Physics: Lecture 11 (Ch. 10-11 Halliday )

1. Physics: Lecture 11(Ch. 10-11 Halliday) • Direction and the right hand rule • Rotational dynamics and torque • Work and energy with example More about rolling

2. Rotational v.s. Linear Kinematics Angular Linear And for a point at a distance R from the rotation axis: • x = Rv = Ra = R

3. ^ r ^ ^   Rotational Dynamics:What makes it spin? • Suppose a force acts on a mass constrained to move in a circle. Consider its acceleration in the (tangential) direction at some instant: • at = r • Now use Newton’s 2nd Law in the direction: • Ft=mat = mr rFt=mr2 F ^ F a m r • Multiply by r : 

4. ^ r ^  Rotational Dynamics:What makes it spin? rFt=mr2 use • Define torque: = rFt. •  is the tangential force Fttimes the lever arm r. • Torque has a direction: • +zif it tries to make the systemspin CCW. • - z if it tries to make the systemspin CW. F F a m r 

6. Since the particles are connected rigidly,they all have the same . Rotational Dynamics:What makes it spin? • So for a collection of many particles arranged in a rigid configuration: i I m4 F1 F4 m1 r1  r4 m3 r2 r3 m2 F2 F3

7. Rotational Dynamics:What makes it spin?  NET=I • This is the rotational analogue of FNET = ma • Torque is the rotational analogue of force: • The amount of “twist” provided by a force. • Moment of inertiaIis the rotational analogue of mass. • If I is big, more torque is required to achieve a given angular acceleration. • Torque has units of kg m2/s2 = (kg m/s2) m = Nm.

8. Fr  F  F  r rp Torque •  = rF • Recall the definition of torque: = r F sin  = rsin F = rpF • Equivalent definitions! rp = “distance of closest approach” or torque arm.

9. Torque =r Fsin  • So if  = 0o, then = 0 • And if  = 90o, then = maximum F r F r

10. In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. L F F (a)case 1 (b)case 2 (c)same L axis case 1 case 2

11. Torque is the same! QuestionSolution • Torque = F x (distance of closest approach) • The applied force is the same. • The distance of closest approach is the same. F F L L case 1 case 2

12. Torque and the Right Hand Rule: • The right hand rule can tell you the direction of torque: • Point your hand along the direction from the axis to the point where the force is applied. • Curl your fingers in the direction of the force. • Your thumb will point in the directionof the torque. F y r x  z

13. B  A C The Cross Product • We can describe the vectorial nature of torque in a compact form by introducing the “cross product”. • The cross product of two vectors is a third vector: AXB=C • The length of C is given by: C = AB sin  • The direction of C is perpendicular to the plane defined by A and B, and inthe direction defined by the right handrule.

14. The Cross Product • Cartesian components of the cross product: C =AXB CX = AY BZ - BY AZ CY = AZ BX - BZ AX CZ = AX BY - BX AY B A C Note: B XA =- A X B

15. Torque Vector Fig. 11-10 (a) A force F, lying in an x-y plane, acts on a particle at point A. (b) This force produces a torque t = r x F on the particle with respect to the origin O. By the right-hand rule for vector (cross) products, the torque vector points in the positive direction of z. Its magnitude is given by in (b) and by in (c).

16. Sample problem Calculations: Because we want the torques with respect to the origin O, the vector required for each cross product is the given position vector r. To determine the angle q between the direction of r and the direction of each force, we shift the force vectors of Fig.a, each in turn, so that their tails are at the origin. Figures b, c, and d, which are direct views of the xz plane, show the shifted force vectors F1, F2, and F3. respectively. In Fig. d, the angle between the directions of and is 90°. Now, we find the magnitudes of the torques to be

17. y x z Torque & the Cross Product: • So we can define torque as:  = rX F = rF sin  X = rY FZ - FY rZ = y FZ - FY z Y = rZ FX - FZ rX = z FX - FZ x Z = rX FY - FX rY = x FY - FX y F  r

18. Comment on=I • When we write =I we are really talking about the z component of a more general vector equation. (Recall that we normally choose the z-axis to be the the rotation axis.) z=Izz • We usually omit the zsubscript for simplicity. z Iz z z

19. Example • To loosen a stuck nut, a (stupid) man pulls at an angle of 45o on the end of a 50 cm wrench with a force of 200 N. • What is the magnitude of the torque on the nut? • If the nut suddenly turns freely, what is the angular acceleration of the wrench? (The wrenchhas a mass of 3 kg, and its shape is that of a thin rod). 45o F = 200 N L = 0.5 m

20. So =/I = (70.7 Nm) / (0.25 kgm2) = 283 rad/s2 Example Wrench w/ bolts • Torque =LFsin = (0.5 m)(200 N)(sin 45) = 70.7 Nm • If the nut turns freely, =I • We know  and we want , so we need to figure out I. 45o F = 200 N L = 0.5m 

21. Work • Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d: • dW = F.dr= FR d cos() = FR d cos(90-) = FR d sin() = FR sin() d  dW =  d • We can integrate this to find: W =  • Analogue of W= F•r • W will be negative if  and  have opposite signs!  F  R dr = R d d axis

22. Work & Kinetic Energy: • Recall the Work/Kinetic Energy Theorem: K = WNET • This is true in general, and hence applies to rotational motion as well as linear motion. • So for an object that rotates about a fixed axis:

23. Connection with CM motion... • So for a solid object which rotates about its center or mass and whose CM is moving: VCM  We will use this formula more in the future.

