Chapter 13: Kinetics Renee Y. Becker Valencia Community College.
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CHAPTER 13 Kinetics of Particles:
Energy and Momentum
Methods
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE 1
Chapter OutLine
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1. Introduction
2. Work of a Force
3. Principle of Work & Energy
4. Applications of the Principle
of Work & Energy
5. Power and Efficiency
6. Potential Energy
7. Conservative Forces
8. Conservation of Energy
10. Motion Under a Conservative
Central Force
11. Principle of Impulse and
Momentum
12. Impulsive Motion
13. Impact
14. Direct Central Impact
15. Oblique Central Impact
16. Problems Involving Energy and
Momentum
Introduction
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• Chapter 12 deals with the motion of particles
through the fundamental equation of motion,
amF
Introduction
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4
Chapter 13: two additional methods of
analysis:
Method of work & energy: directly relates force,
mass, velocity and displacement.
Method of impulse &momentum: directly relates
force, mass, velocity, and time.
Work of a Force
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5
= particle displacement rd
Work of the force is
dzFdyFdxF
dsFrdFdU
zyx
cos
Work is a scalar quantity: has
magnitude & sign but not direction.
Units of work = Length × Force
1 J = (1N)(1m) 1 ft.lb = 1.356 J
Work of a Force
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6
Work of a force during a
finite displacement,
2
1
2
1
2
1
2
1
cos21
A
A
zyx
s
s
t
s
s
A
A
dzFdyFdxFdsF
dsFrdFU
Work = area under Ft vs. s
curve
Work of a Force
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• Work of a constant force in rectilinear motion,
xFU cos21
Work of a Force - gravity
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Work of the force of gravity,
yWyyWdyWU
dyWdzFdyFdxFdU
y
y
zyx
1221
2
1
Work of weight is positive when y < 0, i.e.,
when weight moves down
Work of a Force - spring
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9
Magnitude of force exerted by a spring is
proportional to deflection,
lb/in.or N/mconstant spring
k
kxF
Work of force exerted by spring,
222
1212
121
2
1
kxkxdxkxU
dxkxdxFdU
x
x
Work of a Force - spring
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10
Work of force exerted by spring is positive when
x2 < x1, i.e., when spring is returning to its
undeformed position.
Work of force exerted by spring = negative of area
under F vs. x curve,
xFFU 2121
21
Work of a Force - gravitational
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11
Work of a gravitational force (particle M occupies
fixed position O while particle m follows path
shown),
)11
(12
221
2
2
1
rrGMm
drr
MmGU
drr
MmGFdrdU
r
r
Forces which do not do work (ds = 0 or cos 0:
Weight of a body when its center of gravity
moves horizontally
Reaction at a roller moving along its track
Reaction at frictionless surface when body in
contact moves along surface
Reaction at frictionless pin supporting rotating
body
Work of a Force
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Particle Kinetic Energy: Principle
of Work & Energy
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13
dvmvdsF
ds
dvmv
dt
ds
ds
dvm
dt
dvmmaF
t
tt
• Consider a particle of mass m acted upon by force F
• Integrating from A1 to A2
2
21
1221
2
1212
221
2
1
2
1
mvTTTU
mvmvdvvmdsF
v
v
s
s
t
Particle Kinetic Energy: Principle
of Work & Energy
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14
Work of force = change in kinetic energy
of the particle F
Units of work and kinetic energy are the same:
JmNms
mkg
s
mkg
2
22
21
mvT
Applications of the Principle of
Work and Energy
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15
Determine velocity of bob at A2
Force acts normal to path and does no work P
glv
vg
WWl
TUT
2
2
10
2
22
2211
Velocity found without determining expression for
acceleration and integrating
All quantities are scalars and can be added directly Forces which do no work are eliminated from problem
Applications of the Principle of
Work and Energy
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Principle of work & energy cannot be applied to
directly determine acceleration of bob
Calculating tension in the cord requires application
of Newton’s 2nd law
Applications of the Principle of
Work and Energy
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As the bob passes through A2
Wl
gl
g
WWP
l
v
g
WWP
amF nn
32
22
glv 22
Power and Efficiency
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Power = rate at which work is done
vFdt
rdF
dt
dUPower
Units of power:
W746s
lbft550 hp 1or
s
mN 1
s
J1 (watt) W 1
inputpower
outputpower
input work
koutput worefficiency
Sample Problem 13.1
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An automobile weighing 4000 lb is driven down a 5o
incline at a speed of 60 mph 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:
Evaluate the change in kinetic energy
lbft481000882.324000
sft88s 3600
h
mi
ft 5280
h
mi60
2
212
121
1
1
mvT
v
00 22 Tv
Sample Problem 13.1
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0lb1151lbft481000
2211
x
TUT
ft 418x
Determine the distance required for work to equal
the kinetic energy change.
xxxU lb11515sinlb4000lb150021
Sample Problem 13.1
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Sample Problem 13.2
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Two blocks are joined by an
inextensible cable as shown.
