displacement

6. Chapter 10: Rotational Kinematics and Energy  CCW + Ch2-1 Angular Velocity Average angular velocity av = angular displacement / elapsed timeav = /t Instantaneous angular velocity= lim/tt 0

7. P10.9 (p.308)

8. Chapter 10: Rotational Kinematics and Energy  CCW + Ch2-1 Angular Acceleration Average angular acceleration av = angular velocity / elapsed timeav = /t Instantaneous angular acceleration= lim/tt 0

9. Chapter 10: Rotational Kinematics and Energy  CCW + Ch2-2 Rotational Kinematics

10. CT2: Which equation is correct for the fifth equation? •  = 0 +  • 2 = 02 +  • 2 = 0 +  • 2 = 02 + 2

11. Equations for Constant Acceleration Only • v = v0 + at  = 0 + t • vav = (v0 + v) / 2 av = (0 + ) / 2 • x = x0 + (v0 + v) t / 2  = 0 + (0 + ) t / 2 • x = x0 + v0 t + at2/2  = 0 + 0 t + t2/2 • v2 = v02 + 2a(x – x0) 2 = 02 + 2( – 0) Assuming the initial conditions at t = 0 x = x0 and  = 0 v = v0 and  = 0 and a and  are constant.

12. P10.20 (p.309) P10.22 (p.309) •  = 0 + t • av = (0 + ) / 2 •  = 0 + (0 + ) t / 2 •  = 0 + 0 t + t2/2 • 2 = 02 + 2( – 0)

13. Chapter 10: Rotational Kinematics and Energy Ch2-3 Connections Between Linear and Rotational Quantities s = r vt = r at = r acp = v2/r

14. CT3: A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s linear speed is A. half the ladybug’s. B. the same as the ladybug’s. C. twice the ladybug’s. D. impossible to determine.

15. P10-29 (p.310)

16. CT4: P10.29c The force necessary for Jeff’s centripetal acceleration is exerted by • gravity. • Jeff. • the vine. • air resistance.

17. Chapter 10: Rotational Kinematics and Energy Ch2-4 Rolling Motion v = r if no slipping  = 0 if no friction

18. Rolling Without SlippingConstant v and d = vt2r = vt (2/t)r = v r = v recall that r = vt

19. P10.45 (p.311)

20. CT5: P10.45b If the radius of the tires had been smaller, the angular acceleration of the tires would be • greater. • smaller. • the same.

21. Chapter 10: Rotational Kinematics and Energy Ch2-5 Rotational Kinetic Energy and Moment of Inertia For N particles: I = miri2 and K = I2/2 Recall for translation K = mv2/2 Both translation and rotation: K = mv2/2 + I2/2

22. Kinetic Energy of a Rotating Objectof Arbitrary Shape:Rigid Body of N Particles

23. Table 10-1aMoments of Inertia for Uniform, Rigid Objects of Various Shapes and Total Mass M

24. Table 10-1bMoments of Inertia for Uniform, Rigid Objects of Various Shapes and Total Mass M

25. P10.52 (p.311)

26. CT6: P10.52b If the speed of the basketball is doubled to 2v, the fraction of rotational kinetic energy will • double. • halve. • stay the same.

27. Chapter 10: Rotational Kinematics and Energy Ch2-6 Conservation of Energy WNC = E with K = mv2/2 + I2/2

28. After f Before vf i vi h y=0 P10.60 (p.311) Problem 10-60

29. CT7: P10.60b If the radius of the bowling ball were increased, the final linear speed would • increase. • decrease. • stay the same.

30. CT8: In the race between the hoop and solid disk, which will arrive at the base of the incline first? • hoop. • disk. • neither, it will be a tie.