Physics: Lecture 13 Today’s Agenda

1. Physics: Lecture 13Today’s Agenda • Comment about  = I (not true if I is changing!!) • General expression for the angular momentum of a system • Sliding beam example • Vector considerations of angular momentum • Bike wheel and rotating stool • Gyroscopic Motion • Comments about moving rotation axis

2. Angular momentum, Two Disks • Two different spinning disks have the same angular momentum, but disk 1 has more kinetic energy than disk 2. • Which one has the biggest moment of inertia? (a) disk 1 (b) disk 2 (c) not enough info

3. Solution (using L = I ) If they have the same L, the one with the biggest Iwill have the smallest kinetic energy. w1 w2 I1 < I2 disk 1 disk 2

4. We can’t assume  = Iwhen the moment of inertia is changing! When does  = Inot work ? • Last time we showed that • This is the fundamental equation for understanding rotation. • If we write L = I,then

5. When does  = Inot work? Now suppose EXT = 0: So in this case we can have an without an external torque!

6. Show dA/dt is const and = ½L/m and where l is the angle between and x1 is the component of in the direction of

8. w1 Example... • A puck in uniform circular motion will experience rotational acceleration if its moment of inertia is changed. • Changing the radius changes the moment of inertia, but produces no torque since the force of the string is along the radial direction. (since r X F = 0) I1 > I2 w2 w2> w1 The puck accelerates without external torque!!

9. Review: Angular Momentum • where and • In the absence of external torques Total angular momentum is conserved • This is a vector equation. • Valid for individual components.

10. z  Review... • In general, for an object rotating about a fixed (z) axis we can write LZ = I • The direction of LZ is given by theright hand rule (same as ).

11. Review... • A freely moving particle has a definite angular momentum about any axis. • If no torques are acting on the particle, its angular momentum will be conserved. • In the example below, the direction of L is along the z axis, and its magnitude is given by LZ = pd = mvd. y x d m v

12. Rotations Puck on ice • A puck slides in a circular path on a horizontal frictionless table. It is held at a constant radius by a string threaded through a frictionless hole at the center of the table. If you pull on the string such that the radius decreases by a factor of 2, by what factor does the angular velocity of the puck increase? (a) 2 (b) 4 (c) 8 w

13. L1 = I1w1 = mR2w1 L2 = I2w2 = m w2 = mR2w1 = m R2w2 w1 =w2 w2 = 4w1 m m w1 w2 R R/2 Solution • Since the string is pulled through a hole at the center of rotation, there is no torque: Angular momentum is conserved.

14. r* Rcm r A general expression for L of a system: • For a system of particles we can write • Express position and velocity in terms of the center of mass: ri= Rcm + ri* where ri* and v*i are the position andvelocity measured in the CM frame. vi= Vcm + v*i

15. A general expression for L of a system... • So we can write: Expanding this: Which becomes: Lcm =MV*cm = 0 =MR*cm = 0 L*

16. A general expression for L of a system... • So finally we get the simple expression Where is the angular momentum of the CM and L* is the angular momentum about the CM. • The total angular momentum of a system about a given axis is the sum of the angular momentum of the center of mass about this axis and the angular momentum about an axis through the center of mass. L = Lcm+ L*

17. A general expression for L of a system... • We have just showed that L = Lcm+ L* • Picture it this way: y origin (axis) x d m,I CM  v due to movement of CM due to rotation about CM

18. Example 1 • A rod of length d and mass m1 is sliding on a frictionless surface with speed vo as shown (without rotating). An initially stationary block having mass m2, sticks to the end of the rod as it goes by. • What is the final angular velocity F of the block-rod system? vo d cm F m2 m1 top view: initial top view: final

19. Example 1... • Choose the origin to be at the location of the block before the collision. We can determine the y-position of the center of mass before the collision. y vo d/2 x m1 top view: initial

20. Example 1... • It is best to take z component of angular momentum about the point (0,ycm). The angular momentum before the collision is due entirely to the center of mass motion of the rod since the rod is not rotating. y vo d/2 ycm x m1 top view: initial