If the system is released
from rest, determine the
velocity of block A after it
has moved 2 m. Assume
that the coefficient of
friction between block A
and the plane is mk = 0.25
and that the pulley is
weightless and frictionless.
Sample Problem 13.2
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Solution:
• Apply the principle of work &
energy separately to blocks A & B
2
21
2
21
2211
2
kg200m2N490m2
m2m20
:
N490N196225.0
N1962sm81.9kg200
vF
vmFF
TUT
WNF
W
C
AAC
AkAkA
A
mm
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221
221
2211
2
kg300m2N2940m2
m2m20
:
N2940sm81.9kg300
vF
vmWF
TUT
W
c
BBc
B
Sample Problem 13.2
Sample Problem 13.2
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When the two relations are combined, the work of
the cable forces cancel. Solve for the velocity.
221 kg200m2N490m2 vFC
221 kg300m2N2940m2 vFc
221
221
kg500J 4900
kg300kg200m2N490m2N2940
v
v
sm 43.4v
Sample Problem 13.3
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A spring is used to stop a 60 kg package which is sliding on a
horizontal surface. The spring has a constant k = 20 kN/m and
is held by cables so that it is initially compressed 120 mm. The
package has a velocity of 2.5 m/s in the position shown and the
maximum deflection of the spring is 40 mm. Determine
a. Coefficient of kinetic friction between the package &
surface
b. Velocity of the package as it passes again through the
position shown.
Sample Problem
13.3
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Solution:
Apply principle of work and energy
between initial position and the point
at which spring is fully compressed.
0
J5.187sm5.2kg60
2
2
212
121
1
T
mvT
k
k
kfxWU
m
m
m
J377
m640.0sm81.9kg60 2
21
Sample Problem
13.3
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J0.112
m040.0N3200N2400
N3200
m160.0mkN20
N2400
m120.0mkN20
21
maxmin21
21
0max
0min
xPPU
xxkP
kxP
e
J112J377
212121
k
efUUU
m
0J112J 377-J5.187
:2211
k
TUT
m
20.0km
Sample Problem 13.3
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Apply the principle of work and energy for the
rebound of the package.
232
1232
132 kg600 vmvTT
J36.5J112J377
323232
k
efUUU
m
2
321
3322
kg60J5.360
:
v
TUT
sm103.13 v
Sample Problem 13.4
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A 2000 lb car starts from rest at point 1 and moves
without friction down the track shown. Determine:
a. the force exerted by the track on the car at point 2
b. min safe value of radius of curvature at point 3
Sample Problem 13.4
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Solution:
• Apply principle of work & energy to determine
velocity at point 2.
sft8.50
ft2.32ft402ft402
2
1ft400:
ft40
2
10
2
22
2
2
22211
21
2
2
2
221
21
v
sgv
vg
WWTUT
WU
vg
WmvTT
Apply Newton’s 2nd law to find normal force by the
track at point 2
:nn amF
lbWN
g
g
W
v
g
WamNW n
000,105
ft20
ft402
2
2
2
Sample Problem 13.4
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Sample Problem 13.4
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Apply principle of work & energy to determine
velocity at point 3
sft1.40sft2.32ft252ft252
2
1ft250
323
233311
vgv
vg
WWTUT
Sample Problem 13.4
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Apply Newton’s 2nd law to find min radius of
curvature at point 3 such that a positive normal force
is exerted by the track.
:nn amF
33
23 ft252
g
g
Wv
g
W
amW n
ft503
Sample Problem 13.5
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35
The dumbwaiter D and its load have a
combined weight of 600 lb, while the
counterweight C weighs 800 lb.
