kinetics: f=ma (ch. 3 & 7) reviewrrg.utk.edu/resources/me231/lectures/me231_lecture_32.pdf ·...

Kinetics: F=ma (Ch. 3 & 7) ReviewLecture 32

ME 231: Dynamics

2

Question of the Day

What is the most important concept in mechanics?

ME 231: Dynamics

Free Body Diagram

What is the most important concept in dynamics?

Equations of Motion

aF m��xmmaF xx �����

ymmaF yy �����

zmmaF zz �����

3

Outline for Today

� Question of the day� Where are we in the course?� Inverse vs. forward dynamics� Kinetics: cause of motion� Possible solutions to kinetics problems� Direct application of Newton’s 2nd Law� Plane motion types for rigid bodies� Equations, equations, equations…� Exam 2a breakdown (kinetics: F=ma)

ME 231: Dynamics

� Question of the day

4

Kinematics Kinetics Dynamics

Where are we in the course?

ME 231: Dynamics

Chapters 1, 2, 6 Chapters 3, 5, 7, 8

Relationship among position, velocity, and acceleration

Relationship among forces(and moments)andacceleration

Concept: What is dynamics?

5

F = m a

Where are we in the course?

ME 231: Dynamics

Acceleration. Velocity rate of change with

respect to time

Calculation: How do we use dynamics?

Newton’s 2nd Law

Force. A push or pull exerted on a body, characterized by:�magnitude�direction�point of application

Mass. Measure of the resistance of a body to linear acceleration.

6

Inverse vs. Forward Dynamics

ME 231: Dynamics

aF m�inverse

dtd

dtd

Positions

Velocities

Forces

aF m�

ME 231: Dyna

forward

� � Positions

Velocities

Forces

7

Statics Kinematics Kinetics

Kinetics: Cause of Motion?

ME 231: Dynamics

Chapters 1, 2, 6 Chapters 3, 5, 7, 8

Relationship among position, velocity, and acceleration

Relationship among forces(and moments)andacceleration

Concept: What is kinetics?

ME 202

Relationship among forces (and moments)and equilibrium

8

Possible Solutions to Kinetics Problems

� Direct application of Newton’s 2nd Law – force-mass-acceleration method– Chapters 3 and 7

� Use of impulse and momentum methods– Chapters 5 and 8

� Use of work and energy principles – Chapter 4

ME 231: Dynamics

� Use of impulse and momentum methods– Chapters 5 and 8

� Use of work and energy principles– Chapter 4

9

Step-by-Step Solution Process

1. Kinematics– Identify type of motion– Solve for linear and angular accelerations

2. Diagram– Assign inertial coordinate system– Draw complete free-body diagram– Draw kinetic diagram to clarify equations

3. Equations of motion– Apply 2 linear and 1 angular equations– Maintain consistent sense– Solve for no more than 5 scalar unknowns (3 scalar

equations of motion and 2 scalar relations from the relative-acceleration equation)

ME 231: Dynamics

10

Outline for Today

� Question of the day� Where are we in the course?� Inverse vs. forward dynamics� Kinetics: cause of motion� Possible solutions to kinetics problems� Direct application of Newton’s 2nd Law� Plane motion types for rigid bodies� Equations, equations, equations…� Exam 2a breakdown (kinetics: F=ma)

ME 231: Dynamics

� Question of the day� Where are we in the course?� Inverse vs. forward dynamics� Kinetics: cause of motion� Possible solutions to kinetics problems

11

Direct Application of Newton’s 2nd Law

ME 231: Dynamics

rF ��m�� aF m��or

iim rfF �������iim rfffFFF ���� �������� 321321

xx amF ��

yy amF ��

zz amF ��

12

Rectangular (x-y) Coordinates: Exercise

ME 231: Dynamics

A particle with mass of 10 slugs moving in two-dimensions has a position vector (r) as a function of time (t) with coordinates given by

where r is measured in feet and t is in seconds.

