Holt Physics Chapter 8. Rotational Equilibrium and Dynamics

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1 Holt Physics Chapter 8 Rotational Equilibrium and Dynamics

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

3 Apply two equal and opposite forces acting on a stationary meter stick. F Does the meter stick move? 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.

A torque is produced by a force acting on an extended (not point-like) object. The torque depends on how strong the force is, and where it acts on the object.

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

5 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

6 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.

7 Force is POSITIVE if the 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 F 1 F 2 + or -? + - or -? A

8 net = = = F 1 d 1 + (-F 2 d 2 ) The sign of the net torque will tell you which direction the object will rotate. F 1 F 2 d 1 d 2 A

9 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???

10 F 2 =11 N + or -? - 14cm 7cm F 1 =15 N + - or -?

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

12 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? Draw a diagram!!!

13 49 67N 0.30m 0.40m 34N 35 Signs??? BOTH are POSITIVE!!

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

15 Practice 8A, page 282 Section Review #3 and #4 Same page

16 Happy Thursday! Please take out 8a for me to check.

17 8B 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 10m F T 45m 67kg? 60N 89kg

20 Convert to Newtons! 10m F T 45m 67kg? 60N 89kg

21 10m F T 45m 657N? 60N 873N

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

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

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

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

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 10m F T 12.5m 45m 657N 39.6kg 60N 873N

27 F y = 0, F down = F up, change all to forces Fish is 39.6kgX9.81m/s 2 = 389N F T 10m12.5m 45m -657N -60N -389N 39.6kg -873N

28 F y = 0, F down = F up F T = - F down =-(-657N- 389N -60N-873N) =1980N F T 10m12.5m 45m -657N -389N -60N -873N

29 How to draw a bridge and a car with pillars or a board and a weight lifted by two people F, pillar 1/ person 1 F, pillar 2/ person 2 F w, car or object F w, bridge or board

30 8B (2,3 and 4)page 288

31 Happy Thursday!

32 Write this down 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.

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

34 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 x F R Tx =F T cos y R F Ty=F T sin 1.50 m 53 F T 5.00 m 545 N 315 N

35 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 y R x R 1.50 m 53 F T 5.00 m 315 N 545 N

36 Choose an axis and sum the torques remembering signs for direction of rotation!! R y A 0 R x R 5.00 m 1.50 m N 315 N 53 + Why choose the pin for A? This eliminates R as a variable!! F T

37 =F T L(sin ) F g,b L/2 F g,p d=0 Now substitute and solve for F T. F T = (F g,b L/2 +F g,p d)/(lsin ) F T =315N(2.5m) + 545N(1.5m) R x 5.0mSin53 R y A R 1.50 m m 315 N 545 N F T

38 F T = 1605Nm/4.0m F T = 4.0X10 2 N R x R y A R 1.50 m m 315 N 545 N F T

39 We re not done yet!! Go back to the first condition of equilibrium to solve for Rx, Ry and R. R x R y A R 1.50 m m 315 N 545 N F T

40 R x - F T cos = 0 R 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 + 545N + 315N = 540 N R = (R x2 + R y2 ) = (240N N 2 ) =590 N R x R y A R 1.50 m m 315 N 545 N F T

41 Torque Work Packet

42 Happy Friday! Finish Work packet!

43 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

44 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

45 Which will have the higher average velocity when allowed to roll down a ramp from rest? (refer to pg. 285) 9kg solid ball r=0.50m 8kg hollow ball r=0.40m Solid Ball I = 0.9kgm 2, hollow ball I = 0.85 kgm 2

46 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

47 Example 1 An athlete tosses tennis ball using only the rotation of his forearm to accelerate the ball. The forearm and tennis ball have a combined moment of inertia of kgm 2, and the forearm length is 0.29 m. If the ball has a tangential acceleration of 55m/s 2 just before it is released, what is the torque on the arm and ball?

48 I = kgm 2 a t =55m/s 2 r=0.29m =? =I, where =a t /r =I a t /r = kgm 2 X55m/s 2 /0.29m= 33Nm

49 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?

