1 / 34

Physics 111: Lecture 18 Today’s Agenda

Physics 111: Lecture 18 Today’s Agenda. More about rolling Direction and the right hand rule Rotational dynamics and torque Work and energy with example. Rotational v.s. Linear Kinematics . Angular Linear. And for a point at a distance R from the rotation axis:.

rene
Télécharger la présentation

Physics 111: Lecture 18 Today’s Agenda

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physics 111: Lecture 18Today’s Agenda • More about rolling • Direction and the right hand rule • Rotational dynamics and torque • Work and energy with example

  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. Rolling Motion Roll objects down ramp • Cylinders of different I rolling down an inclined plane: K = - U = Mgh v = 0 = 0 K = 0 R M h v = R

  4. Rolling... • If there is no slipping: v  v 2v v Where v = R In the lab reference frame In the CM reference frame

  5. The rolling speed is always lower than in the case of simple sliding since the kinetic energy is shared between CM motion and rotation. We will study rolling more in the next lecture! Rolling... hoop: c = 1 disk: c = 1/2 sphere: c = 2/5 etc... Use v = R andI = cMR2. c c So: c c

  6. y x z y x z Direction of Rotation: • In general, the rotation variables are vectors (have direction) • If the plane of rotation is in the x-y plane, then the convention is • CCW rotation is in the + zdirection • CW rotation is in the- zdirection

  7. y x z y x z Direction of Rotation:The Right Hand Rule • To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector! • We normally pick the z-axis to be the rotation axis as shown. • = z • = z • = z • For simplicity we omit the subscripts unless explicitly needed.

  8. Realize that  = - 0.5 rad/s2.  Example: • A flywheel spins with an initial angular velocity 0 = 500 rad/s. At t = 0 it starts to slow down at a rate of 0.5 rad/s2. How long does it take to stop?  • Use to find when  = 0 : • So in this case

  9. Lecture 18, Act 1Rotations • A ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is on the ramp? (a)down the ramp (b)into the page (c)out of the page

  10. When the ball is on the ramp, the linear acceleration a is always down the ramp (gravity). • The angular acceleration is therefore counter-clockwise. a a Lecture 18, Act 1Solution • Using your right hand rule, a is out of the page!

  11. ^ 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 direction at some instant: • a = r • Now use Newton’s 2nd Law in the direction: • F=ma = mr rF=mr2 F ^ F a m r • Multiply by r : 

  12. ^ r ^  Rotational Dynamics:What makes it spin? rF=mr2 use • Define torque: = rF. •  is the tangential force Ftimes 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 

  13. 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

  14. 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.

  15. 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”

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

  17. Lecture 18, Act 2Torque • 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. (a)case 1 (b)case 2 (c)same L F F L axis case 1 case 2

  18. Torque is the same! Lecture 18, Act 2Solution • 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

  19. 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

  20. 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.

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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 

  26. 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

  27. 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:

  28. 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?

  29. M R F Disk & String... • The work done is W =  • The torque is = RF (since = 90o) • The angular displacement  is2 rad/rev x 10 rev. • So W = (.1 m)(10 N)(20rad) = 62.8 J  

  30. 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 

  31. Lecture 18, Act 3Work & Energy • 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 ? w2 w1 (a)disk 1 (b)disk 2 (c)same F F

  32. But we know So since I1 = I2 w1=w2 d Lecture 18, Act 3Solution • 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

  33. 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

  34. Recap of today’s lecture • More about rolling (Text: 9-6) • Direction and the right hand rule (Text: 10-2) • Rotational dynamics and torque (Text: 9-2, 9-4) • Work and energy with example (Text: 9-5) • Look at textbook problems Chapter 9: # 21, 23, 25, 49, 91, 119

More Related