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Rotational Mechanics and Simple Machines

-- object must be considered to be an. if the rotation is. crucial to the motion. Rotational Mechanics and Simple Machines. A spinning object rotates about its. center of mass. “here-to-there” motion. translational motion :. --.

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Rotational Mechanics and Simple Machines

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  1. -- object must be considered to be an if the rotation is crucial to the motion. Rotational Mechanics and Simple Machines A spinning object rotates about its... center of mass. “here-to-there” motion translational motion: -- object’s entire mass can be assumed to be at its center of mass point mass. -- object is considered to be a... spinning rotational motion: extended mass

  2. In translational motion, an object’s resistance to changing its state of motion is its... inertia. -- the greater the mass... the greater the inertia. In rotational motion, an object’s resistance to changing its state of motion is its... moment of inertia (I). -- also called rotational inertia -- the more mass that is farther from the axis of rotation... the greater the moment of inertia.

  3. Moment of inertia values have been tabulated for simple shapes, e.g., I =2/5 MR2 for a solid sphere spinning about its central axis. Other shapes for which I is easily found are… I = MR2 -- a point mass M revolving at radius R -- a hollow sphere rotating about its axis I = 2/3 MR2 -- a thin disc rotating about its axis I = 1/2 MR2

  4. ** F and d must be . d use component only Q F torque: involves twisting/rotating t = F d t = torque (N-m) F = applied force (N) d = dist. between force & pivot pt. (m) A 62 N force is applied to end of wrench. Force makes 129o angle w/handle. Distance between force and bolt is 39 cm. Find torque. t = F d = 62 N (cos 39) (0.39 m) = 19 N-m

  5. F F F NOT EQUIL. (St = 0) F Static Equilibrium 1. forces acting on object are balanced AND 2. torques acting on object are balanced EQUIL. (SF and St = 0)

  6. A 4.0 m horiz. beam is supported at ends. Point load of 1.0 x 103 N acts at beam’s center. Find support reactions. Neglect beam’s mass. 1.0 x 103 N 2.0 m 2.0 m RA RB SF = 0 RA + RB – 1000 = 0 StA = 0 – 1000(2.0) RB(4.0) = 0 Solve for… RB = 5.0 x 102 N From SF = 0, RA = 5.0 x 102 N

  7. A 4.0 m horiz. beam is supported at ends. Point load of 1.0 x 103 N acts 3.2 m from left end. Find support reactions. Neglect beam’s mass. 1.0 x 103 N 3.2 m RA 4.0 m RB SF = 0 RA + RB – 1000 = 0 StA = 0 – 1000(3.2) RB(4.0) = 0 Solve for… RB = 8.0 x 102 N From SF = 0, RA = 2.0 x 102 N

  8. A 12 m horiz. beam is supported at ends. point loads of 3.0 x 102 N, 4.0 x 102 N, and 5.0 x 102 N act 3.0 m, 5.0 m, and 7.0 m (respectively) from left. Find support reactions. Neglect beam’s mass. 500 N 400 N 300 N RA 3 m 2 m 2 m 5 m RB SF = 0 RA + RB – 1200 = 0 StA = 0 – 300(3) – 400(5) – 500(7) RB(12) = 0 Solve for… RB = 533 N From SF = 0, RA = 667 N

  9. Find support reactions. Neglect beam’s mass. 850 N 630 N 210 N 3 m 7 m 6 m RA RB SF = 0 RA + RB – 1690 = 0 StA = 0 – 850(13) + 210(3) RB(7) = 0 Solve for… RB = 1490 N From SF = 0, RA = 200 N

  10. A 14 m horiz. beam is supported at 4.0 m and 12.0 m from left end. Point loads of 510 N and 930 N act 2.0 m and 9.0 m (respectively) from left. Beam’s weight acts as a uniformly distributed load of 75 N/m along entire beam. Find support reactions. 930 N 510 N 75 N/m 2 m 2 m 5 m 3 m 2 m RA RB

  11. 75 N/m To deal w/distributed load… Concentrate distributed load into a point load by taking... 75 N/m (14 m) = 1050 N 1050 This _____ N point load acts at center of the span. 1050 N 930 N 510 N 2 m 2 m 5 m 3 m 2 m RA RB SF = 0 RA + RB – 2490 = 0 StA = 0 + 510(2) – 930(5) – 1050(3) RB(8) = 0 RB = 850 N, RA = 1640 N

  12. Newton’s 2nd Law for Rotation St = sum of the torques (N-m) St = Ia I = moment of inertia (kg-m2) a = angular acceleration (rad/s2) SF = ma A football player applies a net torque of 0.082 N-m to a football having a moment of inertia of 5.7 x 10–4 kg-m2. If it starts from rest, what is the angular acceleration of the ball over the time that the quarterback takes to throw it? St = Ia

