Wednesday, October 12, 11. What is Kinetic Energy?

Transcription

1 What is Kinetic Energy?

2 What is Kinetic Energy? The kinetic energy of an object is the energy which it possesses due to its motion.

3 What is Kinetic Energy? The kinetic energy of an object is the energy which it possesses due to its motion. How is it defined?

4 What is Kinetic Energy? The kinetic energy of an object is the energy which it possesses due to its motion. How is it defined? K = 1 2 mv2

5 Rotational energy Just like linear kinetic energy is ½ mv 2, the angular kinetic energy is ½ Iω 2.

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8 The greater the moment of inertia, the greater the kinetic energy of a rigid body rotating with angular speed ω.

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17 Torque, Rolling

18 When we apply a net force to a body what happens?

19 When we apply a net force to a body what happens? We get an acceleration

20 When we apply a net force to a body what happens? We get an acceleration What does it take to get a a stationary body rotating?

21 When we apply a net force to a body what happens? We get an acceleration What does it take to get a a stationary body rotating? A force that gives a twisting or turning action.this is called a torque.

22 Which of these three equal magnitude forces is most likely to loosen the bolt?

23 Which of these three equal magnitude forces is most likely to loosen the bolt?

24 Which of these three equal magnitude forces is most likely to loosen the bolt?

25 Which of these three equal magnitude forces is most likely to loosen the bolt?

26 Which of these three equal magnitude forces is most likely to loosen the bolt? So both the magnitude & direction of the force are important

27 Fa applies a torque about O to the wrench. Fb applies a greater torque about O to the wrench. Fc applies zero torque about O to the wrench.

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29 The Torque depends on the perpendicular distance between the axis of rotation and the line of action of the force. τ = Fl τ 1 = F 1 r 1 τ 2 = F 2 r 2 Physicists usually use the term torque engineers usually say moment.

30 Who needs to apply the greatest force to turn the door?

31 Only a tangential force produces a torque

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35 Q10.1 The four forces shown all have the same magnitude: F 1 = F 2 = F 3 = F 4. F 1 F 3 O Which force produces the greatest torque about the point O (marked by the blue dot)? F 2 F 4 A. F 1 B. F 2 C. F 3 D. F 4 E. not enough information given to decide 2012 Pearson Education, Inc.

36 A10.1 The four forces shown all have the same magnitude: F 1 = F 2 = F 3 = F 4. F 1 F 3 O Which force produces the greatest torque about the point O (marked by the blue dot)? F 2 F 4 A. F 1 B. F 2 C. F 3 D. F 4 E. not enough information given to decide 2012 Pearson Education, Inc.

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38 From Newton's second law a= F m We already know that a=rα, so α= a r = F mr multiplying the top and bottom by r gives α= r r F mr = rf mr 2 = τ I as τ =rf and I=mr 2, so rearranging gives τ =Iα

39 we use a ( ) to represent vectors coming out of the page and a ( ) to represent a vector that points into the page

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43 Q10.3 A plumber pushes straight down on the end of a long wrench as shown. What is the magnitude of the torque he applies about the pipe at lower right? A. (0.80 m)(900 N)sin 19 B. (0.80 m)(900 N)cos 19 C. (0.80 m)(900 N)tan 19 D. none of the above 2012 Pearson Education, Inc.

44 A10.3 A plumber pushes straight down on the end of a long wrench as shown. What is the magnitude of the torque he applies about the pipe at lower right? A. (0.80 m)(900 N)sin 19 B. (0.80 m)(900 N)cos 19 C. (0.80 m)(900 N)tan 19 D. none of the above 2012 Pearson Education, Inc.

45 Newton s Second Law F = ma τ = Iα

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47 Do you think there is a constant α throughout the tornado?

48 Do you think there is a constant α throughout the tornado? Will Στ=Iα apply?

49 Do you think there is a constant α throughout the tornado? Will Στ=Iα apply? Applies only to rigid bodies

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55 F1 and F2 form an action reaction pair (Newton s third law), so F1=-F2. Will they exert a net torque on the body?

56 F1 and F2 form an action reaction pair (Newton s third law), so F1=-F2. Will they exert a net torque on the body? Only external forces affect the bodies rotation.

57 Loosening or tightening a screw requires giving it an angular acceleration and hence applying a torque. Why do you think it helps to use a screwdriver with a large radius handle?

58 Loosening or tightening a screw requires giving it an angular acceleration and hence applying a torque. Why do you think it helps to use a screwdriver with a large radius handle? τ = rf

59 Problem solving: Rotational Dynamics for Rigid Bodies Identify the relevant concepts. Στ=Iα is useful whenever torques act on a rigid body, i.e. forces act on body to change its rotation. Sometimes we can use an energy approach, but if the target variable is a force, a torque, an acceleration, an angular acceleration, an elapsed time, using Στ=Iα is almost always the best approach. 1. Draw a sketch and select body or bodies to be analyzed. 2. For each body draw a free-body diagram and label unknown quantities with algebraic symbols. Must show the shape of the body, including dimensions and angles you will need in the torque calculations. 3. Choose coordinate axes for each body indicating positive direction. If there is a linear acceleration, it is usually simplest to pick its direction as the positive direction. If you know the sense of α in advance it simplifies things to choose this as the positive direction.

60 Problem solving: Rotational Dynamics for Rigid Bodies Execute 1. For each body decide whether it undergoes translational or rotational motion, or both, then apply ΣF=ma, Στ=Iα, or both to the body. Write separate equations for each body. 2. There may be geometrical relationships between the motions of two or more bodies (e.g. a string that unwinds from a pulley, or a wheel that rolls without slipping). Express this in algebraic form, usually between two linear accelerations, or between a linear acceleration and an angular acceleration. 3. Check the number of equations matches the number of unknowns. Then solve to find the target variable(s). Evaluate the answer, check it makes sense.

