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8 Rotational Dynamics • describe/predict rotational behavior: • motion, energy, momentum. • with concepts of: rotational inertia & torque • Homework: • 30, 33, 39, 60, 65, 83, 89, 111. 1 Translation and Rotation • Translation: up, down, left, right, in, out • Rotation: • clockwise (cw), • counterclockwise (ccw) • // 2 Rotational Dynamics: Newton’s 2nd Law for Rotation net rotational action rotational inertia 3 Mass-Distribution Effect Larger Larger Larger radius Speed Effort Rotational Inertia ~ MR2 4 Rotational Inertia ( I ) I m r m r 1 1 2 2 2 2 kg(m)2 Example I (4kg)(3m) 2 (5kg)( 2m) 2 36 20 56 kg m 2 5 I m r m r 1 1 2 2 2 2 Example: Four identical masses. Axis passes symmetrically through center of mass. I m1a m2a m3a m4a 2 2 2 2 I 4ma 2 6 I m r m r 1 1 2 2 2 2 Example: Four identical masses. Axis passes through one end. I m02 m02 m(2a)2 m(2a)2 I 4ma 4ma 8ma 2 2 2 7 You Try It Calculate the Rotational Inertia of this system: 8 Calculated Rot. Inertias, p.273 • rotational inertias of solid objects can be calculated • need to know: a, b, c, d, e, g, h, j, k. • omit: f, i 9 Rotational Kinetic Energy • Rotational K = ½(I)w2. • Example: Constant Power Source has 100kg, 20cm radius, solid disk rotating at 7000 rad/s. • I = ½MR2 = ½(100kg)(0.2m)2 = 2kgm2. • Rot K = ½ (2kgm2)(7000/s)2 = 49 MJ 10 Torque (t) [N·m] F F t F = lever-arm 11 Torque Concept • a small force can cause a large torque • a large force can create no torque 12 Rotationally Effective Application of Forces 13 Rotationally Ineffective Application of Forces 14 • torque = Fsin(f) (r) 15 1. A pair of forces with equal magnitudes and opposite directions is acts as shown. Calculate the torque on the wrench. 16 Example Rod is 3m Long. (3m) cos40 40 t F (5N )(2.3m) 11.5Nm 17 18 Newton’s 2nd Law (Rotation) t Net I FNet ma 19 Concept Review • Torque: rotational action • Rotational Inertia: resistance to change in rotational motion. • Torque ~ force x lever-arm 20 Equilibrium • Translational: net-force = 0 • Rotational: net-torque = 0 21 The drawing shows a person whose weight is 584N. Calculate the net force with which the floor pushes on each end of his body. 22 Rotational Work-Energy Theorem • (Work)rot = tDq. • Example: torque of 50 Nm is applied for one revolution. • rotational work = (50Nm)(2prad) = 314 J • (Rotational Work)net = DKrot. • Krot = ½Iw2. 23 Angular Momentum (L) • analog of translational momentum • L = Iw [kgm2/s] • Example: Disk R = 1m, M = 2kg, w = 10/s • I = ½MR2 = ½(1)(1)2 = 0.5 • L = Iw = (0.5kgm2)(10/s) = 5kgm2/s 24 Conservation of Angular Momentum • For an isolated system • (Iw)before = (Iw)after • Example: Stationary disk M,R is dropped on rotating disk M, R, wi. • (Iw)before = (Iw)after • (½MR2)(wi) = (½MR2 + ½MR2)(wf) • (wf) = ½ (wi) 25 8 Summary • All TRANSLATIONAL quantities; • speed, velocity, acceleration, • force, inertia, energy, and momentum, • have ROTATIONAL analogs. 26 Problem 33 • Pivot at left joint, Fj = ?, but torque = 0. • ccw (Fm)sin15(18) = mg(26) = cw • ccw (Fm)sin15(18) = (3)g(26) = cw • (Fm) = (3)g(26)/sin15(18) = 160N • Note: any point of arm can be considered the pivot (since arm is at rest) 27 #39 • Left force = mg = 30g, Right = 25g • mg = 30g + 25g m = 55kg • ccw mg(xcg) = cw 30g(1.6) • (55)g(xcg) = 30g(1.6) • (55)(xcg) = 30(1.6) • Xcg = (30/55)(1.6) 28 #60, z-axis • Each mass has r2 = 1.52 + 2.52. • I = sum mr2 = (2+3+1+4)(1.52 + 2.52) 29 #65 • First with no frictional torque, then with frictional torque as specified in problem. • M = 0.2kg, R = 0.15m, m1 = 0.4, m2 = 0.8 30 #83 • Pulley M, R. what torque causes it to reach ang. Speed. 25/s in 3rev? • Alpha: use v-squared analog eqn. • Torque = I = (½MR2)() 31 #89, uniform sphere part • Rolling at v = 5m/s, M = 2kg, R = 0.1m • K-total = ½mv2 + ½Iw2. • = ½(2)(5x5) + ½[(2/5)(2)(0.1x0.1)](5/0.1)2. • = 25 + 10 = 35J • Roll w/o slipping, no heat created, mech energy is conserved, goes all to Mgh. • 35 = Mgh h = 35/Mg = 35(19.6) = 1.79m 32 #111 • Ice skater, approximate isolated system • Therefore: • (Iw)before = (Iw)after • (100)(wi) = (92.5)(wf) • (wf) = (100/92.5)(wi) • K-rot increases by this factor squared times new rot. Inertia x ½. 33 Example: Thin rod formulas. 