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Kinetics of Particles: Energy and Momentum Methods
Typology: Lecture notes
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Tenth
Tenth
Edition
Edition
Ferdinand P. Beer
Ferdinand P. Beer
E. Russell Johnston, Jr.
E. Russell Johnston, Jr.
Phillip J. Cornwell
Phillip J. Cornwell
Lecture Notes:
Lecture Notes:
Brian P. Self
Brian P. Self
California Polytechnic State University
California Polytechnic State University
CHAPTER
Kinetics of Particles:
Energy and Momentum
Methods
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 2
Introduction
Work of a Force
Principle of Work & Energy
Applications of the Principle of
Work & Energy
Power and Efficiency
Sample Problem 13.
Sample Problem 13.
Sample Problem 13.
Sample Problem 13.
Sample Problem 13.
Potential Energy
Conservative Forces
Conservation of Energy
Motion Under a Conservative
Central Force
Sample Problem 13.
Sample Problem 13.
Sample Problem 13.
Principle of Impulse and Momentum
Impulsive Motion
Sample Problem 13.
Sample Problem 13.
Sample Problem 13.
Impact
Direct Central Impact
Oblique Central Impact
Problems Involving Energy and
Momentum
Sample Problem 13.
Sample Problem 13.
Sample Problems 13.
Sample Problem 13.
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 4
F ma.
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
2 - 5
Forces and
Accelerations
Velocities and
Displacements
Velocities and
Time
Newton’s Second
Law (last chapter)
Work-Energy
Impulse-
Momentum
1 1 2 2
T U T
G
F ma
2
1
1 2
t
t
mv F dt mv
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 7
2
1
2
1
2
1
2
1
cos
1 2
A
A
x y z
s
s
t
s
s
A
A
F dx F dy F dz
F ds F ds
U F d r
Work is represented by the area under the
curve of F
t
plotted against s.
t
is the force in the direction of the
displacement ds
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 8
1 2
U F cos x
1 2
U F sin x
1 2
1 2
a)
b)
c)
d)
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 10
Magnitude of the force exerted by a spring is
proportional to deflection,
springconstant N/morlb/in.
k
F kx
2
2
2
1
2
1
2
1
1 2
2
1
U kxdx kx kx
dU Fdx kx dx
x
x
when x
2
< x
1
, i.e., when the spring is returning to
its undeformed position.
Work of the force exerted by the spring is equal to
negative of area under curve of F plotted against
x ,
U F F x
1 2
2
1
1 2
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 11
0
1
2
o
Displacement is
in the opposite
direction of the
force
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
2 - 13
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 14
Forces which do not do work (ds = 0 or cos :
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 16
from rest at position A
1
.
Determine the velocity
of the pendulum bob at
A
2
using work & kinetic
energy.
work.
P
v gl
v
g
W
Wl
T U T
2
2
1
0
2
2
2
1 1 2 2
expression for acceleration and integrating.
directly.
Forces which do no work are eliminated
from the problem.
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 17
applied to directly determine the acceleration
of the pendulum bob.
supplementing the method of work and
energy with an application of Newton’s
second law.
2
,
W
l
gl
g
W
P W
l
v
g
W
P W
F m a
n n
3
2
2
2
v 2 gl
2
If you designed the rope to hold twice the weight of the bob, what would happen?
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 19
An automobile weighing 4000 lb is
driven down a 5
o
incline at a speed of
60 mi/h when the brakes are applied
causing a constant total breaking force
of 1500 lb.
Determine the distance traveled by the
automobile as it comes to a stop.
SOLUTION:
work to equal the kinetic energy
change.
Vector Mechanics for Engineers: Dynamics
Vector Mechanics for Engineers: Dynamics
13 - 20
SOLUTION:
4000 32. 2 88 481000 ft lb
88 ft s
3600 s
h
mi
5280 ft
h
mi
60
2
2
1
2
1
2
1
1
1
T mv
v
481000 ft lb 1151 lb 0
1 1 2 2
x
T U T
x 418 ft
to equal the kinetic energy change.
x
U x x
1151 lb
1500 lb 4000 lb sin 5
1 2
0 0
2 2
v T