50 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

51 = -1.2 rad/s 2 =I = 1/2mr 2 = =1/2(.025kg)(.07) 2 (-1.2rad/s 2 ) =-7.35X10-5 Nm

52 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.

53 Fire Drill and Lock down changes If there is a fire drill during a class change, report to the waiting area outside for the teacher you are going to. Lock down: before hiding in the back, we will blockade the door with a table and then consider, if there is a breach, what we have to fight with.

54 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, and ω =v t /r (you will need to use ch. 7 formulas again)

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

56 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 merrygo-round is initially 0.20 rad/sec, what is its angular speed when the student reaches a point 0.50m from the center?

57 m m = 525 kg r i,s = r m =2.00m m s =65 kg r 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

58 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 m2 i +M s R s,i2 i =½M m R m2 f +M s R s,f2 f

59 ½M m R m2 i +M s R s,i2 i = ½M m R m2 f +M s R s,f2 f ½525kg(2.00m) 2 (0.20rad/s)+65kg(2.00m) 2 (0.20rad/s)= ½525kg(2.00) 2 f +65kg(0.50m) 2 f =1050 f f Plug it into the calculator and solve for f f =0.245rad/s

60 Example 2 A comet has a speed of 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 X10 4 m/s?

61 L i =L f Point Mass 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 = X10 4 m/s MR I i i2 = Iv i /r f i = MR f2 v f /r f R i v i = R f v f R f =R i v i / v f

62 R f = 4.95X10 10 m X X 10 4 m/s R f = 6.95X10 10 m X10 4 m/s

63 8 D, page 294

66 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.

67 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? v 2.00m 30

68 h i = 2.00 m R = m m = 4.10 kg v i = 0.0 m/s = 30.0 h f = 0 m v f =? I What = 2/5MR will 2 I be??? v 2.00m 30

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

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

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

72 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) kgm 2 /s 2 = 2.05kg v f kg v f kg v f 2 = kgm 2 /s 2 v f 2 =28 m 2 /s 2 v f = 5.29 m/s

73 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?

74 m p = 3.5kg, v i,p = 5.4m/s θ=30 h f =? Hypotenuse (slope dist.)=? ½ mv 2 i + ½ I (v 2 i +mgh i = ½ mv 2 f + ½ I 2 t,i /r) 2 f + 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 =( )/ mg h f =2.08m sin30=opp/hyp Hyp=opp/sin30 (v t,i /r) 2 D=2.08m/sin30=4.2m 30 height

75 Kinetic Energy Practice 8E, page 297

76 Review Problems: 6,7,8,9, 11,15,18,20,21,24,27,35,37,50

78 Simple Machines A machine is any device that transmits or modifies force. There are six simple machines Lever, inclined plane, wheel and axle, wedge, pulley and screw

79 Using Simple Machines Machines usually give a mechanical advantage (ratio of output force to input force) Formula MA = output force = F out input force F in Because Torque input = Torque output, the ratio of the distances also gives MA. (see page 299) MA = d in /d out

80 Efficiency Because Work in = Work out, if a machine is frictionless, F 1 d 1 =F 2 d 2 (see page 299 again) Efficiency is a measure of how much input energy is lost (heat or sound because of friction) compared with how much energy is used to perform work on an object. eff = W out /W in

81 Happy Friday!! Lab Today! Build a device using at least 3 simple machines (see page 298) that will lift a 200 gram mass at least 6 cm and determine the MA and the efficiency of the device (see page 301) The most efficient device that meets the criteria wins!

82 Section Review page 301 Review for Test Review Problems page ,6-9,11,12,14,15,18-20,21,24,27,31,32,34,35,3 7,40,43,72,75

83 8B Example with a different axis of rotation for the torque 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.

84 L=5.00 m F g,beam =315 N =53 F g,person =545 N F T =? R=? d=1.00m Record distances, put weight of object at center of mass and position all forces. R 1.50 m 53 F T 5.00 m 545 N 315 N

85 The unknowns are 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 x F R Tx =F T cos y R F Ty=F T sin 1.50 m 53 F T 5.00 m 545 N 315 N

86 R x - F T cos = 0 R y +F T sin -F g,p F g,b =0 R x =F T cos R y =-F T sin +F g,p +F g,b Because there are too many unknowns, pause and go to the second condition of equilibrium. R x R y R 1.50 m 53 F T 5.00 m 545 N 315 N

87 Choose an axis and sum the torques remembering signs for direction of rotation!! R y R x R 5.00 m 1.50 m A 545 N 315 N 53 We have to put the axis at the center of mass!! This does not eliminate R as a variable!! F T But it does Eliminate the 315N force.