21. Example 1... • The z component of angular momentum about the point (0,ycm) after the collision is due to rotation about the center of mass of the rod+block: 0 y Icm (0,ycm) vF F x top view: final

22. Example 1... • We need to know the moment of inertia Icm about the center of mass of the system. Irod (using || axis thm.) Iblock m1 cm of rod d/2 - ycm cm of block-rod system d/2 ycm m2

23. Example 1... • Using conservation of angular momentum: And plugging in for Icm and ycm: Icm y vF ycm F x top view: final

24. Example 1... • Suppose m1 = 2m2 = 2m initial final Icm vo d cm F m 2m

25. 1992 Mech2 Two identical spheres, each of mass Mand negligible radius, are fastened to opposite ends of a rod of negligible mass and length 2l. This system is initially at rest with the rod horizontal, as shown above, and is free to rotate about a frictionless, horizontal axis through the center of the rod and perpendicular to the plane of the page. A bug, of mass 3M, lands gently on the sphere on the left. Assume that the size of the bug is small compared to the length of the rod. Express your answers to all parts of the question in terms of M, l, and physical constants. a. Determine the torque about the axis immediately after the bug lands on the sphere. b. Determine the angular acceleration of the rod‑spheres‑bug system immediately after the bug lands.

26. Mech 1992 conti1 The rod‑spheres‑bug system swings about the axis. At the instant that the rod is vertical, as shown above, determine each of the following. c. The angular speed of the bug. 1) Conservation of energy gives: 2) But the loss in potential energy of the left sphere is equal to the gain in potential energy of the left sphere, therefore: 3) The change in potential energy is now: 4) The gain in kinetic energy is: 5) Setting them equal to each other we get:

27. Mech 1992 conti The rod‑spheres‑bug system swings about the axis. At the instant that the rod is vertical, as shown above, determine each of the following: d. The angular momentum of the system.

28. Mech 1992 conti The rod‑spheres‑bug system swings about the axis. At the instant that the rod is vertical, as shown above, determine each of the following. d. The magnitude and direction of the force that must be exerted on the bug by the sphere to keep the bug from being thrown off the sphere N-3Mg = Fcp N = 3Mg + Fcp

29. 1998 Mech • A space shuttle astronaut in a circular orbit around the Earth has an assembly consisting of two small dense spheres, each of mass m, whose centers are connected by a rigid rod of length l and negligible mass. The astronaut also has a device that will launch a small lump of clay of mass m at speed v0 . Express your answers in terms of m, v0l. and fundamental constants.

30. 1998 Mech 2 Conti • a. Initially, the assembly is "floating" freely at rest relative to the cabin, and the astronaut launches the clay lump so that it perpendicularly strikes and sticks to the midpoint of the rod, as shown above. i. Determine the total kinetic energy of the system (assembly and clay lump) after the collision. ii. Determine the change in kinetic energy as a result of the collision.

31. a)

32. a) ii. Determine the change in kinetic energy as a result of the collision.

33. b. The assembly is brought to rest, the clay lump removed, and the experiment is repeated as shown above, with the clay lump striking perpendicular to the rod but this time sticking to one of the spheres of the assembly. • i. Determine the distance from the left end of the rod to the center of mass of the system (assembly and clay lump) immediately after the collision. (Assume that the radii of the spheres and clay lump are much smaller than the separation of the spheres.)

34. bii. On the figure above, indicate the direction of the motion of the center of mass immediately after the collision.

35. iii. Determine the speed of the center of mass immediately after the collision.

36. iv. Determine the angular speed of the system (assembly and clay lump) immediately after the collision

37. v. Determine the change in kinetic energy as a result of the collision.

38. Angular momentum is a vector!Demo: Turning the bike wheel. • A student sits on the rotatable stool holding a bicycle wheel that is spinning in the horizontal plane. She flips the rotation axis of the wheel 180o, and finds that she herself starts to rotate. • What’s going on?

