1. Try to lift the meter stick about an axis at the end by lifting on the far end. KEEP THE “AXIS” ( ) ON THE TABLE! Is it easy to lift? F Now change the position of your force to the middle. Now change the position to about 20 cm from the axis. How is the force required changing? FF

2. Try to lift the meter stick about an axis at the end by lifting on the far end. KEEP THE “AXIS” ( ) ON THE TABLE! Is it easy to lift? F Now change the position of your force to the middle. Now change the position to about 20 cm from the axis. How is the force required changing? FF

3. Holt Physics Chapter 8 Rotational Equilibrium and Dynamics

4. Apply two equal and opposite forces acting at the center of mass of a stationary meter stick. Does the meter stick move? F2F2 F1F1 F ext = 0. F 1 =F 2

5. Apply two equal and opposite forces acting on a stationary meter stick. Does the meter stick move? F F The center of mass of the meter stick does not accelerate, so it does not undergo translational motion. However, the meter stick would begin to rotate about its center of mass.

6. A torque is produced by a force acting on an extended (not point-like) object. F The torque depends on how strong the force is, and where it acts on the object. Torques cause changes in rotational motion. not a force Torque is a vector. It is not a force,* but is related to force. never *So never set a force equal to a torque! A You must always specify your reference axis for calculation of torque. By convention, we indicate that axis with the letter “A” and a dot.

7. Torque Torque is a quantity that measures the ability of a force to rotate an object around some axis. Torque depends on force and the lever arm.  Lever arm (moment arm)is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force.  See figure 8-3, page 279

8. The most torque is produced when the force is perpendicular to the object.  Formula   =Fd(sin  )  d is the lever arm and  is the angle between the lever arm and the force. (If 90º, then sin  =1).  See figure 8-5, page 280.

9. Force is POSITIVE if rotation is counterclockwise.  If there is more than one force, add the two resultant torques, using the appropriate signs.  EX: Wishbone…sum the two torques A F2F2 F1F1

10.   net =  =  1 +  2 = F 1 d 1 + (-F 2 d 2 )  The sign of the net torque will tell you which direction the object will rotate. A F2F2 F1F1 d1d1 d2d2

11. Example 8A, page 281 A basketball is being pushed by two players during tip-off. One player(to the right) exerts a downward force of 11 N at a distance of 7.0 cm from the axis of rotation. The second player(to the left) applies an upward fore of 15 N at a perpendicular distance of 14cm from the axis of rotation. Find the net torque acting on the ball.  Direction and sign??? Units???

12. F 1 = -15 NF 2 = -11N d 1 = 0.14md 2 = 0.070m  net =?  net =  1 +  2 = F 1 d 1 + F 2 d 2 (-15N X 0.14m)+(-11N X 0.070m) = -2.9Nm

13. Example with an Angle An upward 34N force is exerted on the right side of a meter stick 0.30 m from the axis of rotation at an angle of 35 degrees. A second downward force of 67N is exerted at an angle of 49 degrees to the meter stick 0.40m to the left of the axis of rotation. What is the net torque?

14. 34N 67N 35 49 0.30m 0.40m Signs??? BOTH are POSITIVE!!

15. F 1 = 34N F 2 = 67N d 1 = 0.30md 2 = 0.40m  1 = 35   2 = 49   net =?  net =  1 +  2 = F 1 d 1 sin  1 + F 2 d 2 sin  2 (34NX.30m)sin35  +(67NX.40m)sin49  = 26Nm

16. Practice 8A, page 282

17. Equilibrium Complete equilibrium requires zero net force and zero net torque.  Translational equilibrium: net force in x and y direction = 0  Called 1 st condition of equilibrium  ∑F x = 0, ∑Fy = 0  Rotational equilibrium: net torque=0  Called 2 nd condition of equilibrium  ∑  = 0

18. A 45.0m beam that weighs 60.0N is supported in the center by a cable. The beam is in equilibrium and supports three masses. A 67.0kg mass is on one end, an 89.0kg mass is on the other. A fish is hanging 10.0m from the 67.0 kg mass. What is the mass (kg!) of the fish and what is the tension (force!) in the cable.

