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Chapter 8: Torque and Angular Momentum

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Chapter 8: Torque and Angular Momentum

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    1. Chapter 8: Torque and Angular Momentum Concept. Questions: 2, 4. Problems: 5,13,18, 27, 39, 43, 55, 69, 73. Rotational Kinetic Energy Torque & Angular Acceleration Torque & Angular Momentum (Vector Nature of)

    2. 2 Rotational Dynamics: Newton’s 2nd Law for Rotation

    3. Rotational Inertia & Energy

    4. Central Axis 4

    5. Axis on End 5

    6. 6 Calculated Rot. Inertias rotational inertias of solid objects can be calculated The calculated values are listed in your textbook on p.263 /

    7. Ex: Rotational Inertia: A 0.3kg meter stick is held horizontally from one end. Its rotational inertia about one end is: 7

    8. Torque = lever-arm x force meter-newton ft-lb

    9. torque lever-arm is the shortest distance from axis to line of the force Torque (giam7-11)

    10. Ex: Zero and Non-Zero Torque Zero Torque Large Torque

    12. Ex: Torque due to Gravity: A 0.3kg meter stick is held horizontally from one end. The torque due to gravity about the end is: 12

    13. 13 Newton’s 2nd Law (Rotation)

    14. Ex: Angular Acceleration: A 0.3kg meter stick is held horizontally from one end. Its angular acceleration when released is: 14

    15. Ex: A merry-go-round has a rotational inertia of 100kgm^2 and a radius of 1.0 meter. A force of 250 N is applied tangentially at its edge. The angular acceleration is: 15

    16. 16 Equilibrium Problems Equilibrium is state when: Net force = 0 & Net torque = 0 You can choose the axis anywhere, so we choose it where an unknown force acts. 1st Step: Storque-ccw = Storque-cw 2nd Step: Sforce-up = Sforce-down /

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    18. 18 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

    19. 19 Rotational Work-Energy Theorem (Work)rot = tDq. Example: torque of 50 mN is applied for one revolution. rotational work = (50Nm)(2prad) = 314 J (Rotational Work)net = DKrot. /

    20. 20 Angular Momentum (L) analog of translational momentum L = Iw [kgm2/s] Example: Disk R = 1m, M = 1kg, w = 10/s I = ½MR2 = ½(1)(1)2 = 0.5 kgm2 L = Iw = (0.5kgm2)(10/s) = 5kgm2/s

    21. 21 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)

    22. 22 8 Summary Mass ? Rotational Inertia Force ? Torque Rotational KE Angular Acceleration Work and Energy Angular Momentum

    23. 23 Concept Review Torque: rotational action Rotational Inertia: resistance to change in rotational motion. Torque = force x lever-arm

    24. 24 Mass-Distribution

    25. 25 Torque (t) [m·N]

    26. Rotational Inertia ( I )

    27. 27

    28. 28 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)

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    30. 30 #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)

    31. 31 #60, z-axis Each mass has r2 = 1.52 + 2.52. I = sum mr2 = (2+3+1+4)(1.52 + 2.52)

    32. 32 #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

    33. 33 #83 Pulley M, R. what torque causes it to reach ang. Speed. 25/s in 3rev? Alpha: use v-squared analog eqn. Torque = Ia = (½MR2)(a)

    34. 34 #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

    35. 35 #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.

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    37. 37 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.

    38. 38 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.

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    50. 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 =

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