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Chapter 12 Rotation of a Rigid Body. Vector (or cross) Product Cross Product is a vector perpendicular to the plane of vectors A and B.

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Presentation on theme: "Chapter 12 Rotation of a Rigid Body. Vector (or cross) Product Cross Product is a vector perpendicular to the plane of vectors A and B."— Presentation transcript:

1 Chapter 12 Rotation of a Rigid Body

2 Vector (or cross) Product Cross Product is a vector perpendicular to the plane of vectors A and B

3 Cross Product of Vectors AP Physics C 3 Right hand rule: Curl your right hand around the center of rotation with the fingers going from the first vector to the second vector and the thumb will be pointing in the torque direction

4 Cross Product Problem AP Physics C 4 C D 110° Csin(110)

5 1.Clockwise. 2.Counter-clockwise. 3.Not at all. 4.Not sure what will happen. Which way will it rotate once the support is removed? Test your Understanding

6 Torque AP Physics C 6 If the forces are equal, which will open the heavy door more easily?

7 Interpretation of torque Measures tendency of any force to cause rotation Torque is defined with respect to some origin – must talk about torque of force about point X, etc. Torques can cause clockwise (+) or anticlockwise rotation (-) about pivot point

8 Torque cont AP Physics C 8

9 Torque cont AP Physics C 9

10 Definition of Torque: where is the vector from the reference point (generally either the pivot point or the center of mass) to the point of application of the force. If r and F are not perpendicular then: where is the angle between the vectors and.

11 Definition of Torque:

12 Torque Problem AP Physics C 12 Adrienne (50 kg) and Bo (90kg) are playing on a 100 kg rigid plank resting on the supports seen below. If Adrienne stands on the left end, can Bo walk all the way to the right end with out the plank tipping over? If not, How far can he get past the support on the right?

13 Torque Problem cont AP Physics C 13 2m3m4m 50kg 100kg90kg N1N1 N2N2 x

14 Moments AP Physics C 14 M1M1 M1M1 M2M2 M2M2 d1d1 d2d2 Suppose we have masses m 1 and m 2 on the seesaw at distances d 1 and d 2, respectively, from the fulcrum, when does the seesaw balance? By Archimedes Law of the lever, this occurs when m 1 d 1 + m 2 d 2 = 0

15 Moments cont AP Physics C15 M1M1 M1M1 M2M2 M2M2 x1x1 x2x2 If we place a coordinate system so that 0 is at the fulcrum and if we let x i be the coordinate at which is placed then: m 1 x 1 + m 2 x 2 = m 1 d 1 + m 2 d 2 = 0

16 Moments cont AP Physics C16 More generally, if we place masses m 1, m 2, …, m r at points x 1, x 2, …., x r, respectively, then the see saw balances with the fulcrum at the origin, if and only if m 1 x 1 + m 2 x 2 + …+ m r x r = 0 M1M1 M1M1 M4M4 M4M4 x1x1 x4x4 M2M2 M2M2 M3M3 M3M3 x2x2 x3x3 M5M5 M5M5 x5x5

17 Moments cont AP Physics C17 Now, suppose that we place masses m 1, m 2, …, m r at points x 1, x 2, … x r, respectively, then where should we place the fulcrum so that the seesaw balances? The answer is that we place the fulcrum at x-bar where: m 1 (x 1 - (x-bar) ) + m 2 (x 2 - (x-bar) ) + …+ m r (x r - (x-bar) )= 0 M1M1 M1M1 M4M4 M4M4 x1x1 x4x4 M2M2 M2M2 M3M3 M3M3 x2x2 x3x3 M5M5 M5M5 x5x5

18 Moments cont AP Physics C18 m 1 (x 1 - (x-bar) ) + m 2 (x 2 - (x-bar) ) + …+ m r (x r - (x-bar) ) is called the moment about x-bar. Moment is from the Greek word for movement, not time. If positive, movement is counter-clockwise, negative it is clockwise. M1M1 M1M1 M4M4 M4M4 x1x1 x4x4 M2M2 M2M2 M3M3 M3M3 x2x2 x3x3 M5M5 M5M5 x5x5 x-bar

19 Moments cont AP Physics C19 M1M1 M1M1 M4M4 M4M4 x1x1 x4x4 M2M2 M2M2 M3M3 M3M3 x2x2 x3x3 M5M5 M5M5 x5x5 x-bar Suppose that We want to solve for

20 Center of Mass AP Physics C20 M1M1 M1M1 M4M4 M4M4 x1x1 x4x4 M2M2 M2M2 M3M3 M3M3 x2x2 x3x3 M5M5 M5M5 x5x5 x-bar

21 Center of Mass cont AP Physics C 21 M1M1 M1M1 M4M4 M4M4 M2M2 M2M2 M3M3 M3M3 Suppose m 1, m 2, …, m r are masses located at points (x 1, y 1 ), (x 2, y 2 ), …, (x r, y r ). The moment about the y-axis is: The moment about the x-axis is: Center of Mass is

