Download Principle of Work & Energy

ENGR 214
Chapter 13
Kinetics of Particles:
Energy & Momentum Methods
All figures taken from Vector Mechanics for Engineers: Dynamics, Beer and Johnston, 2004
1
Work of a Force

• Particle displacement: dr
• Work of force F corresponding

to displacement dr :
dU  F dr
 F ds cos 
 Fx dx  Fy dy
Work is a scalar quantity of units [Nm]
If F and
If F and

dr

dr
have same direction
have opposite directions
If F is perpendicular to

dr
dU  Fds
dU   Fds
dU  0
2
Work of a Force
Work of a force during a finite displacement:
A2
U12   F dr
A1
s2
s2
s1
s1
   F cos   ds   Ft ds
Ft = Tangential component
Work is represented by the area under
the curve of Ft plotted versus s.
If we use rectangular components:
U12 
A2
  F dx  F dy 
x
y
A1
3
Work of a Force
Work of a constant force in rectilinear motion:
U12  F cos  x
Work of the force of gravity:
dU  Fx dx  Fy dy  W dy
y2
U12    W dy
y1
 W  y2  y1   W y
• Work done by weight = product of weight
W and vertical displacement y
• Work done by weight is +ve when y < 0,
i.e. when body moves down
4
Work of a Force
• Magnitude of the force exerted by a spring is
proportional to deflection,
F  kx
k  spring constant N/m or lb/in.
• Work of the force exerted by spring,
dU   F dx  kx dx
x2
U12    kx dx  12 kx12  12 kx22
x1
• Work of the force exerted by spring is positive
when x2 < x1, 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,
U12   12  F1  F2  x
5
Work of a Force
Forces that do no work:
• Reaction at frictionless pin supporting rotating body
• Reaction at frictionless surface when body in
contact moves along surface
• Reaction at a roller moving along its track
• Weight of a body when its center of gravity
moves horizontally.
6
Particle Kinetic Energy: Principle of Work & Energy

• Consider a particle of mass m acted upon by force F
dv
Ft  mat  m
dt
dv ds
dv
m
 mv
ds dt
ds
F t ds  mv dv
• Integrating from A1 to A2 ,
s2
v2
s1
v1
 Ft ds  m  v dv  12 mv2  12 mv1
2
2
U12  T2  T1
T  12 mv 2  kinetic energy

work of F = change in kinetic energy
T1  U12  T2
Principle of Work & Energy
• Units of work and kinetic energy are the same:
2
m
 m


2
T  12 mv  kg    kg 2 m  N  m  J
s
 s 
7
Applications of the Principle of Work and Energy
Determine velocity of pendulum bob at
A2. Consider work & kinetic energy.

Force P acts normal to path and
does no work:
T1  U12  T2
1 2
0  mgl  mv2
2
v2  2 gl
• Velocity found without determining acceleration and integrating.
• All quantities are scalars
• Forces which do no work are eliminated
8
Applications of the Principle of Work and Energy
• Principle of work and energy cannot be
applied to directly determine the acceleration
of the pendulum bob.
• Calculating the tension in the cord requires
supplementing the method of work and energy
with an application of Newton’s second law.
• As the bob passes through A2 ,
F
n
W=mg
v2  2 gl
 m an
v22
P  mg  m
l
P  mg  m
2 gl
 3mg
l
9
Power and Efficiency
• Power  rate at which work is done.
 
dU F  dr


dt
dt
 
 F v
• Dimensions of power are work/time or force*velocity.
Units for power are
J
m
ft  lb
1 W (watt)  1  1 N 
or 1 hp  550
 746 W
s
s
s
•   efficiency
output work