Determine the power delivered by the
electric motor M when the dumbwaiter
a. is moving up at a constant speed of 8
ft/s
b. has an instantaneous velocity of 8 ft/s
and an acceleration of 2.5 ft/s2, both
directed upwards.
Sample Problem 13.5
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In first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium.
Free-body C:
:0 yF
lb 4000lb8002 TT
Sample Problem 13.5
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slbft1600
sft8lb 200
DFvPower
hp 91.2slbft550
hp 1slbft1600
Power
Free-body D:
:0 yF
lb 200lb 400lb 600lb 600
0lb 600
TF
TF
Sample Problem 13.5
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2nd case: both bodies are accelerating.
Apply Newton’s 2nd law to each body to
determine the required motor cable force.
2212 sft25.1sft5.2 DCD aaa
Free-body C: :CCy amF
lb5.38425.12.32
8002800 TT
Sample Problem 13.5
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Free-body D: :DDy amF
lb 1.2626.466005.384
5.22.32
600600
FF
TF
slbft2097
sft8lb 1.262
DFvPower
hp 81.3
slbft550
hp 1slbft2097
Power
Home Work Assignment 13-1
1, 8, 15, 22, 28, 43, 50
Due Saturday 8/3/2014
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40
Potential Energy Friday, February 28, 2014
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41
2121 yWyWU
Work of force of gravity , W
Work is independent of path
followed; depends only on
initial and final values of Wy.
WyVg
2121 gg VVU
Potential Energy Friday, February 28, 2014
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42
Choice of datum from which elevation y is measured
is arbitrary.
Potential Energy Friday, February 28, 2014
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43
For a space vehicle, variation of the force of
gravity with distance from the center of earth
should be considered.
Work of a gravitational force,
1221
r
GMm
r
GMmU
Potential energy Vg when the
variation in the force of gravity
can not be neglected,
r
WR
r
GMmVg
2
Potential Energy Friday, February 28, 2014
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44
Work of force exerted by a spring
depends only on initial & final
deflections of the spring,
222
1212
121 kxkxU
2121
221
ee
e
VVU
kxV
Conservative
Forces
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45
If the work of force is
independent of the path
followed by its point of
application.
22211121 ,,,, zyxVzyxVU
Such forces are described as
conservative forces.
For any conservative force
applied on a closed path,
0 rdF
Conservative
Forces
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46
Elementary work corresponding
to displacement between two
neighboring points,
zyxdV
dzzdyydxxVzyxVdU
,,
,,,,
Vz
V
y
V
x
VF
dzz
Vdy
y
Vdx
x
VdzFdyFdxF zyx
grad
Conservation of
Energy
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47
Work of a conservative force,
2121 VVU
Concept of work & energy,
1221 TTU
constant
2211
VTE
VTVT
WVT
WVT
11
11 0
WVT
VWgg
WmvT
22
2222
12 02
2
1
Conservation of
Energy
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48
When a particle moves under the action of
conservative forces, total mechanical energy is
constant.
Friction forces are not conservative. Total
mechanical energy of a system involving friction
decreases.
Mechanical energy is dissipated by friction into
thermal energy. Total energy is constant.
Motion Under a Conservative Central Force
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49
Both the principle of conservation of angular
momentum & the principle of conservation of
energy may be applied.
sinsin 000 rmvmvr
r
GMmmv
r
GMmmv
VTVT
221
0
202
1
00
Sample Problem 13.6
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50
A 20 lb collar slides without
friction along a vertical rod as
shown. The spring attached to
the collar has an undeflected
length of 4 in. and a constant
of 3 lb/in.
If the collar is released from
rest at position 1, determine its
velocity after it has moved 6
in. to position 2.
Sample Problem 13.6
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Solution:
Apply principle of conservation of energy between
positions 1 & 2.
Position 1:
0 lb,ft20lbin.24
lbin.24in. 4in. 8in.lb3
11
2
212
121
TVVV
kxV
ge
e
Sample Problem 13.6
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52
Position 2:
22
22
222
12
2
2
212
221
311.02.32
20
2
1
lbft 5.5lbin. 6612054
lbin. 120in. 6lb 20
lbin.54in. 4in. 01in.lb3
vvmvT
VVV
WyV
kxV
ge
g
e
Conservation of Energy:
lbft 5.50.311lbft 20 22
2211
v
VTVT
sft91.42v
Sample Problem 13.7
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53
The 0.5 lb pellet is pushed against the spring and
released from rest at A. Neglecting friction,
determine the smallest deflection of the spring for
which the pellet will travel around the loop and
remain in contact with the loop at all times.
