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Which object will win? All moments of inertia in the previous table can be expressed as;

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Presentation on theme: "Which object will win? All moments of inertia in the previous table can be expressed as; "β€” Presentation transcript:

1 Which object will win? All moments of inertia in the previous table can be expressed as; 𝐼 π‘π‘š =𝛽𝑀 𝑅 2 =𝑐𝑀 𝑅 2 (c - number) Compare; a) For a thin walled hollow cylinder 𝛽=1 For a solid cylinder 𝛽= 1 2 etc. Conservation of energy; 0+π‘€π‘”β„Ž= 1 2 𝑀 𝑣 𝐼 𝑀 2 = 1 2 𝑀 𝑣 𝛽𝑀 𝑅 2 ( 𝑣 𝑅 ) 2 = 1 2 (1+𝛽)𝑀 𝑣 2 𝒗= πŸπ’ˆπ’‰ 𝟏+𝛽 Small 𝛽 bodies wins over large 𝛽 bodies.

2 Finding the moment of inertia for common shapes

3 Chapter 10: Dynamics of Rotational Motion
Β© 2016 Pearson Education, Inc.

4 Dynamics of Rotational Motion
(Conservation of angular momentum)

5 Goals for Chapter 10 Torque: β€œangular force”
To see how torques cause rotational dynamics (just as linear forces cause linear accelerations) To examine the combination of translation and rotation To calculate the work done by a torque To study angular momentum and its conservation To relate rotational dynamics and angular momentum To study how torques add a new variable to equilibrium To see the vector nature of angular quantities.

6 Rotational Dynamics: I

7 Definition of Torque – Figure 10.1
Torque ( ) is defined as the force applied multiplied by the moment arm. The moment arm is the perpendicular distance from the point of force application to the pivot point. Β© 2016 Pearson Education, Inc.

8 Torque Note: F(rad)has no torque with respect to O

9 There Is a Sign Convention – Figure 10.3
A counterclockwise force is designated as positive (+). A clockwise force is designated as negative (βˆ’). Β© 2016 Pearson Education, Inc.

10

11 Clicker question

12 A Plumbing Problem to Solve – Example 10.1
Β© 2016 Pearson Education, Inc.

13 Q10.4 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 B. (0.80 m)(900 N)cos C. (0.80 m)(900 N)tan D. (0.80 m)(900 N)/(sin ) E. none of the above Answer: B

14 Note: t = F R sinq

15 Note: sign of t

16 Clicker question Torques on a Rod
𝑏) 𝜏 1 > 𝜏 3 > 𝜏 2 𝑐) 𝜏 2 > 𝜏 3 > 𝜏 1 𝑐) 𝜏 1 > 𝜏 3 > 𝜏 2 𝑐) 𝜏 1 > 𝜏 3 > 𝜏 2 𝜏 2 > 𝜏 3 > 𝜏 1 Β© 2016 Pearson Education, Inc.

17

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19 Why Do Acrobats Carry Long Bars?
Refer to the photo caption on the bottom of page 287. Β© 2016 Pearson Education, Inc.

20 Rotating Cylinders Clicker Question
Relationship of torque and angular acceleration Σ𝜏=𝐼𝛼 rotational analog of Newton’s law Σ𝐹=π‘šπ‘Ž Which choice correctly ranks the magnitudes of the angular accelerations of the three cylinders? A. 𝛼 3 > 𝛼 2 > 𝛼 B. 𝛼 1 > 𝛼 2 > 𝛼 3 C. 𝛼 2 > 𝛼 1 > 𝛼 D. 𝛼 1 = 𝛼 2 = 𝛼 3 Β© 2016 Pearson Education, Inc.

21 Unwinding a Winch (Again) – Example 10.2
=9 N (0.6m) Β© 2016 Pearson Education, Inc.

22 A Bowling Ball Rotates on a Moving Axis – Example 10.5
Β© 2016 Pearson Education, Inc.

23 B. the normal force exerted by the ramp
Q10.8 A solid bowling ball rolls down a ramp. Which of the following forces exerts a torque on the bowling ball about its center? A. the weight of the ball B. the normal force exerted by the ramp C. the friction force exerted by the ramp D. more than one of the above E. depends on whether the ball rolls without slipping Answer: C

24 Work and power in rotational motion

25 Angular momentum

26 Nifty Sculpture Β© 2016 Pearson Education, Inc.

27 Calculate the angular momentum and kinetic energy of a solid uniform sphere with a radius of 0.12 m and a mass of 14.0 kg if it is rotating at 6.00 rad/s about an axis through its center 𝐼= 2 5 𝑀 𝑅 2 = 2 5 βˆ— 14π‘˜π‘” π‘š 2 = π‘˜π‘” π‘š 2 𝐿=𝐼𝑀= π‘˜π‘” π‘š 2 βˆ—6.0 rad=0.484 kg π‘š 2 𝑠 𝐾= 1 2 𝐼 𝑀 2 = 1 2 𝐿𝑀= kg π‘š 2 𝑠 βˆ—6 rad=1.45 J

28 Angular Momentum Is Conserved – Figure 10.19
The first figure shows the figure skater with a large moment of inertia. In the second figure, she has made the moment much smaller by bringing her arms in. Since L is constant, must increase. Β© 2016 Pearson Education, Inc.

29 Conservation of angular momentum
L=r x p |L|=(rp)sin90=(r)(mv) v=r 5 kg Conservation of L

30 Angular Momentum – Figure 10.15
The diver changes rotational velocities by changing body shape. The total angular momentum is the moment of inertia multiplied by the angular velocity. Β© 2016 Pearson Education, Inc.