34 K = Ktranslation + Krotation 35 Angular Momentum • Symbol: L Unit: kg·m2/s • L = mvr = m(rw)r = mr2w = Iw. • v is perpendicular to axis • r is perpendicular distance from axis to line containing v. 36 Angular Momentum • Symbol: L Unit: kg·m2/s • L = mvr = m(rw)r = mr2w = Iw. • v is perpendicular to axis • r is perpendicular distance from axis to line containing v. 37 13) Consider a bus designed to obtain its motive power from a large rotating flywheel (1400. kg of diameter 1.5 m) that is periodically brought up to its maximum speed of 3600. rpm by an electric motor at the terminal. If the bus requires an average power of 12. kilowatts, how long will it operate between recharges? Answer: 39. minutes Diff: 2 Var: 1 Page Ref: Sec. 8.4 38 6) A 82.0 kg painter stands on a long horizontal board 1.55 m from one end. The 15.5 kg board is 5.50 m long. The board is supported at each end. (a) What is the total force provided by both supports? (b) With what force does the support, closest to the painter, push upward? 39 28) A 4.0 kg mass is hung from a string which is wrapped around a cylindrical pulley (a cylindrical shell). If the mass accelerates downward at 4.90 m/s2, what is the mass of the pulley? A) 10.0 kg B) 4.0 kg C) 8.0 kg D) 2.0 kg E) 6.0 kg 40 19) A solid disk with diameter 2.00 meters and mass 4.0 kg freely rotates about a vertical axis at 36. rpm. A 0.50 kg hunk of bubblegum is dropped onto the disk and sticks to the disk at a distance d = 80. cm from the axis of rotation. (a) What was the moment of inertia before the gum fell? (b) What was the moment of inertia after the gum stuck? (c) What is the angular velocity after the gum fell onto the disk? (a) 2.0 kg-m2 (b) 2.3 kg-m2 41 (c) 31. rpm 42 3. The drawing shows the top view of two doors. The doors are uniform and identical. The mass of each door is M and width as shown below is L. How do their rotational accelerations compare? 43 A Ring, a Solid-Disk, and a Solid-Sphere are released from rest from the top of an incline. Each has the same mass and radius. Which will reach the bottom first? 44 5. The device shown below is spinning with rotational rate wi when the movable rods are out. Each moveable rod has length L and mass M. The central rod is length 2L and mass 2M. Calculate the factor by which the angular velocity is increased by pulling up the arms as shown. 45 How large is the height h in the drawing? The ball’s initial speed is 4.0 m/s. Does it depend on M and R? solid ball rolls without slipping 46 Rotational Review (angles in radians) s rq Dq wavg Dt vt rw Dw avg at r Dt a c rw a a c at 2 + 4 kinematic equations 47 Graph showing mt2 is more constant than mt Rotation Disk-Hanging Mass 2500 2000 mt and mt2/10 1500 mt mt2 1000 500 0 0 200 400 600 mass (grams) 48 Angular Momentum Calculation L = Iw Example: Solid Disk M = 2kg R = 25cm Spins about its center-of-mass at 35 rev/s L Iw 1 (2kg)(.25m) 2 )(35rev / s)(2prad / rev) ) 2 13.7kgm2 / s 49 4. A one-meter-stick has a mass of 480grams. a) Calculate its rotational inertia about an axis perpendicular to the stick and through one of its ends. b) Calculate its rotational inertia about an axis perpendicular to the stick and through its center-of-mass. c) Calculate its angular momentum if spinning on axis (b) at a rate of 57rad/s. 50 Conservation of Angular Momentum • Example: 50 grams of putty shot at 3m/s at end of 200 gram thin 80cm long rod free to rotate about its center. • Li = mvr = (0.050kg)(3m/s)(0.4m) • Lf = Iw = {(1/12)(0.200kg)(0.8m)2 + (0.050kg)(0.4m)2}(w) • final rotational speed of rod&putty = 51 Dq wavg Dt Dw avg Dt 52 53 54 55 56 57 58

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posted: | 2/1/2013 |

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