Sample Problem 13.7
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54
Solution:
Setting the force exerted by the loop to zero, solve
for min velocity at D.
:nn maF
222
2
sft4.64sft32.2ft 2
rgv
rvmmgmaW
D
Dn
Apply principle of conservation of energy
between points A & D.
0 ,18ftlb36
0
1
22
21
2
21
1
Txx
kxVVV ge
Sample Problem 13.7
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55
lbft5.0
sft4.64sft2.32
lb5.0
2
1
lbft2
ft4lb5.00
22
2
2
21
2
2
D
ge
mvT
WyVVV
25.0180 2
2211
x
VTVT
in. 47.4ft 3727.0 x
Home Work Assignment 13-2
55, 61, 67, 76, 95
Due Wednesday
12/3/2014
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56
221121 , of impulse 2
1
vmvmFdtF
t
t
ImpImp
12
2
1
, vmvmdtFvmddtF
t
t
Principle of Impulse & Momentum
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57
From Newton’s 2nd law,
vmvmdt
dF
linear momentum
Principle of Impulse & Momentum
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58
Final momentum of particle = vector sum of its
initial momentum and the impulse of force during
the time interval.
Units for the impulse of a force are
smkgssmkgsN 2
Impulsive Motion
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59
Impulsive force: Force acting on a particle during
a very short time interval that is large enough to
cause a significant change in momentum
When impulsive forces act on a particle,
21 vmtFvm
Impulsive Motion
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60
21 vmtFvm
When a baseball is struck by a bat, contact occurs
over a short time interval but force is large
enough to change sense of ball motion.
Non-impulsive forces: forces for which the
impulse is small and therefore, may be neglected.
Sample Problem 13.10
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61
An automobile weighing 4000 lb is driven down a 5o
incline at a speed of 60 mph when the brakes are
applied, causing a constant total braking force of
1500 lb.
Determine the time required for the automobile to
come to a stop.
Sample Problem 13.10
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62
Solution:
Apply the principle of impulse & momentum
2211 vmvm
Imp
015005sin4000sft882.32
4000
05sin1
tt
FttWmv
Taking components parallel to the incline,
s49.9t
Sample Problem 13.11
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63
A 4 oz baseball is pitched
with a velocity of 80 ft/s.
After the ball is hit by the
bat, it has a velocity of
120 ft/s in the direction
shown. If the bat and
ball are in contact for
0.015 s, determine the
average impulsive force
exerted on the ball during
the impact.
Sample
Problem 13.11
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64
Solution:
Apply the principle of impulse
& momentum in terms of
horizontal & vertical component
equations.
2211 vmvm
Imp
x
y
x component:
lb89
40cos1202.32
16415.080
2.32
164
40cos21
x
x
x
F
F
mvtFmv
Sample
Problem 13.11
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65
x
y
y component:
lb9.39
40cos1202.32
16415.0
40sin0 2
y
y
y
F
F
mvtF
lb5.97,lb9.39lb89 FjiF
Sample Problem 13.12
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Dr. Mohammad Suliman Abuhaiba, PE
66
A 10 kg package drops from a
chute into a 24 kg cart with a
velocity of 3 m/s. Knowing
that the cart is initially at rest
and can roll freely, determine
a. the final velocity of the cart
b. the impulse exerted by the
cart on the package
c. the fraction of the initial
energy lost in the impact
Sample Problem 13.12
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Dr. Mohammad Suliman Abuhaiba, PE
67
Solution:
Apply the principle of impulse & momentum to
package-cart system to determine the final velocity.