19. 67kg 89kg 60N 45m 10m ? FTFT

20. 67kg 89kg 60N Convert to Newtons! 45m 10m ? FTFT

21. 657N 873N 60N 45m 10m ? FTFT

22. 657N 873N 60N Choose Axis Choose center to eliminate a variable!! 45m 10m ? FTFT

23. 657N 873N 60N How far is the fish from the axis? 45m 10m ? 12.5m FTFT

24. 657N 873N 60N Calculate Torques and sum to zero 45m 10m ? 12.5m FTFT

25. 657N 873N 60N Assign sign to torques 45m 10m ? 12.5m + +-0 FTFT

26. ∑  =0= 657N(22.5m)+W f (12.5m)–873N(22.5m)=0 W f =4860Nm/(12.5m) = 388.8N m f =388.8N/9.81m/s 2 =39.6kg 657N 873N 60N 45m 10m 39.6kg 12.5m FTFT

27. ∑F y = 0, ∑F down = ∑F up, change all to forces Fish is 39.6kgX9.81m/s 2 = 389N -657N -873N-60N 45m 10m -389N 12.5m FTFT 39.6kg

28. ∑F y = 0, ∑F down = ∑F up F T = - ∑F down =-(-657N- 389N -60N-873N) =1980N -657N -873N-60N 45m 10m -389N 12.5m FTFT

29. F, pillar 2 F w, bridge How to draw a bridge with pillars… F, pillar 1 F w, car

30. Begin Worksheet

31. Example 8B, page 287-8 A uniform 5.00 m long horizontal beam that weighs 315N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53 degrees with the horizontal, and a 545 N person is standing 1.5 m from the pin. Find the force in the cable, F T, and the force exerted on the beam by the wall, R, if the beam is in equilibrium.

32. L=5.00 m F g,beam =315 N  =53  F g,person =545 N d=1.50m F T =?R=? Record distances, put weight of object at center of mass and position all forces. R FTFT 315 N 1.50 m 53  5.00 m 545 N

33. The unknowns are R ( R x, R y ), and F T Because we have equilibrium, ∑F x = 0, ∑F y = 0 R x - F T cos  = 0 R y + F T sin  - F g,p – F g,b = 0 R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx F Ty =F T sin  F Tx =F T cos 

34. R x - F T cos  = 0 R y + F T sin  - F g,p – F g,b = 0 Because there are too many unknowns, pause (don’t panic) and go to the second condition of equilibrium. R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx

35. Choose an axis and sum the torques…remembering signs for direction of rotation!! R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx A Why choose the pin for A? This eliminates R as a variable!! 0 + - -

36.  =F T L(sin  ) – F g,b L/2 –F g,p d=0 Now substitute and solve for F T. R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx A

37. F T 5.00m(sin53) – (315N)(5.00m)/2 –(545N)(1.5m)=0 F T = 1606Nm/4.0m F T = 4.0X10 2 N R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx A

38. We’re not done yet!! R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx A

39. Now we can substitute force in the wire (F T ) into the R x and R y equations to find R x and R y, and then solve for R… R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx A

40. R x - F T cos  = 0R x = F T cos  R x = 400N X cos53  = 240N R y +F T sin  -F g,p –F g,b =0 R y = -F T sin53  + F g,p +F g,b Ry = -3.2X10 2 N + 860 N = 540 N R =  (R x 2 + R y 2 ) =  (240N 2 + 540N 2 ) =590 N R FTFT 315 N 1.50 m 53  5.00 m 545 N RyRy RxRx A

41. Finish Worksheet! Homework

42. Moment of Inertia The moment of inertia is the resistance of an object to changes in rotational motion about some axis.  Similar to mass…mass ( simple inertia) is the measure of resistance to translational motion…

