Unit 2C - Rigid Body Rotation

# Unit 2C - Rigid Body Rotation

## Unit 2C - Rigid Body Rotation

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript

1. Unit 2C - Rigid Body Rotation

2. Objectives: After completing this module, you should be able to: • Define and calculate the moment of inertia for simple systems. • Define and apply the concepts of Newton’s second law, rotational kineticenergy, rotational work, rotational power, and rotational momentum to the solution of physical problems. • Apply principles of conservation of energy and momentum to problems involving rotation of rigid bodies.

3. Linear Inertia, m F = 20 N 24 N 4 m/s2 a = 4m/s2 m = = 5kg F = 20 N Rotational Inertia, I R = 0.5 m t a (20 N)(0.5 m) 4 m/s2 a = 4 m/s2 I = = = 2.5kg m2 Inertia of Rotation Consider Newton’s second law for the inertia of rotation to be patterned after the law for translation. Force does for translation what torque does for rotation: 1 rad/s = 60/2π rpm exactly = 9.55 rpm approx.

4. v = wR m m4 w m3 m1 m2 axis Object rotating at constant w. Rotational Kinetic Energy Consider tiny mass m: K = ½mv2 K = ½m(wR)2 K = ½(mR2)w2 Sum to find K total: Rotational Inertia Defined: K = ½(SmR2)w2 I = SmR2 (½w2 same for all m )

5. 3 kg 3 m 2 kg w 1 m 2 m 1 kg Example 1:What is the rotational kinetic energy of the device shown if it rotates at a constant speed of 600 rpm? First: I = SmR2 I = (3 kg)(1 m)2 + (2 kg)(3 m)2 + (1 kg)(2 m)2 w = 600rpm = 62.8 rad/s I = 25 kg m2 K = ½Iw2= ½(25 kg m2)(62.8 rad/s) 2 K = 49,300 J

6. L L R R R I = mR2 I = ½mR2 Hoop Disk or cylinder Solid sphere Common Rotational Inertias

7. R I = 0.120 kg m2 I = 0.0600 kg m2 I = mR2 Hoop R I = ½mR2 Disk Example 2:A circular hoop and a disk each have a mass of 3 kg and a radius of 30 cm. Compare their rotational inertias.

8. x f t A resultant torque tproduces angular acceleration a of disk with rotational inertia I. A resultant force Fproduces negative acceleration a for a mass m. w wo = 50 rad/s t = 40 N m R 4 kg Important Analogies For many problems involving rotation, there is an analogy to be drawn from linear motion. m I

9. How many revolutions required to stop? F wo = 50 rad/s R = 0.20 m F = 40 N w R t = Ia 4 kg 0 Newton’s 2nd Law for Rotation FR = (½mR2)a 2aq = wf2 - wo2 q = 12.5 rad = 1.99 rev a = 100rad/s2

10. R = 50 cm M 6 kg a = ? 2 kg aR a = aR; a = but R = 50 cm 6 kg aR T T = ½MR( ) ; T T +a 2 kg mg a = 3.92m/s2 Example 3:What is the linear accel- eration of the falling 2-kg mass? Apply Newton’s 2nd law to rotating disk: t = Ia TR = (½MR2)a T = ½MRa T = ½Ma and Apply Newton’s 2nd law to falling mass: mg - T = ma ½Ma mg - = ma (2 kg)(9.8 m/s2) - ½(6 kg) a = (2 kg) a 19.6 N - (3 kg) a = (2 kg) a

11. 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

12. 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 4: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)

13. 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

14. 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

15. 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

16. 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

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

18. 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:

19. Example (a): Find velocity vof a disk if given its total kinetic energy E. Total energy: E = ½mv2 + ½Iw2

20. Example (b) Find angularvelocityof a disk given its total kinetic energy E. Total energy: E = ½mv2 + ½Iw2

21. Draw and label a sketch of the problem. Strategy for Problems • List givens and state what is to be found. • Write formulas for finding the moments of inertia for each body that is in rotation. • Recall concepts involved (power, energy, work, conservation, etc.) and write an equation involving the unknown quantity. • Solve for the unknown quantity.

22. w Two kinds of energy: w v v Kr = ½Iw2 KT = ½mv2 vR w = Example 5: 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

23. 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?

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

25. 20 m v = 16.2 m/s Example 7: 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;

26. 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

27. 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 8: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

28. 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

29. 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 9: 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

30. 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

31. Summary – Rotational Analogies

32. Analogous Formulas

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