2211 vmmvm cpp
Imp
x
y
x components: 2
21
kg 25kg 1030cosm/s 3kg 10
030cos
v
vmmvm cpp
m/s 742.02 v
Sample Problem 13.12
Friday, February 28, 2014
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68
Apply same principle to package alone to determine
impulse exerted on it from change in its momentum.
x
y
2211 vmvm pp
Imp
x components:
2
21
kg 1030cosm/s 3kg 10
30cos
vtF
vmtFvm
x
pxp
sN56.18 tFx
Sample Problem 13.12
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69
x
y
y components:
030sinm/s 3kg 10
030sin1
tF
tFvm
y
yp
sN15 tFy
sN 9.23
sN 51sN 56.1821
tF
jitF
Imp
Sample Problem 13.12
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70
To determine the fraction of energy lost,
J 63.9sm742.0kg 25kg 10
J 45sm3kg 10
2
212
221
2
2
212
121
1
vmmT
vmT
cp
p
786.0J 45
J 9.63J 45
1
21
T
TT
Home Work Assignment 13-3
119, 129, 136, 149, 154
Due Monday 18/3/2014
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71
Impact Friday, February 28, 2014
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72
• Impact: Collision between two
bodies which occurs during a
small time interval and during
which the bodies exert large
forces on each other.
• Line of Impact: Common
normal to the surfaces in
contact during impact.
Direct Central Impact
Oblique Central Impact
Impact Friday, February 28, 2014
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73
• Central Impact: Impact for
which the mass centers of the
two bodies lie on the line of
impact; otherwise, it is an
eccentric impact..
• Direct Impact: Impact for
which the velocities of the two
bodies are directed along the
line of impact.
Direct Central Impact
Oblique Central Impact
Impact Friday, February 28, 2014
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74
• Oblique Impact: Impact for
which one or both of the bodies
move along a line other than the
line of impact.
Direct Central Impact
Oblique Central Impact
Direct Central
Impact
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75
Bodies moving in same
straight line, vA > vB .
Upon impact the bodies
undergo a period of
deformation, at end of which,
they are in contact and moving
at a common velocity.
A period of restitution follows
during which the bodies either
regain their original shape or
remain permanently deformed.
Direct Central
Impact
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
76
Total momentum of the two
body system is preserved,
BBAABBAA vmvmvmvm
A second relation between
final velocities is required.
Direct Central Impact
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
77
• Period of deformation: umPdtvm AAA
• Period of restitution: AAA vmRdtum
10
e
uv
vu
Pdt
Rdtnrestitutio of tcoefficien e
A
A
Direct Central Impact
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
78
A similar analysis of particle B yields B
B
vu
uv e
Combining the relations leads to the desired
second relation between the final velocities.
BAAB vvevv
Perfectly plastic impact, e = 0: vvv AB
vmmvmvm BABBAA
Perfectly elastic impact, e = 1:
Total energy and total momentum conserved.
BAAB vvvv
Oblique Central Impact
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
79
No tangential impulse component; tangential
component of momentum for each particle is
conserved. tBtBtAtA vvvv
Normal component of total momentum of the two
particles is conserved.
nBBnAAnBBnAA vmvmvmvm
Oblique Central Impact
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
80
Final velocities are unknown in magnitude &
direction.
Four equations are required.
Normal components of relative velocities before
and after impact are related by the coefficient of
restitution.
nBnAnAnB vvevv
Oblique Central Impact
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
81
Block constrained to move along horizontal surface.
Impulses from internal forces along the n
axis and from external force exerted by
horizontal surface and directed along the vertical to
the surface.
FF
and
extF
Oblique Central Impact
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
82
Tangential momentum of ball is conserved. tBtB vv
Total horizontal momentum of block & ball is
conserved.
xBBAAxBBAA vmvmvmvm
Normal component of relative velocities of block
and ball are related by coefficient of restitution.
nBnAnAnB vvevv
Problems Involving Energy and
Momentum
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
83
Three methods for analysis of kinetics problems:
1. Direct application of Newton’s 2nd law
2. Method of work & energy
3. Method of impulse & momentum
Sample Problem 13.14
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
84
A ball is thrown against a
frictionless, vertical wall.
Immediately before the ball
strikes the wall, its velocity
has a magnitude v and forms
angle of 30o with the
horizontal. Knowing that
e = 0.90, determine the
magnitude and direction of
the velocity of the ball as it
rebounds from the wall.
Sample Problem 13.14
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
85
Component of ball momentum
tangential to wall is conserved.
vvv tt 500.0
Solution:
Resolve ball velocity into components
parallel & perpendicular to wall
vvv
vvv
t
n
500.030sin
866.030cos
n
t
Sample Problem 13.14
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
86
Apply coefficient of restitution
relation with zero wall velocity.
vvv
vev
n
nn
779.0866.09.0
00
n
t
7.32500.0
779.0tan926.0
500.0779.0
1vv
vvv tn
Sample Problem 13.15
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
87
The magnitude & direction of the velocities of two
identical frictionless balls before they strike each other
are as shown. Assuming e = 0.9, determine the
magnitude & direction of the velocity of each ball after
the impact.