22 Center of Mass cont AP Physics C 22 Now lets find the center of mass of a thin plate with uniform density, ρ. First we need the mass of the plate:

23 Center of Mass cont AP Physics C23 Next we need the moments of the region: f(x) g(x) x To find the center of mass we divide by mass:

24 Center of Mass Problem AP Physics C 24 Determine the center of mass of the region bounded by y = 2 sin (2x) and y = 0 on the interval, [0, π/2] Given the symmetry of the curve it is obvious that x-bar is at π/4. First find the area. Using the table of integrals:

25 Moment of Inertia AP Physics C 25 M CR r F

26 Moment of Inertia cont AP Physics C 26

27 Calculating Moment of Inertia AP Physics C 27

28 Calculating Moment of Inertia AP Physics C 28

29 Moment of Inertia cont AP Physics C 29

30 Parallel Axis Theorem AP Physics C 30

31 ||-axis Theorem Proof AP Physics C 31

32 Rotational Kinetic Energy AP Physics C 32

33 Rotational Dynamics AP Physics C 33

34 Rotation About a Fixed Axis AP Physics C 34

35 Bucket Problem AP Physics C 35 A 2.0 kg bucket is attached to a mass-less string that is wrapped around a 1.0 kg, 4.0 cm diameter cylinder, as shown. The cylinder rotates on an axel through the center. The bucket is released from rest 1.0 m above the floor, How long does it take to reach the floor

36 Bucket Problem cont AP Physics C 36

37 Bucket Problem Cont AP Physics C 37

38 Static Equilibrium AP Physics C 38

39 Statics Problem AP Physics C 39 A 3.0 m ladder leans against a frictionless wall at an angle of 60°. What is the minimum value of μ s, that prevents the ladder from slipping?

40 Statics Problem AP Physics C 40

41 Balance and Stability AP Physics C 41 θcθc h t/2 θcθc h

42 Rolling Motion AP Physics C 42

43 Rolling Motion cont AP Physics C 43

44 Rolling Motion cont AP Physics C 44

45 Rolling Motion cont AP Physics C 45

46 Rolling Kinetic Energy AP Physics C 46

47 Great Downhill Race AP Physics C 47

48 Downhill Race cont AP Physics C 48

49 Angular Momentum AP Physics C 49

50 Angular Momentum AP Physics C 50

51 Angular Momentum AP Physics C 51

52 Conservation of Angular Momentum AP Physics C 52 Two equal masses are at the ends of a mass- less 500 cm long rod. The rod spins at 2.0 Rev/s about an axis through its midpoint. If the rod lengthens to 160 cm, what is the angular velocity

53 Testing Understanding AP Physics C 53 There is no torque on the buckets so angular momentum is conserved. Increased mass in buckets increases inertia so angular velocity must decrease.

54 Problem 1 AP Physics C 54 An 18 cm long bicycle crank arm with a pedal at one end is attached to a 20 cm diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 60 rpm to 90 rpm in 10 s. a. What is the tangential acceleration of the pedal? b. Want length of chain passes over the top of the sprocket during this interval?

55 Problem 1 cont AP Physics C 55 a. Since at = rα, find α first. With 60 rpm = 6.28 rad/s and 90 rpm = 9.43 rad/s: b. Since L = rθ, find rθ. The angular acceleration of the sprocket and chain are the same. The length of chain which has passed over the top of the sprocket is

56 Problem 2 AP Physics C 56 A 100 g ball and a 200 g ball are connected by a 30 cm mass-less rigid rod. The balls rotate about their center of mass at 120 rpm. What is the speed of the 100g ball?

57 Problem 3 AP Physics C 57 A 300 g ball and a 600 g ball are connected by a 40 cm mass-less rigid rod, The structure rotates about its center of mass at 100 rpm. What is its rotational kinetic energy?

58 Problem 4 AP Physics C 58 A 25 kg solid door is 220 cm tall, 91 cm wide. What is the doors moment of inertia for (a) rotation on its hinges (b) rotation about a vertical axis inside the door, 15 cm from one edge.

59 Problem 5 AP Physics C 59 An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.o kg. What is the magnitude of the torque about his shoulder if he holds the ball (a) straight out to his side, parallel to the floor and (b) straight, but 45 degrees below horizontal.

60 Problem 5 cont AP Physics C 60

61 Problem 6 AP Physics C 61 Starting from rest, a 12 cm diameter compact disk takes 3.0 s to reach its operating angular velocity of 2000 rpm. Assume that the angular acceleration is constant. The disks moment of inertia is 2.5x10 -5 kgm 2. (a) How much torque is applied? (b) How many revolutions does it make before reaching full speed?

62 Problem 6 cont AP Physics C 62 a. Using the rotational kinematic equation b.

63 Problem 7 AP Physics C 63 The two objects in the figure are balanced on the pivot. What is the distance d?


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