AP Physics 2 Plan for Today: Magnetism, Momentum, and Moment of Inertia Derivation

3. Plan for Today (AP Physics 2) • B Testers • Go over Magnetism FR and additional problem • C Testers • Go over momentum problems • Discuss Moment of Inertia and Derive Equations

4. Moment of Inertia • Use symbol I • Rotational inertia, analog to mass for linear • Must specify with respect to an axis of rotation

5. Equations of Linear Motion Transformed

6. 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 = 2 rad/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:

7. 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 )

8. Moment of Inertia for a Point Mass • I = mr^2 • Use this as a basis for other shapes

9. 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

10. More Generally • Moment of inertia defined with respect to a specific rotation axis • Moment of Inertia for a point mass is: • Can extend that definition for any object

12. General Form of Moment of Inertia

13. Note: “General Form” • General Form is really only for principal axes – along axes of symmetry • More complicated for arbitrary axes

14. Moment of Inertia of a Rod About center of rod

15. Moment of Inertia of a Rod Continued

16. Moment of Inertia of a Cylinder • Build up from moment of inertia of thin cylindrical shells

18. Continuing Moment of Inertia • Mass of Small Cylindrical Shell

19. Now Plugging In

20. Moment of Inertia for a Sphere • Sum small disks

21. Moment of Inertia of a Sphere

22. Sphere Again

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

24. More Common Moments of Inertia

25. 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.

26. 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

27. 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

28. 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

29. 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

30. 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)

31. 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

32. 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

33. 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

34.  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

35. 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:

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

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

38. 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

39. 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?

40. 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

41. 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;

42. 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

43. 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

44. 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

45. 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

46. 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

47. Summary – Rotational Analogies

48. Analogous Formulas

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