Chapter 11

1. Chapter 11 Rotation

2. v a R Uniform Circular Motion Recall: v is tangent to the circle The magnitude of the centripetal acceleration is, pointing towards the center of the circle of radius R Period (the time to make one revolution):

3. Polar and Cartesian Coordinates Cartesian to Polar: y q x Polar to Cartesian:

4. r q r Radians If we take a line of length r and stretch it along the edge of the circle, the angle swept out is 1 radian p radians = 180°

5.  Ds r Linear Motion Along Circle s is the arclength along the circle s = r Only true for q in radians Linear speed along circle: v = ds/dt = r dq/dt Define angular velocity: w = dq/dt, then: v = rw

6. Linear Motion Along Circle v = rw If an extended object is rotating, we have different v at different r

7. Uniform Circular Motion Easier in Polar Coordinates radians  angular position  angular velocity radians/second  = v/r where v is the linear speed around the circle Similar to before (t) = t + 0 x(t) = vxt + x0

8. Angular Acceleration Plays same role in rotational motion as acceleration in linear motion Example: uniform circular motion but

9. Angular Acceleration Can break linear acceleration into two parts: Tangential: atan Radial: arad r Then:

10. x, v, a q, w, a Vectors? Analogous to 1D kinematics Vectors point in direction of right-hand rule Clockwise is into page Counter-clockwise is out of page

11. Rotational Motion Analogous to motion in a straight line Translation Rotation position x  velocity v = dx/dt  = d/dt acceleration a = dv/dt  = d/dt However, linear acceleration is never zero for rotational motion. There is always a radial acceleration (otherwise it wouldn’t be rotational motion!)

12. Rotational Kinematics We will be primarily dealing with “1-D” rotations, so the vector direction will be indicated by the sign and we can drop the vector notation for simplicity If a is constant: As with linear kinematics, we can use these two equations to solve most constant a problems!

13. Constant Angular Acceleration We can derive variants of these equations which are also useful:

14. Fake Gravity Spin ship to simulate Earth-like gravity Rotation: half turn in about 13 s   =  rad / 13 s = 0.24 rad/s Inner radius: about 18 ladder rungs  r = 18 ft = 5.5 m Centripetal acceleration: ac = v2/r = 2 r = (0.24 s-1)2  5.5 m = 0.32 m/s2 ac = 3.3% g !

15. w0 Example: Hampster on a Wheel • A hamster is running in place on a spinning wheel with R =0.5 m, • w0 = -1 rad/s • a = + 0.5 rad/s2 At what time will the wheel have w = 2 rad/s?

16. w0 Example: (continued) How far will the wheel have rotated from its position at t = 0?

17. w0 Example: (continued) What is his displacement at that time? What is his linear acceleration? Radial: Tangential:

18. w r Rotational Kinetic Energy (Point Particle) K = ½mv2 K = ½m(wr)2 K = ½ (mr2) w2 I I = mr2 is the rotational inertia or moment of inertia for a point particle K = ½ I w2 I is therotational equivalent of mass

19. 4 kg 4 kg 1 m 2 m Rotational Inertia of Point Mass For a single particle I = mr2 (all mass at same r) The same particle farther out A single particle I = (4 kg)(1 m)2 = 4 kg m2 I = (4 kg)(2 m)2 = 16 kg m2 Four times the rotational “mass”

20. State College Sydney Rigid Bodies A rigid body is one where all the particles maintain their relative position As the body rotates Each particle moves Relative positions don’t change Cities on the earth are always moving But they don’t get closer together

21.  vi ri Kinetic Energy (Rigid Body) Divide rigid body into bits: m1, m2, m3, … For a rigid body: vi = riw (same w for every part) Where I = Smiri2 is “rotational inertia” of body

22. Rotational Inertia: Extended Objects For an extended object, we break it into differential pieces and sum them up: z dm y x

23. Rotational Inertia aka “Moment of Inertia” Plays the role of mass for rotational motion I = Smiri2 = r2dm ri is the distance of mi to the axis of rotation For a single particle I = mr2 For a rigid body, I depends on how the mass is distributed in an object relative to the axis of rotation