61 A cable is wrapped several times around a uniform solid cylinder which can rotate about its axis. The cylinder has a diameter of m and mass 50 kg. The cable is pulled with a force of 9N. Assuming that the cable unwinds without stretching or slipping, what is its acceleration?

62 A cable is wrapped several times around a uniform solid cylinder which can rotate about its axis. The cylinder has a diameter of m and mass 50 kg. The cable is pulled with a force of 9N. Assuming that the cable unwinds without stretching or slipping, what is its acceleration? Target variable is acceleration

63 Find the acceleration using relationship between the motion of the cable and the motion of the cylinder rim. τ = Iα α = τ I = Fr Mr 2 2 = 2F Mr = = 6.0rad / s2 a = rα = = 0.36m / s 2

64 Find the acceleration of the block of mass, m.

65 Find the acceleration of the block of mass, m.

66 Find the acceleration of the block of mass, m. For the cylinder the weight and normal force of bearings balance, so the only torque is due to the cable tension, T

67 Find the acceleration of the block of mass, m. For the cylinder the weight and normal force of bearings balance, so the only torque is due to the cable tension, T The acceleration of the cable is the same as the tangential acceleration of a point on the rim. a y = a tan = Rα z already know that RT = 1 2 MR2 α z, cancelling R T = 1 2 MRα z = 1 2 Ma y

68 Find the acceleration of the block of mass, m. For the cylinder the weight and normal force of bearings balance, so the only torque is due to the cable tension, T The acceleration of the cable is the same as the tangential acceleration of a point on the rim. a y = a tan = Rα z already know that RT = 1 2 MR2 α z, cancelling R T = 1 2 MRα z = 1 2 Ma y Substitute expression for T into Newton's second law mg 1 2 Ma y = ma y g a y = 1+ M 2m

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70 a y = g 1 + M 2m What do you notice about ay?

71 a y = g 1 + M 2m What do you notice about ay? The acceleration is less than g as the cable is holding the block back

72 Rigid Rotation About a Moving Axis Combined translation & rotation. Every possible motion of a rigid body can be represented as a combination of translational motion of the center of mass and rotation about an axis through the center of mass.

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75 Rolling Without Slipping A wheel is symmetrical, so its center of mass is its geometric center. If we consider an inertial frame where the surface is at rest, then the point of contact is also at rest (i.e does not slip).

76 Rolling Without Slipping If at point of contact v 1=0, then ' v 1 = v cm If the wheel has radius R and angular speed ω v cm = Rw is the condition for rolling without slipping

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79 What is the energy of a yo-yo after it has dropped distance h? U 1 = Mgh, and U 2 = 0 K 1 = 0, and K 2 = 1 2 Mv 2 cm Iω 2 since v cm =Rω, and I= 1 2 MR2 K 2 = 1 2 Mv 2 cm MR2 v cm R 2 K 2 = 3 4 Mv 2 cm From conservation of energy K 1 +U 1 =K 2 +U Mgh = 3 4 Mv 2 cm + 0 so v cm = 4 3 gh

80 What is the energy of a yo-yo after it has dropped distance h? We found that K 2 = 3 4 Mv 2 cm What would it have been if the yo-yo was not rotating?

81 What is the energy of a yo-yo after it has dropped distance h? We found that K 2 = 3 4 Mv 2 cm What would it have been if the yo-yo was not rotating? So 2/3 of the total energy is translational, and 1/3 is rotational.

82 Which one gets to the bottom first?

83 Which one gets to the bottom first? Let us say I = cmr 2 where c depends on the shape From conservation of energy K 1 +U 1 =K 2 +U Mgh = 1 2 Mv 2 cm + 1 v cm 2 cmr2 R = 1 2 (1 + c)mv 2 cm v cm = 2gh 1 + c

84 Which one gets to the bottom first? Let us say I = cmr 2 where c depends on the shape From conservation of energy K 1 +U 1 =K 2 +U Mgh = 1 2 Mv 2 cm + 1 v cm 2 cmr2 R All solid cylinders have the same speed at the bottom = 1 2 (1 + c)mv 2 cm v cm = 2gh 1 + c

85 Which one gets to the bottom first? Let us say I = cmr 2 where c depends on the shape From conservation of energy K 1 +U 1 =K 2 +U Mgh = 1 2 Mv 2 cm = 1 2 (1 + c)mv 2 cm + 1 v cm 2 cmr2 R All solid cylinders have the same speed at the bottom Small c bodies beat high c bodies v cm = 2gh 1 + c because they have less kinetic energy tied up in rotation

86 Consider the acceleration of a rolling sphere

87 Consider the acceleration of a rolling sphere with friction

88 Balancing see-saw W w A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground. Given that her heavier child of weight W is sitting a distance L to the left of the pivot, at what distance L1 must she place her second child of weight w on the right side of the pivot to balance the seesaw?

89 L 1 = LW w Balancing see-saw W w A mother is helping her children, of unequal weight, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child simply sinking to the ground. Given that her heavier child of weight W is sitting a distance L to the left of the pivot, at what distance L1 must she place her second child of weight w on the right side of the pivot to balance the seesaw? LW = L 1 w

90 Balancing see-saw W w Find τ, the torque about the pivot due to the weight w of the smaller child on the seesaw.

91 Balancing see-saw W w Find τ, the torque about the pivot due to the weight w of the smaller child on the seesaw. τ = L 1 w

92 Balancing see-saw W w w The smaller child has an identical twin of the same weight, what must L be now?

93 Balancing see-saw W w w The smaller child has an identical twin of the same weight, what must L be now? τ = 0 = LW (L + L )w 2 3 L = (L 2 + L 3 )w W