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Work and Power for Rotation

s. q. F. F. q t. tq t. Workt. w =. Power = =. s = R q. Power = t w. Power = Torque x average angular velocity. Work and Power for Rotation. t = FR. Work = Fs = FR q. Work = tq. s. q. F. 6 kg. 2 kg. F=W. 20 m 0.4 m. sR. q = = = 50 rad. s = 20 m.

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Work and Power for Rotation

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  1. s q F F q t tq t Workt w = Power = = s = Rq Power = t w Power = Torque x average angular velocity Work and Power for Rotation t = FR Work = Fs = FRq Work = tq

  2. s q F 6 kg 2 kg F=W 20 m 0.4 m sR q = = = 50 rad s = 20 m Power = 98 W Work = 392 J Workt 392 J 4s Power = = Example 1:The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power if the 2-kg mass is lifted 20 m in 4 s. Work = tq = FR q F = mg = (2 kg)(9.8 m/s2); F = 19.6 N Work = (19.6 N)(0.4 m)(50 rad)

  3. Recall for linear motion that the work done is equal to the change in linear kinetic energy: Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy: The Work-Energy Theorem

  4. F wo = 60 rad/s R = 0.30 m F = 40 N What work is needed to stop wheel rotating: w R Work= DKr 4 kg 0 Work = -648 J Applying the Work-Energy Theorem: First find I for wheel: I = mR2 = (4 kg)(0.3 m)2 = 0.36 kg m2 Work = -½Iwo2 Work = -½(0.36 kg m2)(60 rad/s)2

  5. vcm First consider a disk sliding without friction. The velocity of any part is equal to velocity vcmof the center of mass. vcm vcm Now consider a ball rolling without slipping. The angular velocity  about the point P is same as  for disk, so that we write:  v R P Or Combined Rotation and Translation

  6. Kinetic Energy of Translation: K = ½mv2 v R Kinetic Energy of Rotation: P K = ½I2 Total Kinetic Energy of a Rolling Object: Two Kinds of Kinetic Energy

  7. Displacement: Velocity: Angular/Linear Conversions In many applications, you must solve an equation with both angular and linear parameters. It is necessary to remember the bridges:

  8. Translation or Rotation? If you are to solve for a linear parameter, you must convert all angular terms to linear terms: If you are to solve for an angular parameter, you must convert all linear terms to angular terms:

  9. w Two kinds of energy: w v v Kr = ½Iw2 KT = ½mv2 vR w = Example 2: A circular hoop and a circular disk, each of the same mass and radius, roll at a linear speed v. Compare the kinetic energies. Total energy: E = ½mv2 + ½Iw2 Disk: E = ¾mv2 Hoop: E = mv2

  10. mghf ½Iwf2 ½mvf2 mgho ½Iwo2 ½mvo2 = Conservation of Energy The total energy is still conserved for systems in rotation and translation. However, rotation must now be considered. Begin: (U + Kt + KR)o = End: (U + Kt + KR)f Height? Rotation? velocity? Height? Rotation? velocity?

  11. R = 50 cm 6 kg 2 kg h = 10 m mghf ½Iwf2 ½mvf2 mgho ½Iwo2 ½mvo2 = v = 8.85 m/s Example 3:Find the velocity of the 2-kg mass just before it strikes the floor. 2.5v2 = 196 m2/s2

  12. 20 m v = 16.2 m/s Example 4: A hoop and a disk roll from the top of an incline. What are their speeds at the bottom if the initial height is 20 m? Hoop: I = mR2 mgho = ½mv2 + ½Iw2 mgho = ½mv2 + ½mv2; mgho = mv2 v = 14 m/s Hoop: mgho = ½mv2 + ½Iw2 Disk: I = ½mR2;

  13. v = wr m m4 w m3 m1 m2 axis L = mvr Object rotating at constant w. L = Iw Angular Momentum Defined Consider a particle m moving with velocity v in a circle of radius r. Define angular momentum L: Substituting v= wr, gives: Since I = Smr2, we have: L = m(wr) r = mr2w For extended rotating body: L = (Smr2) w Angular Momentum

  14. L = 2 m m = 4 kg 1 12 1 12 For rod: I = mL2 = (4 kg)(2 m)2 L = 1315 kg m2/s Example 5:Find the angular momentum of a thin 4-kg rod of length 2 m if it rotates about its midpoint at a speed of 300 rpm. I = 1.33 kg m2 L = Iw = (1.33 kg m2)(31.4 rad/s)2

  15. Recall for linear motion the linear impulse is equal to the change in linear momentum: Using angular analogies, we find angular impulse to be equal to the change in angular momentum: Impulse and Momentum

  16. D t = 0.002 s wo = 0 rad/s R = 0.40 m F = 200 N w R F 2kg 0 wf= 0.5 rad/s Example 6: A sharp force of 200 N is applied to the edge of a wheel free to rotate. The force acts for 0.002 s. What is the final angular velocity? I = mR2 = (2 kg)(0.4 m)2 I = 0.32 kg m2 Applied torque t = FR Impulse = change in angular momentum t Dt = Iwf- Iwo FR Dt = Iwf

  17. 0 Ifwf = Iowo Io = 2 kg m2; wo = 600 rpm If = 6 kg m2; wo = ? wf = 200 rpm Conservation of Momentum In the absence of external torque the rotational momentum of a system is conserved (constant). Ifwf- Iowo= t Dt

  18. Summary – Rotational Analogies

  19. Analogous Formulas

  20. Height? Rotation? velocity? mghf ½Iwf2 ½mvf2 mgho ½Iwo2 ½mvo2 Height? Rotation? velocity? = Summary of Formulas: I = SmR2

  21. CONCLUSION: Angular Motion

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