39. Turning the bike wheel... • Since there are no external torques acting on the student-stool system, angular momentum is conserved. • Initially: LINI = LW,I • Finally: LFIN = LW,F+LS LS LW,I LW,I = LW,F+LS LW,F

40. Lecture 23, Act 3Rotations • A student is initially at rest on a rotatable chair, holding a wheel spinning as shown in (1). He turns it over and starts to rotate (2). If he keeps twisting, turning the wheel over again(3), his rotation will: (a) stop (b) double (c) stay the same ?? (1) (2) (3)

41. Lecture 23, Act 3Solution LNET LNET LNET not turning LW LW LS LW [1] [2] [3]

42. Door jamb Example Door Below is a closed swinging door leading to a restaurant kitchen, seen from above. A meatball has been thrown at the door. The door has a mass of M = 4.0 kg, a length of L = 1.0 m, and a rotational inertia of I = ⅓ML2 about the hinge. The m = 0.2 kg meatball is about to strike the door at a point ¾ down the length, also shown below. The meatball is moving horizontally toward the door and has a speed of 15.0 m/s on impact. (Neglect its vertical motion due to gravity.) The meatball strikes the door and sticks to it. Both the meatball and the door fly open. Where applicable, provide a numerical answer with appropriate units.

43. Door jamb Example Door Door: M = 4.0 kg, a lL = 1.0 m, and a rotational inertia of I = ⅓ML2 Meatball: m = 0.2 kg strikes the door at a point ¾ down the length, with speed of 15.0 m/s . a. What are the initial linear momentum, angular momentum about the hinge, and kinetic energy of the meatball? (Before it strikes the door.) p = mv = (0.2 kg)(15 m/s) = 3 kg∙m/s L = r×p = rmv = (3L/4)(m)(v) = (0.75 m)(0.2 kg)(15 m/s) = 2.25 kg∙m2/s K = ½mv2 = (0.5)(0.2 kg)(15 m/s)2 = 22.5 J

44. Door jamb Example Door Door: M = 4.0 kg, a lL = 1.0 m, and a rotational inertia of I = ⅓ML2 Meatball: m = 0.2 kg strikes the door at a point ¾ down the length, with speed of 15.0 m/s . b. Which of these (p, L, K) are conserved in the collision? Explain. Only angular momentum is conserved. There is no net torque here. There is a net force applied by the hinge, and there is energy transferred out of the system to heat, sound, deformation, etc.

45. Door jamb Example Door Door: M = 4.0 kg, a lL = 1.0 m, and a rotational inertia of I = ⅓ML2 Meatball: m = 0.2 kg strikes the door at a point ¾ down the length, with speed of 15.0 m/s . c. What is the angular velocity of the door/meatball combination after the collision? Li = Lf = Ifωf = (Imeatball + Idoor)ωf =[(0.2 kg)(0.75 m)2 + 1/3(4 kg)(1 m)2)]ωf = (1.45 kg∙m2)ωf So ωf = Li / If = (2.25 kg∙m2/s) / (1.45 kg∙m2) = 1.6 rad/s. d. How long after the collision is it until the door hits the door jamb (that is, until θ = 90°)? θ = ωt (no angular acceleration) so t = θ / ω = (π/2) / (1.6 rad/s) = 1 second.

46. Gyroscopic Motion: • Suppose you have a spinning gyroscope in the configuration shown below: • If the left support is removed, what will happen?? pivot support  g

47. Gyroscopic Motion... • Suppose you have a spinning gyroscope in the configuration shown below: • If the left support is removed, what will happen? • The gyroscope does not fall down! pivot  g

48. Gyroscopic Motion... Bicycle wheel • ... instead it precesses around its pivot axis ! • This rather odd phenomenon can be easily understood using the simple relation between torque and angular momentum we derived in Lecture 22. pivot 

49. Gyroscopic Motion... • The magnitude of the torque about the pivot is  = mgd. • The direction of this torque at the instant shown is out of the page (using the right hand rule). • The change in angular momentum at the instant shown must also be out of the page! d L pivot  mg

50. Gyroscopic Motion... • Consider a view looking down on the gyroscope. • The magnitude of the change in angular momentum in a time dt is dL = Ld. • So where is the “precession frequency” L(t) dL d pivot L(t+dt) top view