1 / 28

Physics 151: Lecture 21 Today’s Agenda

Physics 151: Lecture 21 Today’s Agenda. Topics Moments of Inertia Ch. 10.5 Torque Ch. 10.6, 10.7. ?. Lecture 22, ACT 1 Rotational Definitions.

duman
Download Presentation

Physics 151: Lecture 21 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 151: Lecture 21Today’s Agenda • Topics • Moments of Inertia Ch. 10.5 • Torque Ch. 10.6, 10.7

  2. ? Lecture 22, ACT 1Rotational Definitions • Your goofy friend likes to talk in physics speak. She sees a disk spinning and says “ooh, look! There’s a wheel with a negative w and with antiparallel w and a !!” Which of the following is a true statement ? (a)The wheel is spinning counter-clockwise and slowing down. (b) The wheel is spinning counter-clockwise and speeding up. (c)The wheel is spinning clockwise and slowing down. (d) The wheel is spinning clockwise and speeding up

  3. See text: 10.1 Example: • A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s2. The diameter of the wheel is 40 cm. What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s2? a. 39 cm/s b. 42 cm/s c. 45 cm/s d. 35 cm/s e. 53 cm/s

  4. See text: 10.4 Moment of Inertia • So where • Notice that the moment of inertia I depends on the distribution of mass in the system. • The further the mass is from the rotation axis, the bigger the moment of inertia. • For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). • We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics !

  5. See text: 10.5 Parallel Axis Theorem • Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass is known, = ICM • The moment of inertia about an axis parallel to this axis but a distance R away is given by: IPARALLEL = ICM + MR2 • So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis.

  6. We know So which agrees with the result from the board. See text: 10.5 Parallel Axis Theorem: Example • Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod. IPARALLEL = ICM + MD2 D=L/2 M CM x L ICM IEND

  7. y x z y x z Direction of Rotation: • In general, the rotation variables are vectors (have a 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

  8. y x z y x z See text: 10.1 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.

  9. ^ r ^ ^   See text: 10.6 and 10.7 Rotational Dynamics:What makes it spin? video • 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 F ^ F a m r • Multiply by r : rF=mr2 

  10. ^ r ^  See text: 10.6 and 10.7 Rotational Dynamics:What makes it spin? rF=mr2 use • Define torque: = rF. •  is the tangential force Ftimes the lever arm r. F F a m r • Torque has a direction: • +zif it tries to make the systemspin CCW. • - z if it tries to make the systemspin CW. 

  11. i I • Since the particles are connected rigidly,they all have the same . See text: 10.6 and 10.7 Rotational Dynamics:What makes it spin? • So for a collection of many particles arranged in a rigid configuration: m4 F1 F4 m1 r1  r4 m3 r2 r3 m2 F2 F3

  12. See text: 10.6 and 10.7 Rotational Dynamics:What makes it spin?  TOT=I • This is the rotational version of FTOT = ma • Torque is the rotational cousin of force: • The amount of “twist” provided by a force. • Moment of inertiaIis the rotational cousin 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.

  13. Fr  F  F  r r = “distance of closest approach” r See text: 10.6, 10.7, 11.2 Torque • Recall the definition of torque: • = rF • = r Fsin  • = r sinf F See Figure 10.13

  14. See text: 10.1 Example: • You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is : a. 0. b. mgr. c. 2mgr. d. a function of the angular velocity. e. small at first, then increasing as the Frisbee loses the torque given it by your hand.

  15. See text: 10.6 and 10.7 Torque =r Fsin  • So if  = 0o, then = 0 • And if  = 90o, then = maximum F r F r See Figure 10.13

  16. Lecture 21, Act 1Torque • 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

  17. Lecture 21, ACT 2 • A uniform rod of mass M = 1.2kg and length L = 0.80 m, lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. If a force (F = 5.0 N, q = 40°) acts as shown, what is the resulting angular acceleration about the pivot point ? a. 16 rad/s2 b. 12 rad/s2 c. 14 rad/s2 d. 10 rad/s2 e. 33 rad/s2

  18. See text: 11.2 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 DIRECTION: t =rXF MAGNITUDE: t=rF sin  r x  z

  19. y x z See text: 11.2 Torque & the Cross Product: • So we can define torque as:  = r x F = r F sin  X = y FZ - z FY Y = z FX - x FZ Z = x FY - y FX F f  r

  20. See text: 10.8 How Much WORK is Done ? • 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= FRdcos() = FRdcos(90-) = FRdsin() = FRsin() d  dW = d • We can integrate this to find: W =  • Analogue of W= F•r • W will be negative if  and  have opposite sign !  F  R dr=Rd d axis

  21. See text: 10.8 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:

  22. M R = 792.5 rad/s F See text: 10.8 Example: Disk & String • A massless string is wrapped 10 timesaround a disk of mass M=40 g and radius R=10cm. The disk is constrained to rotate without friction about a fixed axis though its center. The string is pulled with a force F=10N 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? See example 10.15

  23. Lecture 21, ACT 2 • 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 are made of identical material (i.e. their density r = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest. w2 Which disk has the biggest angular velocity after the pull ? (a)disk 1 (b)disk 2 (c)same w1 F F

  24. 1.7 s 11m/s 32 N Example 2 • A rope is wrapped around the circumference of a solid disk (R=0.2m) of mass M=10kg and an object of mass m=10 kg is attached to the end of the rope 10m above the ground, as shown in the figure. • How long will it take for the object to hit the ground ? • What will be the velocity of the object when it hits the ground ? • What is the tension on the cord ? w M T m h =10 m

  25. a) w = 7.67 rad/s a = 30 rad/s2 b) c) a = 15 m/s2 See text: 10.8 Example: Rotating Road • A uniform rod of length L=0.5m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is a) angular speed when it reaches the lowest point ? b) initial angular acceleration ? c) initial linear acceleration of its free end ? L m See example 10.14

  26. Lecture 22, ACT 2 • A campus bird spots a member of an opposing football team in an amusement park. The football player is on a ride where he goes around at angular velocity w at distance R from the center. The bird flies in a horizontal circle above him. Will a dropping the bird releases while flying directly above the person’s head hit him? a. Yes, because it falls straight down. b. Yes, because it maintains the acceleration of the bird as it falls. c. No, because it falls straight down and will land behind the person. d. Yes, because it mainatins the angular velocity of the bird as it falls. e. No, because it maintains the tangential velocity the bird had at the instant it started falling.

  27. See text: 10.1 Example: • A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a moment of inertia I = 0.090 kg m2. The cord does not slip on the pulley. What is the magnitude of the acceleration of m? a. 2.4 m/s2 b. 2.8 m/s2 c. 3.2 m/s2 d. 4.2 m/s2 e. 1.7 m/s2

  28. Recap of today’s lecture • Chapter 10, • Calculating moments of inertia • Tourque • Right Hand Rule

More Related