Determine the magnitude of the net force (F) accelerating the particle at time t = 3 s.

x(t) = t2 – 4t + 20 , y(t) = 3 sin(2t)

13

Polar (r-��) Coordinates: Exercise

Tube A rotates about the vertical O-axis with constant angular velocity � and contains a small cylinder B of mass m whose radial position is controlled by a cord passing through the tube and wound around a drum of radius b.

ME 231: Dynamics

� � ���� eea r 2 2 ������� rrrr ���

Determine the tension T in the cord and ��-component of force F� if the drum has a constant angular rate of rotation of �� as shown.

14

Normal and Tangential (n-t) Coordinates: Exercise

A 1500-kg car enters an s-curve and slows down from 100 km/h at A to a speed of 50 km/h as it passes C.

Determine the total horizontal force exerted by the road on the tires at positions A, B, and C.

ME 231: Dynamics

tn eea 2

vv���

�

15

Outline for Today

� Question of the day� Where are we in the course?� Inverse vs. forward dynamics� Kinetics: cause of motion� Possible solutions to kinetics problems� Direct application of Newton’s 2nd Law� Plane motion types for rigid bodies� Equations, equations, equations…� Exam 2a breakdown (kinetics: F=ma)

ME 231: Dynamics

� Question of the day� Where are we in the course?� Inverse vs. forward dynamics� Kinetics: cause of motion� Possible solutions to kinetics problems� Direct application of Newton’s 2nd Law

16

B

A

B

A

B

A

B

A

B’

A’

B’

A’

B’

A’

B

APlane Motion Types for Rigid Bodies

� Translation

� Fixed-axis rotation

� General plane motion

ME 231: Dynamics

BBBBBBBBBB

A

B’BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

A’AAAA

B’

A’

17ME 231: Dynamics

Rigid-Body Translation

curvilinearrectilinear

aF m��

0

0

����

���

A

P

GG

MmadM�IM

00

��

�

BtB

AnA

GG

dmaMdmaM�IM

����

��� 0

18

Rigid-Body Translation: Exercise

A cleated conveyor belt transports solid cylinders up a 15º incline. The diameter of each cylinder is half its height.

Determine the maximum acceleration for the belt without tipping the cylinders as it starts.

ME 231: Dynamics

19

Fixed-Axis Rotation

ME 231: Dynamics

free-bodydiagram

kineticdiagram

� Mass center’s circular motion easily expressed in n-t coordinates

� Plane-motion equations: aF m�� �M GG I�� �M OO I��

20

Fixed-Axis Rotation: Exercise

Determine the angular acceleration and the force on the bearing at O for (a) the narrow ring of mass m and (b) the flat circular disk of mass m immediately after each is released from rest with OC horizontal.

ME 231: Dynamics

21

General Plane Motoin:Combined Translation and Rotation

ME 231: Dynamics

aF m�� �M GG I��

mad�IM GP ���

PPP mI a��M ����

22

General Plane Motion: Exercise

A truck has a mass of 2030 kg and carries a 1500-mm-diameter spool of cable with a massof 0.75 kg per meter of length. There are 150 turns on the full spool. The empty spool has a mass of 140 kg with radius of gyration of 530 mm.

Determine the tension Tin the cable when the truck starts from rest with an acceleration of 0.2g.