  13. A compact disc has mass 8.0 g and radius 6.0 cm. Starting from rest, the CD speeds up to 220 rpm in 1.6 s. What net torque does the disc player apply to the CD? St = Ia ½ MR2 I = ? = ½ (0.008)(0.06)2 = 1.44 x 10–5 kg-m2 St = Ia = 1.44 x 10–5 = 2.1 x 10–4 N-m

  14. p = mv L = Iw Angular Momentum L = angular momentum (kg-m2/s) I = moment of inertia (kg-m2) w = angular velocity (rad/s) If the mass of the Earth is 6.0 x 1024 kg and its mean radius is 6.4 x 106 m, find the angular momentum of the Earth as it spins on its axis. = 9.83 x 1037 kg-m2 L = Iw = 7.1 x 1033 kg-m2/s = 9.83 x 1037 (7.27 x 10–5)

  15. 3.2 m = 1178 kg-m2 A 68 kg man sits 3.2 m from the axis of a spinning wooden plank of mass 95 kg and length 7.8 m. The axis goes through the center of the plank, which spins at 1.3 rad/s. Find the angular momentum of this system. 68 kg 1.3 rad/s 95 kg L = Iw 7.8 m = 1178 kg-m2 = 482 + 696 I = ? (man) (plank) L = Iw

  16. Law of Conservation of Angular Momentum When... the net external torque acting on a system is zero, ...the angular momentum of the system is conserved. Lf = Li If wf = Ii wi

  17. If wf = Ii wi A merry-go-round of mass 115 kg and radius 2.0 m is spins at 2.6 rad/s while a 65 kg student stands at the edge. Find the new angular speed after the student has moved to a distance of 0.50 m from the axis. Lf = Li Li = Ii wi = (Imgr + Is) wi = (½ MmgrRmgr2 + MsRs2)wi = [ ½(115)(2)2 + 65(2)2 ] (2.6) = 1274 kg-m2/s [ ½(115)(2)2 + 65(0.5)2 ] (wf) = 246.25 wf Lf = 246.25 wf = 1274 wf = 5.2 rad/s

  18. I = 9.83 x 1037 kg-m2 w = 7.27 x 10–5 rad/s KE = ½ mv2 Rotational Kinetic Energy KErot = rotational kinetic energy (J) KErot = ½ I w2 I = moment of inertia (kg-m2) w = angular velocity (rad/s) If the mass of the Earth is 6.0 x 1024 kg and its mean radius is 6.4 x 106 m, find the rotational kinetic energy of the Earth as it spins on its axis. From p. 4 of the notes… KErot = ½ I w2 = ½ (9.83 x 1037)(7.27 x 10–5)2 KErot = 2.6 x 1029 J

  19. Simple Machines inclined plane lever wheel & axle wedge screw pulley

  20. Fe Fe Fe Fr Fr Fr de dr de dr dr de Levers effort force de = Fe = effort distance resistance distance Fr = dr = resistance force 3rd Class 2nd Class 1st Class

  21. Fr and dr Fe and de The Math of Simple Machines work input: Win = Fe de work output: Wout = Fr dr “Effort” deals w/force you apply; “resistance” refers to force from load.

  22. For ideal machines… Win = Wout Fe de = Fr dr i.e., The ideal mechanical advantage (IMA) is… Win > Wout For real machines… and actual mechanical advantage (AMA) < IMA (Note that, for ideal machines, IMA = AMA.) No machine... is ideal.

  23. Efficiency (Eff) tells how closely a real machine comes to being an ideal machine. It is a %.

  24. For object being pushed up an inclined plane: …effort distance de = ramp length L, and …resistance distance dr = ramp height h. Therefore… L = de h = dr As IMA , Fe and de .

  25. First class lever is 2.13 m long. Effort force is 320 N and is 1.60 m from fulcrum. 85 kg mass to be lifted is 0.53 m from fulcrum. Find IMA, AMA, and efficiency of lever. Fe m de dr = 3.0 = 2.6  87%

  26. Ramp of length 3.2 m allows 54 kg mass to be pushed up 73 cm above floor. If force req’d to push mass up ramp is 230 N, find IMA, AMA, and efficiency of ramp. m L h Fe = 4.4 = 2.3  52%

  27. 2000 N 1500 N 800 N 2 m 3 m 2 m 4 m RA RB (i.e., ) Find reactions. Ignore beam’s weight. SF = 0 RA + RB – 2700 = 0 StA = 0 + 800(5) – 1500(2) – 2000(11) RB(7) = 0 Solve for… RB = 3000 N From SF = 0, RA = –300 N

  28. 25 kg 6.5 m 1.3 m Fe = ? Eff = 76% = 5.0 = 0.76 (5.0) = 3.8 = 65 N

  29. Win = Fe de t = F d Wout = Fr dr St = Ia L = Iw Lf = Li If wf = Ii wi KErot = ½ I w2

  30. h h

  31. h h

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