43. Moment of Inertia depends on the object’s mass and the distribution of mass around the axis of rotation.  The farther the center of mass from the axis of rotation, the more difficult it is to rotate the object, and therefore, the higher the moment of inertia.  Use Table 8-1, page 285

44. Newton’s 2 nd Law F=ma can be translated to rotational motion.   net = I  = Ia t /r  I = moment of inertia   = angular acceleration   net = net torque  a t =tangential acceleration  r=radius

45. Example page 291 A student tosses dart using only the rotation of her forearm to accelerate the dart. The forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart have a combined moment of inertia of 0.075 kgm 2 about the axis, and the length of the forearm is 0.26 m. If the dart has a tangential acceleration of 45m/s 2 just before it is released, what is the net torque on the arm and dart?

46. I = 0.075 kgm 2 a t =45m/s 2 r=0.26m  =?  =I , where  =a t /r  =I a t /r  = 0.075 kgm 2 X45m/s 2 /0.26m= 13Nm

47. Example 2 A 25 g CD (radius =7.0 cm) is rotating at 100 rev/min. If it stops in 8.5 sec, what is the angular acceleration of the CD? How much torque is required to stop the CD?

48. r=0.07m, m=0.025kg,  t=8.5 sec,  i = 100rev/min=100x2  60=10.5rad/s  f = 0  =  /  t = (0-10.5rad/s)/8.5s

49. r=0.07m, m=0.025kg,  t=8.5 sec,  i = 100rev/min=100x2  60=10.5rad/s  f = 0  =  /  t = (0-10.5rad/s)/8.5s  = -1.2 rad/s 2  =I  …What is I? Look in table on page 285 Rotating disk is 1/2mr 2

50.  = -1.2 rad/s 2  =I  = 1/2mr 2  = =1/2(.025kg)(.07) 2 -1.2rad/s 2 =-7.35X10 -5 Nm

51. 8C, page 291 Be ready to use table 8-1 on page 285 to calculate I and use old chapter 7 formulas for quantities like , ,  s, a t and .

52. Momentum and Rotation Linear momentum can be translated to angular momentum  L = I   L = angular momentum  I = moment of inertia – look in the table again (page 285!!)   = angular speed (you may need to use ch. 7 formulas again)

53. Conservation of Angular Momentum As in linear momentum, angular momentum is also conserved. L i = L f

54. Example page 293 A 65 kg student is spinning on a merry-go-round that has a mass of 5.25X10 2 kg and a radius of 2.00 m. She walks from the edge of the merry-go-round toward the center. If the angular speed of the merry- go-round is initially 0.20 rad/sec, what is its angular speed when the student reaches a point 0.50m from the center?

55. m m = 525 kg m s =65 kg r i,s = r m =2.00mr f,s = 0.50 m  i =0.20 rad/s  f =? Use conservation of momentum L i = L f L m.i + L s,i = L m,f + L s,f Need moments of inertia!! Because L = I  = Iv t /r

56. The merry-go-round is a _____ The Student is a _____ Merry-go-round (I = ½ MR 2 ) Student (I = MR 2 ) L m.i + L s,i = L m,f + L s,f ½M m R m 2  i +M s R s,i 2  i = ½M m R m 2  f +M s R s,f 2  f

57. ½525kg(2.00m) 2 (0.20rad/s) +65kg(2.00m) 2 (0.20rad/s) = ½525kg(2.00) 2  f +65kg(0.50m) 2  f Plug it into the calculator and solve for  f  f =0.2435rad/s =0.24 rad/s Keep ch. 7 formulas handy too! Esp. v t =r 

58. Example 2 A comet has a speed of 7.056 X 10 4 m/s at a distance of 4.95X10 10 m. At what distance from the sun would the comet have a speed of 5.0278X10 4 m/s?