88 =F T L/2(sin ) R y L/2 +F g,p d=0 Now substitute previous equation for R y (R y =-F T sin +F g,p +F g,b ) into the torque eauation and solve for F T. R x R y R 1.50 m A 53 F T 5.00 m 545 N 315 N

89 =F T L/2(sin ) R y L/2 +F g,p d=0 =F T 5.00m/2(sin53) R y 5.00m/2 +545N(1.0m)=0 R x R y R A 1.50 m 53 F T 5.00 m 545 N 315 N

90 =F T 5.00m/2(sin53) (-F T sin +F g,p +F g,b )5.00m/2 +545N(1.0m)=0 F T 5.00m/2(sin53) (-F T sin )5.00m/2 +545N(1.0m)=0 F T (1.997m) + F T (.799)(2.5m)-860N(2.5m) +545Nm= m F T = 2150Nm-545Nm F T = 1605Nm/3.995m=402N=4.0X10 2 N R x R y R A 1.50 m 53 F T 5.00 m 545 N 315 N

91 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 x R y A R 1.50 m m 315 N 545 N F T

92 R x - F T cos = 0 R 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 N = 540 N R = (R x2 + R y2 ) = (240N N 2 ) =590 N R x R y A R 1.50 m m 315 N 545 N F T

93 A bridge 20.0 m long and weighing 4.00X10 5 N is supported by two pillars located 3.00 m from each end. If a 1.96X10 4 N car is parked 8.00 m from one end of the bridge, how much force does each pillar exert?

94 L=20.0 m F g,bridge =4.00X10 5 N d pillars = 3.0m from end F g,car =1.96X10 4 N d car =8.00m F p1 =? F p2 =? Record distances, put weight of object at center of mass and position all forces. F P2 3.0 m 8.0 m 20.0 m 4.00X10 5 N 1.96X10 4 N F P1 3.0 m

95 The unknowns are F p1 and F p2. Because we have equilibrium, F x = 0, F y = 0 F p2 + F p1 -F car -F bridge = 0 F p2 = F c +F b -F p1 = 4.2X F p1 F P2 3.0 m 8.0 m 20.0 m 4.00X10 5 N 1.96X10 4 N F P1 3.0 m

96 Now lets look at torque Choose the axis to be a pillar to eliminate an unknown F P2 A 3.0 m 8.0 m 20.0 m 4.00X10 5 N 1.96X10 4 N F P1 3.0 m

97 Now lets look at torque =F p1 (14.0m) F c (5.0m) -F b (7.0m)=0 =F p1 (14.0m) (1.96X10 4 N)(5.0m) - (4.00X10 5 N) (7.0m)=0 F p1 =2.07X10 5 N F p2 = 4.2X F p1 = 4.2X X10 5 N F p2 = 2.13X10 5 N F P2 A 3.0 m 8.0 m 20.0 m 4.00X10 5 N 1.96X10 4 N F P1 3.0 m

98 1m 3m F r1 F r2 700N 200N Window washer.problem 3, page 288 F x = 0, F y = 0 F r1 + F r2-200n -700N= 0 =F r2 (3.00m) 700N(1.0m)-200N(1.5m)=0 F r2 = 333N F r1 = 900N- F r2 = 900N 333 = 567N

3. A bicycle tire of radius 0.33 m and a mass 1.5 kg is rotating at 98.7 rad/s. What torque is necessary to stop the tire in 2.0 s?

3. A bicycle tire of radius 0.33 m and a mass 1.5 kg is rotating at 98.7 rad/s. What torque is necessary to stop the tire in 2.0 s? Practice 8A Torque 1. Find the torque produced by a 3.0 N force applied at an angle of 60.0 to a door 0.25 m from the hinge. What is the maximum torque this force could exert? 2. If the torque required