31 Q10.12 Clicker question A spinning figure skater pulls his arms in as he rotates on the ice. As he pulls his arms in, what happens to his angular momentum L and kinetic energy K? A. L and K both increase. B. L stays the same; K increases. C. L increases; K stays the same. D. L and K both stay the same. E. None of the above. Answer: B

32 Clicker question The block in the figure slides in a circle on a nearly frictionless table with a speed v. You hold it by means of a string passing through a hold in the table. If you raise your hand so that the radius of the block's circle doubles, the block's final speed will be 2v. 1/2v. 1/4v. d) v. Answer: b) 1/2Ξ½. Β© 2016 Pearson Education, Inc.

33 Clicker question a) Β½ T b) 1/8 T c) 1/4 T d) 1/16 T
The block in the figure slides in a circle on a nearly frictionless table with a speed v. You hold it by means of a string passing through a hole in the table. If you raise your hand so that the radius of the block's circle doubles, the strings tension T will be a) Β½ T b) 1/8 T c) 1/4 T d) 1/16 T Β© 2016 Pearson Education, Inc.

34 Equilibrium condition for equilibrium
Center of gravity is the turning point

35

36 A New Equilibrium Condition – Figure 10.24
Now, in addition to , we also must add Β© 2016 Pearson Education, Inc.

37 Balancing on a Teeter-Totter – Figure 10.26
The heavier child must sit closer to balance the torque from the smaller child. Your torque is 𝜏 π‘¦π‘œπ‘’ =+ 90 π‘˜π‘” 𝑔π‘₯ Your friend is 3.0 mβˆ’π‘₯ from the pivot, and her torque is: 𝜏 π‘“π‘Ÿπ‘–π‘’π‘›π‘‘ =βˆ’ 60 π‘˜π‘” ×𝑔 3.0 mβˆ’π‘₯ . For equilibrium, Σ𝜏=0. 𝜏 π‘¦π‘œπ‘’ + 𝜏 π‘“π‘Ÿπ‘–π‘’π‘›π‘‘ =0 + 90 kg 𝑔π‘₯βˆ’ 60 kg 𝑔 3.0 mβˆ’π‘₯ =0 𝒙=𝟏.𝟐 𝐦 Β© 2016 Pearson Education, Inc.

38 Walking the plank

39 Clicker question

40 A fishy wind chime

41 Q11.2 A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. How far from the left end of the stick should the triangular object be placed so that the combination of meter stick and rock is in balance? less than 0.25 m 0.25 m C. between 0.25 m and 0.50 m D m E. more than 0.50 m Answer: B

42 The Firefighter on the Ladder – Figure 10.27
This problem is "classic" and appears in one form or another on most standardized exams. Refer to worked example on page 304. Β© 2016 Pearson Education, Inc.

43 Balanced Forces During Exercise
Β© 2016 Pearson Education, Inc.

44 Hanging a farm gate Adjust T so horizontal force on A =0 𝐴=πœ‹ π‘Ÿ 2 𝜏 -57
𝐴=πœ‹ π‘Ÿ 2 𝜏  -57 𝜏 𝐡 𝐻 𝐴𝑣 𝐻 𝐴𝑣

45 Carrying a box up the stairs
F

46 How a car’s clutch work The clutch disk and the gear disk is pushed into each other by two forces that do not impart any torque, what is the final angular velocity when they come together?

47 Clicker question

48 Problem 10.66 What is the original kinetic energy Ki of disk A? 𝐼 𝐴 = 1 3 𝐼 𝐡 β†’ 𝐼 𝐡 =3 𝐼 𝐴 Initially; 𝐿 𝑖 = 𝑀 π‘œ 𝐼 𝐴 and 𝐾 𝑖 = 1 2 𝐼 𝐴 𝑀 π‘œ 2 Final; 𝐼=𝐼 𝐴 +3 𝐼 𝐴 =4 𝐼 𝐴 (after coupling together) Conservation of angular momentum; 𝐿 𝑖 = 𝑀 π‘œ 𝐼 𝐴 = 𝐿 𝑓 =π‘ŠπΌ=π‘Š4 𝐼 𝐴 β†’π‘Š= 𝑀 π‘œ 4 Final Kinetic energy; 𝐾 𝑓 = 1 2 𝐼 π‘Š 2 = 4 2 𝐼 𝐴 𝑀 π‘œ = 1 4 𝐾 𝑖 βˆ΄βˆ†πΎ= 𝐾 𝑓 βˆ’ 𝐾 𝑖 =βˆ’ 3 4 𝐾 𝑖 =βˆ’2400 𝐽 (thermal energy developed during coupling) So, original energy; 𝐾 𝑖 =βˆ’ 3 4 βˆ’2400 =3200 𝐽

49 Honda 600RR Who races this bike? Why can anybody race it, if he just
dares to go fast? The oval track of the Texas World Speedway allows speeds of 250 mph.

50 Angular Quantities Are Vectors – Figure 10.29
The "right-hand rule" gives us a vector's direction. Β© 2016 Pearson Education, Inc.

51 Vector nature of angular quantities

52 A non rotating and rotating gyroscope

53 Precesion Angular Velocity Ξ©
Ξ” 𝐿 = 𝜏 Δ𝑑 𝐿 =𝐼 πœ” 𝜏 = Ξ” 𝐿 Δ𝑑 Analogous to 𝐹 = Ξ” 𝑃 Δ𝑑 Ξ”πœ™= Ξ” 𝐿 𝐿 Ξ©= Ξ”πœ™ Δ𝑑 = Ξ” 𝐿 𝐿 Δ𝑑 = 𝜏 𝐿 = π‘Šπ‘Ÿ πΌπœ” rad/s π‘Š is the weight of the flywheel, and π‘Ÿ is the distance from the pivot point of the center of mass of the flywheel.

54

55 The Yo-Yo Rotates on a Moving Axis
Β© 2016 Pearson Education, Inc.


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