24. Example: Four masses are arranged in a square as shown. Find I for each of the given axes. m1 m1 = 10 kg m2 = 20 kg m2 1 d=4m m2 m1 3 2

25. 4 m m2 m1 d1= 8 1 m2 m1 Example: The moment of inertia around an axis going through the center is: m1 = 10 kg m2 = 20 kg I1 = 2m1d12 + 2m2d12 = 2(10kg)(8 m2)+2(20 kg)(8 m2) = 480 kg m2

26. d2=2 m m2 m1 m2 m1 2 Example: The moment of inertia around axis 2 is: I2= 2m1d22 + 2m2d22 = 2(10 kg)(4 m2)+2(20 kg)(4 m2) = 240 kg m2

27. Example d=4 m m2 m1 The moment of inertia around axis 3 is: m2 m1 3 I3= 2m1d2 + 2m2×0 = 2(10 kg)(4 m)2 + 0 = 320 kg m2

28. ℓ b a Rotational Inertia: Common Objects ℓ R R More in the textbook (pg 227)

29. Rotational Inertia Example A rod with blocks: Mrod = Mblock = M 7 times larger!!!

30. Example: Solid Cylinder How do we get: I = mR2/2? h dV = dx dy dz = (rdθ) dr dz

31. Parallel-Axis Theorem I about CM is “easy” to calculate I about other axis of rotation is not! D Bar rotating about CoM Parallel Axis Theorem: Bar rotating about one end I = ICM + MD2 I about CM Object of Mass M distance D

32. L Faster Than a Falling Rock The rod starts at rest and falls over How fast is it going when it hits the table? Conservation of Energy: Ki+Ui = Kf+Uf Initial: Ui = mg(L/2) and Ki = 0 Final: Uf = 0 and Kf = ½Iw2

33. L Faster Than a Falling Rock I = mL2/3 Ei = Ef A little algebra: Moment of inertia for the bar:

34. L Faster Than a Falling Rock Center of mass: End of the rod: The end of the rod is falling faster than a dropped stone! Dropped Rock:

35. Torque on a Point Particle at How does an applied tangential force lead to rotational acceleration? r Ft Ft = mat= m(ra) Want to replace m with I = mr2 r Ft = mr2a = Ia Define torque t = r Ft (simplified definition) t = Ia rotational equivalent of F = ma

36. r1 r2,3 Torque: Rotational “Force” Door on a hinge: How do you open? F3 F1 F2 Define a new vector quantity: Torque r = vector from origin to point where force is applied Usually take origin to be on axis of rotation

37. “Cross” or “Vector” Product Recall: Magnitude only…

38. Right Hand Rule • Point your fingers in the direction of the first vector • Curl them in the direction of the second vector • Your thumb now points in the direction of the cross product • Sanity check your answers!

39. Torque: Key Points 1) t is a vector with unit Nm  J 2) t is defined around a certain origin 3) tnet = Ia (just like Fnet = ma) ttot and a direction both given by right-hand-rule

40. Torque: How to Think of It F F r r sin θ θ q r r sinq is the component of r perpendicular to F “Moment arm”

41. Torque on a Door: Example F4 Top View q=30° F3 F1 F2 A 3N force is applied to different places on 50 cm door. Find the magnitude of the torques? 1 = 0 (r=0) 2 = 0.5m  3N = 1.5 Nm out of screen 3 = 0 (θ=180°) 4 = 0.5m  3N  sin(150°) = 0.75 Nm into screen

42. A Falling Stuntman Stuntmen sometimes need to fall large distances without getting hurt! But it has to look like they fall Cylinder (R) The rope is around the spindle Spindle (r) M m Reduced gravity!

43. Torque and Work Recall: 2(θf – θi) = f2 - i2 Rearrange and multiply by I I(θf – θi ) = ½If2 - ½Ii2 tot Dθf = DK= W Can also write:

44. Translation Rotation Position x  Velocity v = dx/dt  = d/dt Acceleration a = dv/dt  = d/dt Mass m I = Smiri2 Newton’s 2nd Law F = ma  = I Kinetic Energy K=½mv2 K = ½Iw2 Rotational Analogy to Linear Motion