input work
power output

power input
10
Sample Problem 13.1
A car weighing 1814.4 kg is driven down a 5o incline at a
speed of 96.54 km/h when the brakes are applied causing a
constant total braking force of 6672 N.
Determine the distance traveled by the car as it comes to a
stop. Use the principle of work and energy.
11
Sample Problem 13.1
v1=96.54 km/h
v2=0
T1  U12  T2
v1  96.54 km / h  26.817 m / s
T1  mv 
2
1
1
2
1
2
1814.4  26.817 
2
 652398 Nm
v2  0
N
6672 N
T2  0
U12   6672  x   mg sin 5  x
   5120.7  x
mg
T1  U12  0
652398   5120.7  x  0
x  127.4 m
12
Sample Problem 13.2
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. Apply the principle of work and
energy.
13
T1  U12  T2
x
T1  0
m mA g
Blocks move by equal amounts
U12   m mA gx  mB gx
   m mA  mB  gx
x
mB g
T2  12 mAv 2  12 mB v 2
 12  mA  mB  v 2  250 v 2
  0.25  200  300  9.81 2
 4905 Nm
4905  250 v 2
v  4.43 m / s
READ SOLUTION IN BOOK
14
Sample Problem 13.3
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) the coefficient of kinetic friction between the
package and surface and (b) the velocity of the package as it
passes again through the position shown.
15
Sample Problem 13.3
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
T1  12 mv12  12 60 kg 2.5 m s 2  187.5 J
U12  f
  m kW x

T2  0

  m k 60 kg  9.81m s 2 0.640 m   377 J m k
Pmin  kx0  20 kN m 0.120 m   2400 N
Pmax  k  x0  x   20 kN m 0.160 m   3200 N
U12 e   12 Pmin  Pmax x
  12 2400 N  3200 N 0.040 m   112.0 J
U12  U12  f  U12 e  377 J m k  112 J
T1  U12  T2 :
187.5 J - 377 J m k  112 J  0
mk  0.20
16
Sample Problem 13.3
• Apply the principle of work and energy for the rebound
of the package.
T2  0
T 3 12 mv32  12 60kg v32
U 23  U 23  f  U 23 e  377 J m k  112 J
 36.5 J
T2  U 23  T3 :
0  36.5 J  12 60 kg v32
v3  1.103 m s
17
Sample Problem 13.4
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, and
b) the minimum safe value of the radius of curvature at point 3.
18
Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to determine
velocity at point 2.
T1  0
T2  12 mv22 
U12  W 40 ft 
T1  U12  T2 :
1W 2
v2
2g
0  W 40 ft  

v22  240 ft g  240 ft  32.2 ft s 2

1W 2
v2
2g
v2  50.8 ft s
• Apply Newton’s second law to find normal force by
the track at point 2.
   Fn  m an :
W v22 W 240 ft g
 W  N  m an 

g  2 g 20 ft
N  5W
N  10000 lb
19
Sample Problem 13.4
• Apply principle of work and energy to determine
velocity at point 3.
T1  U13  T3
0  W 25 ft  
v32  225 ft g  225 ft 32.2 ft s 
1W 2
v3
2g
v3  40.1ft s
• Apply Newton’s second law to find minimum radius of
curvature at point 3 such that a positive normal force is
exerted by the track.
   Fn  m an :
W  m an
W v32 W 225 ft g


g 3 g
3
3  50 ft
20
Sample Problem 13.5
D has a weight of 600 lb, and C
weighs 800 lb. Determine the
power delivered by the electric
motor M when D:
(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.
21
Sample Problem 13.5
• In the first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium.
Free-body C:
   Fy  0 :
2T  800 lb  0
T  400 lb
Free-body D:
   Fy  0 : F  T  600 lb  0
F  600 lb  T  600 lb  400 lb  200 lb
Power  Fv D  200 lb8 ft s 
 1600ft  lb s
Power  1600ft  lb s 
1 hp
 2.91 hp
550 ft  lb s
22
Sample Problem 13.5
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.
a D  2.5 ft s 2 
aC   12 a D  1.25 ft s 2 
Free-body C:
   Fy  mC aC : 800  2T 
800
1.25
32.2
T  384.5 lb
Free-body D:
600
2.5
32.2
F  384.5  600  46.6
   Fy  mD a D : F  T  600 
F  262.1 lb
Power  Fv D  262.1 lb 8 ft s   2097 ft  lb s
Power  2097 ft  lb s 
1 hp
 3.81 hp
550 ft  lb s
23
Potential Energy

