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Ch 10 Rotation of a Rigid Object About a Fixed Axis PowerPoint Presentation
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Ch 10 Rotation of a Rigid Object About a Fixed Axis

Ch 10 Rotation of a Rigid Object About a Fixed Axis

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Ch 10 Rotation of a Rigid Object About a Fixed Axis

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  1. 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.

  2. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.1 Angular position, displacement, velocity and acceleration  CCW +

  3. Angular Position Counterclockwise is positive

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

  5. Angular Displacement

  6. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.1 Angular position, displacement, velocity and acceleration  CCW + Angular Velocity Average angular velocity av = angular displacement / elapsed timeav = /t Instantaneous angular velocity= lim/tt 0

  7. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.1 Angular position, displacement, velocity and acceleration  CCW + Angular Acceleration Average angular acceleration av = angular velocity / elapsed timeav = /t Instantaneous angular acceleration= lim/tt 0

  8. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.2 Rotational kinematics: The Rigid Object Under Constant Angular Acceleration  CCW +

  9. Equations for Constant Acceleration Only • vf = vi + axt f = i + t • xf = xi + (vi + vf) t / 2 f = i + (i + f) t / 2 • xf = xi + vi t + axt2/2 f = i + i t + t2/2 • vf2 = vi2 + 2ax(xf – xi) f2 = i2 + 2(f – i) Assuming the initial conditions at t = 0 x = xi and  = i v = vi and  = i and a and  are constant.

  10. P10.5 (p.300) P10.6 (p.300) • f = i + t • f = i + (i + f) t / 2 • f = i + i t + t2/2 • f2 = i2 + 2(f– i)

  11. 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.

  12. CT4: A ladybug sits at the outer edge of a merry-go-round, that is turning and slowing down. At the instant shown in the figure, the radial component of the ladybug’s (Cartesian) acceleration is A. in the +x direction. B. in the –x direction. C. in the +y direction. D. in the –y direction. E. in the +z direction. F. in the –z direction. G. zero.

  13. CT5: A ladybug sits at the outer edge of a merry-go-round, that is turning and slowing down. At the instant shown in the figure, the tangential component of the ladybug’s (Cartesian) acceleration is A. in the +x direction. B. in the –x direction. C. in the +y direction. D. in the –y direction. E. in the +z direction. F. in the –z direction. G. zero.

  14. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.3 Angular and Translational Quantities v Connections s = r vt = r at = r ar = v2/r P10.17 (p.301) r   

  15. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.4 Rotational Kinetic Energy mi About fixed axis O I = miri2 K = I2/2 P10.17 (p.301) vi ri    O

  16. P10.22 (p.301)

  17. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.5 Calculation of Moments of Inertia

  18. IO = ICM + MD2 Fig. 10.12, p.305

  19. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.6 Torque

  20. CT7: You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? A B C D

  21. rsin

  22. P10.32 (p.302) d = 2sin57 57 33 F┴ = 100cos33  = Fr = (100cos33N)(2m) = 168 Nm  = Fd = (100N)(2sin57m) = 168 Nm  = Frsin = (100N)(2m)(sin57) = 168 Nm

  23. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.7 The Rigid Object Under a Net Torque  = I P10.36 (p.303)

  24. CT9: Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia is I = mR2. In order to impart identical angular accelerations, how large must F2 be? A. 0.25 N B. 0.5 N C. 1 N D. 2 N E. 4 N

  25. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.8 Energy Considerations in Rotational Motion W =  = d K = mv2/2 + I2/2 for a system P10.37 (p.303)

  26. P10.49 (p.304)

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

  28. Ch 10 Rotation of a Rigid Object About a Fixed Axis 10.9 Rolling Motion of a Rigid Object K = Mvcm2/2 + Icm2/2 for rolling without slipping scm = R vcm = R acm = R P10.56 (p.305)

  29. CT10: A sphere (S), a cylinder C, and a hollow cylinder (HC) with the same outer radius and mass start from rest at the same position at the top of an incline. In what order do the three reach the bottom of the incline - which is first, second, and third? Think about I2/2 and energy conservation. • S-1st,C-2nd,HC-3rd • S-1st,HC-2nd,C-3rd • C-1st,HC-2nd,S-3rd • C-1st,S-2nd,HC-3rd • HC-1st,S-2nd,C-3rd • HC-1st,C-2nd,S-3rd • 3 way tie

  30. CT11: Does the answer to the previous race depend on the radii of the objects? • Yes • No • CT12: Does the answer to the previous race depend on the masses of the objects? • Yes • No

  31. Table 10.3, p.314