ME 231: Dynamics

23

Outline for Today

� Question of the day� Where are we in the course?� Inverse vs. forward dynamics� Kinetics: cause of motion� Possible solutions to kinetics problems� Direct application of Newton’s 2nd Law� Plane motion types for rigid bodies� Equations, equations, equations…� Exam 2a breakdown (kinetics: F=ma)

ME 231: Dynamics

� Question of the day� Where are we in the course?� Inverse vs. forward dynamics� Kinetics: cause of motion� Possible solutions to kinetics problems� Direct application of Newton’s 2nd Law� Plane motion types for rigid bodies

24

Lecture Equations

18. Newton 2nd Law19. Eqs. of Motion20. Rectilinear

21. Curvilinear

27. Lin. Imp. Mom.

28. Ang. Imp. Mom.

29. Sys. Imp. Mom.

Equations, Equations, Equations…Particle Kinetics: F=ma

ME 231: Dynamics

aF m��xmmaF xx �����

ymmaF yy �����zmmaF zz �����

rr maF ��

�� maF ��nn maF ��

tt maF ��

GF ���vG m� 21 2

1

GFG ��� � dtt

t0G ��

vrH mO ��

OO HM ���� � 21 2

1O

t

t OO dt HMH ��� �0H �� O

vG m�� iiiO m vrH ���

v�HH mGP ���� iiiG m ��H ����

GG HM ��� a�HM mGP ���� �

27. Lin. Imp. Mom.

28. Ang. Imp. Mom.

29. Sys. Imp. Mom.

GF ���vG m� 21

2

1

GFG ��� �1

dtt

��t��0G ��

vrH mO ��

OO HM ���� � 21

2

1O

t

t OO dt HMH ��� �1

tt

tt0H �� O

vG m�� iiiO m vriH ���

v�HH mGP ���� iiiG m ��H ����

GG HM ��� a�HM mGP ���� �

25

Lecture Equations

18. Newton 2nd Law22. Gen. Eqs. Mot. I23. Gen. Eqs. Mot. II

24. Fixed-Axis Rot.

25. Gen. Plane Mot. I

26. Gen. Plane Mot. II

31. Body Imp. Mom.

Equations, Equations, Equations…Rigid Body Kinetics: F=ma

ME 231: Dynamics

GG HM ���iiG F�H ����

mad�IM GP ���PPP mI a��M ����

2mrII GO ��aF m�� �M GG I�� �M OO I��

aF m�� �M GG I��mad�IM GP ���PPP mI a��M ����

vG m�

GF ���

21 2

1

GFG ��� � dtt

t

GG HM ���

� � 21 2

1G

t

t GG dt HMH ��� �

�H GG I� mvd�IH GP ��

PP HM ��� OO HM ���

�IH OO �

2OO mkI �

31. Body Imp. Mom.

vG m�

GF ���

212

1

GFG ��� �1

dtt

��t��

GG HM ���

� � 212

1G

t

t GG dt HMH ��� �1

tt

tt

�H GG I� mvd�IH GP ��

PP HM ��� OO HM ���

�IH OO �

26

Exam 2a Breakdown (particle kinetics: F=ma)

ME 231: Dynamics

0

5

10

15

20

25

30

35

40

18. Newton 2nd Law

19. Eqs. of Motion

20. Rectilinear 21. Curvilinear

27. Lin. Imp. Mom.

28. Ang. Imp. Mom.

29. Sys. Imp. Mom.

3440

0

3426

30

40

nu

mb

er

of

po

ints

lecture

27. Lin. Imp. Mom.

28. Ang. Imp.Mom.

29. Sys. Imp.

2630

Mom.

40

27

Exam 2 Breakdown (rigid body kinetics: F=ma)

ME 231: Dynamics

0

10

20

30

40

50

60

70

18. Newton 2nd Law

22. Gen. Eqs. of Motion I

23. Gen. Eqs. of Motion II

24. Fixed-Axis Rotation

25. Gen. Plane Mot. I

26. Gen. Plane Mot. II

31. Body Imp. Mom.

34

70

40 4026

0

40

nu

mb

er

of

po

ints

lecture

31. Body Imp. Mom.

40

28

For Next Time…

� Review Chapters 3 & 7� Review Lectures slides

– http://rrg.utk.edu/resources/ME231/lectures.html

� Review Examples from class– http://rrg.utk.edu/resources/ME231/examples.html

� Exam #2a on Friday (11/9)

ME 231: Dynamics