Kinetics: F=ma (Ch. 3 & 7) Reviewrrg.utk.edu/resources/ME231/lectures/ME231_lecture_32.pdf · Possible solutions to kinetics problems Direct application of Newton’s 2 nd Law Plane
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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
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