59. So, I = MR 2 Mass is constant ω= v t /r R i = 4.95X10 10 m V i =7.056 X 10 4 m/s R f = ? V f = 5.0278X10 4 m/s MR i 2 v i /r i = MR f 2 v f /r f R i v i = R f v f R f =R i v i / v f L i =L f Point Mass

60. R f = 4.95X10 10 m X 7.056 X 10 4 m/s 5.0278X10 4 m/s R f = 6.95X10 10 m

61. 8 D, page 294

62. Happy Friday! All labs in? Today we will cover Conservation of Mechanical Energy. We will skip simple machines and start review problems on Monday for a Wednesday Test.

63. Kinetic Energy Rotational Kinetic energy (KE rot ) is the kinetic energy associated with their angular speed.  Formula KE rot = ½ I  2 = ½ I(v t /r) 2  Conservation of Kinetic Energy also applies…  KE trans + KE rot + PE i = KE trans + KE rot + PE f ½ mv i 2 + ½ I  i 2 + mgh i = ½ mv f 2 + ½ I  f 2 + mgh f Be sure to keep track of initial and final conditions as well as angular vs. translational speeds and moments of inertia.

64. Example page 296 A solid ball with a mass of 4.10 kg and a radius of 0.050m starts from rest at a height of 2.00 m and rolls down a 30  slope. What is the translational speed of the ball when it leaves the incline? 2.00m 30  v 

65. h i = 2.00 m m = 4.10 kg R = 0.050 mv i = 0.0 m/s  = 30.0  h f = 0 m v f = ? 2.00m 30  v  I = 2/5MR 2 What will I be???

66. h i = 2.00 mm = 4.10 kg R = 0.050 mv i = 0.0 m/s  = 30.0  h f = 0 mv f = ? ½ mv i 2 + ½ I  i 2 + mgh i = ½ mv f 2 + ½ I  f 2 + mgh f 2.00m 30  v  WE have two variables So we need to find one In terms of the other… REMEMBER!!!  =v t /r

67. h i = 2.00 mm = 4.10 kg R = 0.050 mv i = 0.0 m/s  = 30.0  h f = 2.00 mv f = ? ½ mv i 2 + ½ I  i 2 + mgh i = ½ mv t,f 2 +½I  f 2 +mgh f 2.00m 30  v  Substitute v t /r into The equation for  f (v t,f /r)

68. 4.10kg(9.81m/s 2 ) (2.00m) = ½ 4.10kgv f 2 +½I (v f /0.050m) 2 I = 2/5MR 2 4.10kg(9.81m/s 2 ) (2.00m) = ½ 4.10kgv f 2 +½(2/5MR 2 ) (v f /0.050m) 2 PLUG IN THE NUMBERS

69. 80.442kgm 2 /s 2 = 2.05kg v f 2 + 0.82kg v f 2 2.87kg v f 2 = 80.442kgm 2 /s 2 v f 2 =28 m 2 /s 2 v f = 5.29 m/s 4.10kg(9.81m/s 2 ) (2.00m) =½ 4.10kg v f 2 +½(2/5)(4.10kg)(0.050m) 2 (v f /0.050m) 2

70. Example 2 A 3.5 kg spherical potato (radius.070m) is kicked up a 30 degree slope at a speed of 5.4 m/s. What distance along the slope did the potato roll before it stopped?

71. m p = 3.5kg, v i,p = 5.4m/s θ=30° h f =? Hypotenuse (slope dist.)= ? ½ mv i 2 + ½ I  i 2 +mgh i = ½ mv f 2 + ½ I  f 2 + mgh f *Substitute v t /r for ω ½ mv i 2 + ½ I  i 2 = mgh f *Solve for h f h f =½ mv i 2 + ½ I(v t /r) 2 =(51.03 + 20.412)/34.335 mg h f =2.08m D=2.08/sin30=4.2m (v t,i /r) 2 distance height 30

72. Kinetic Energy  Practice 8E, page 297