Solution:
Resolve ball velocities into components normal and
tangential to the contact plane.
sft0.2630cos AnA vv
sft0.1530sin AtA vv
sft0.2060cos BnB vv
sft6.3460sin BtB vv
Sample Problem 13.15
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
88
• Tangential component of momentum for each ball is
conserved. sft0.15tAtA vv
sft6.34tBtB vv
Total normal component of
momentum of the two ball system
is conserved.
0.6
0.200.26
nBnA
nBnA
nBBnAAnBBnAA
vv
vmvmmm
vmvmvmvm
Sample Problem 13.15
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
89
Sample Problem 13.15
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
90
Normal relative velocities of
the balls are related by the
coefficient of restitution.
4.41
0.200.2690.0
nBnAnBnA vvevv
Solve the last two equations simultaneously for
the normal velocities of the balls after the impact.
sft7.17nAv sft7.23
nBv
Sample Problem 13.15
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
91
6.557.23
6.34tansft9.41
6.347.23
3.407.17
0.15tansft2.23
0.157.17
1
1
B
ntB
A
ntA
v
v
v
v
t
n
Sample Problem 13.16
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
92
Ball B is hanging from an
inextensible cord. An identical ball
A is released from rest when it is
just touching the cord and acquires
a velocity v0 before striking ball B.
Assuming perfectly elastic impact
(e = 1) and no friction, determine
the velocity of each ball
immediately after impact.
Sample Problem
13.16
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
93
Solution:
Determine orientation of
impact line of action.
30
5.02
sin
r
r
Momentum component of
ball A tangential to the
contact plane is
conserved.
0
0
5.0
030sin
vv
vmmv
vmtFvm
tA
tA
AA
Sample Problem 13.16
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
94
0
0
433.05.0
30sin30cos5.00
30sin30cos0
vvv
vvv
vmvmvm
vmvmtTvm
BnA
BnA
BnAtA
BAA
Momentum x-component of the two ball system is
conserved.
Sample Problem 13.16
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
95
Relative velocities along the
line of action before and after
the impact are related by the
coefficient of restitution.
0
0
866.05.0
030cos30sin
vvv
vvv
vvevv
nAB
nAB
nBnAnAnB
Solve last two expressions
for the velocity of ball A
along the line of action and
the velocity of ball B which
is horizontal.
00 693.0520.0 vvvv BnA
0
10
00
693.0
1.16301.46
1.465.0
52.0tan721.0
520.05.0
vv
vv
vvv
B
A
ntA
Sample Problem 13.16
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
96
Sample Problem 13.17
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
97
A 30 kg block is dropped
from a height of 2 m onto
the 10 kg pan of a spring
scale. Assuming the impact
to be perfectly plastic,
determine the maximum
deflection of the pan. The
spring constant is k = 20
kN/m.
Sample Problem 13.17
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
98
Solution:
Apply principle of conservation of energy to
determine velocity of block at instant of impact.
sm26.6
030 J 5880
030
J 588281.9300
2
2
221
2211
2
2
2212
221
2
11
A
A
AAA
A
v
v
VTVT
VvvmT
yWVT
Sample Problem 13.17
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
99
Determine velocity after impact: total momentum of
the block and pan is conserved.
sm70.41030026.630 33
322
vv
vmmvmvm BABBAA
Sample Problem
13.17
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
100
Initial spring deflection
due to pan weight:
m1091.4
1020
81.910 3
33
k
Wx B
Apply principle of conservation
of energy to determine
maximum deflection of spring.
2
43
213
4
24
3
21
34
242
14
4
233
212
321
3
2
212
321
3
10201091.4392
1020392
0
J 241.01091.410200
J 4427.41030
xx
xxx
kxhWWVVV
T
kx
VVV
vmmT
BAeg
eg
BA
Sample Problem 13.17
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
101
m 230.0
10201091.43920241.0442
4
24
3
213
4
4433
x
xx
VTVT
m 1091.4m 230.0 334
xxh m 225.0h
Home Work Assignment 13-4
155, 162, 168, 175, 189
Due Saturday 22/3/2014
Friday, February 28, 2014
Dr. Mohammad Suliman Abuhaiba, PE
102