24. M R F Example: Disk & String • A massless string is wrapped 10 timesaround a disk of mass M = 40 g and radius R = 10 cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F = 10 N until it has unwound. (Assume the string does not slip, and that the disk is initially not spinning). • How fast is the disk spinning after the string has unwound?

25. M R F Disk & String... • The work done is W = F x d • The displacement is2 x 0.1m x 10 rev = 2  m • So W = (10 N)(2) = 62.8 J F d

26. Recall thatIfor a disk about its central axis is given by: So = 792.5 rad/s Disk & String... Flywheel, pulley, & mass WNET = W = 62.8 J = K M R 

27. Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same distance. Disk 1 has a bigger radius, but both have the same moment of inertia. Both disks rotate freely around axes though their centers, and start at rest.Which disk has the biggest angular velocity after the pull ? (a)disk 1 (b)disk 2 (c)same w2 w1 F F d

28. But we know So since I1 = I2 w1=w2 d Solution • The work done on both disks is the same! • W = Fd • The change in kinetic energy of each will therefore also be the same since W = DK. w2 w1 F F

29. In this case, I = 1 kg - m2 W = mgh = (2 kg)(9.81 m/s2)(1 m) = 19.6 J  = 6.26 rad/s ~ 1 rev/s Spinning Disk Demo: I • We can test this with our big flywheel. negligiblein this case m

30. Applications of Newton’s Second Law for Rotation Examples

32. Radius sprocket rs= 7cm. Find ω after t = 5 s. 1. ω = αt 2. 4 3. 5.

33. Known L, M. Find: • α immediately following the release • The force FA a. 3 1 2 b. 2. Mg-FA=Macmy 3. at = rα 1 4. acmy = rcmα = (L/2)α 6. FA = (1/4)Mg 5. From 2, 3 and 4:

34. Nonslip Conditions A string wrapped around a rotating wheel must move with a tangential velocity vt that is equal to the tangential velocity of the rim of the wheel, provided that the string remains taut and does not slip. By differentiating the previous equation we get:

35. Problem Solving Guide

36. Falling weight & pulley... An object of mass m is tied to a light string wound around a pulley that has the moment of inertia I and the radius R. In non slipping conditions, find the tension in the string and the acceleration of the object. 1. Draw a free body diagram, drawing each force in its point of application. 2. The only force causing a torque is T:

37. Falling weight & pulley... 3. Draw a free body diagram for the suspended object and apply Newton’s second law: 4. Relate the linear and angular accelerations: 5. α from 2) and at from 3) in 4) gives: so 6. Substitute T into 3) and solve for at:

38. Falling weight & pulley... • Using 1-D kinematics (Lecture 1) we can solve for the time required for the weight to fall a distance L: I  R T m mg a L where

39. Power • The work done by a torque  acting through a displacement  is given by: • The power provided by a constant torque is therefore given by:

41. Review: Torque and Angular Acceleration  NET=I • This is the rotational analogue of FNET = ma • Torque is the rotational analogue of force: • The amount of “twist” provided by a force. • Moment of inertiaIis the rotational analogue of mass • If I is big, more torque is required to achieve a given angular acceleration.

42. F2 F1 Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radius of the other. Forces F1 and F2 are applied as shown. What is F2 / F1 if the angular acceleration of the wheels is the same? • A) 1 • B) 2 • C) 4

43. and but so Since R2 = 2R1 F2 F1 Solution We know

44. Rotation around a moving axis. • A string is wound around a puck (disk) of mass M and radius R. The puck is initially lying at rest on a frictionless horizontal surface. The string is pulled with a force F and does not slip as it unwinds. • What length of string L has unwound after the puck has moved a distance D? M R F Top view

45. The distance moved by the CM is thus • So the angular displacement is Rotation around a moving axis... • The CM moves according to F= MA • The disk will rotate about its CM according to =I M A  R F

46. Rotation around a moving axis... • So we know both the distance moved by the CM and the angle of rotation about the CM as a function of time: (a) (b) The length of string pulled out is L = R: Divide (b) by (a):  F F D L

47. M A  R F Comments on CM acceleration: • We just used =I for rotation about an axis through the CM even though the CM was accelerating! • The CM is not an inertial reference frame! Is this OK??(After all, we can only use F = ma in an inertial reference frame). • YES! We can always write =Ifor an axis through the CM. • This is true even if the CM is accelerating. • We will prove this when we discuss angular momentum!

48. Chapter 11(Halliday) Rolling, Torque, and Angular Momentum

49. Rolling Objects Rolling Without Slipping

50. Rolling as Translational and Rotation Combined Although the center of the object moves in a straight line parallel to the surface, a point on the rim certainly does not. This motion can be studied by treating it as a combination of translation of the center of mass and rotation of the rest of the object around that center.