• Work of the force of gravity W :
U12  W y1  W y2
• Work is independent of path followed; depends
only on the initial and final values of Wy.
V g  Wy
 potential energy of the body with respect
to force of gravity.
 1  Vg 2
U12  V g
• Choice of datum from which the elevation y is
measured is arbitrary.
• Units of work and potential energy are the same:
Vg  Wy  N  m  J
24
Potential Energy
• Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
U12  12 kx12  12 kx22
• The potential energy of the body with respect
to the elastic force,
Ve  12 kx 2
U12  Ve 1  Ve 2
• Note that the preceding expression for Ve is
valid only if the deflection of the spring is
measured from its undeformed position.
25
Conservative Forces
 A force is conservative if its work U 1 2 is independent of the
path followed by the particle as it moves from A1 to A2.
U12  V  x1 , y1   V  x2 , y2 
 Gravity forces and elastic spring forces are conservative.
 Friction forces are non-conservative.
26
Conservation of Energy
• Work of a conservative force:
U12  V1  V2
• Principle of work and energy:
U12  T2  T1
• It follows that:
T1  V1  T2  V2
E  T  V  constant
T1  0 V1  W
T1  V1  W
T2  12 mv22 
T2  V2  W
1W
2 g   W V2  0
2g
• When a particle moves under the action of
conservative forces, the 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.
27
Conservation of Energy
•
•
•
•
Particle moving along path.
In the absence of friction, total mechanical energy is constant.
Potential energy depends only on elevation.
Particle speed is the same at any given elevation (A, A', A")
28
Sample Problem 13.6
A 20 lb collar slides without friction along a vertical rod. 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.
29
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: Ve  12 kx12  12 3 lb in.8 in.  4 in.2  24 in.  lb
V1  Ve  Vg  24 in.  lb  0  2 ft  lb
T1  0
Position 2: Ve  12 kx22  12 3 lb in.10 in.  4 in.2  54 in.  lb
Vg  Wy  20 lb  6 in.  120 in.  lb
V2  Ve  Vg  54  120  66 in.  lb  5.5 ft  lb
T2  12 mv22 
1 20 2
v2  0.311v22
2 32.2
Conservation of Energy:
T1  V1  T2  V2
0  2 ft  lb  0.311v22  5.5 ft  lb
v2  4.91ft s 
30
Sample Problem 13.7
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.
• For pellet to remain in contact, force
exerted on pellet must be greater
than or equal to zero.
• Apply principle of conservation of
energy between A & D.
• Set force exerted by loop on pellet to
zero & solve for the minimum
velocity at D.
31
Sample Problem 13.7
SOLUTION:
• Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.
2
   Fn  man : W  man
mg  m vD
r
2
vD
 rg  2 ft 32.2 ft s   64.4 ft 2 s 2
• Apply the principle of conservation of energy between
points A and D.
V1  Ve  Vg  12 kx2  0  12 36 lb ft x 2  18 x 2
T1  0
V2  Ve  Vg  0  Wy  0.5 lb 4 ft   2 ft  lb
2
T2  12 mvD



1 0.5 lb
2 2
64
.
4
ft
s  0.5 ft  lb
2 32.2 ft s 2
T1  V1  T2  V2
0  18 x 2  0.5  2
x  0.3727 ft  4.47 in.
32
Principle of Impulse and Momentum
• Newton’s second law:
 d


F  mv 
mv  linear momentum
dt


Fdt  d mv 
t2



F
dt

m
v

m
v

2
1
t1
t2


 Fdt  Imp12  impulse of the force F
t1


mv1  Imp12  mv2
• Final momentum = initial momentum +
impulse of force during time interval
• Units of impulse is [Ns]=[kg m/s]
33
Principle of Impulse and Momentum
t2
Principle of impulse & momentum:
mv1   Fdt  mv2
t1
Can be resolved
into components:
t2
t2
 mvx 1   Fx dt   mvx 2 ,  mv y 1   Fy dt   mv y 2
t1
t2
t1
When several forces act on a particle: mv1    Fdt  mv2
t1
When more than 1 particle is involved, we can either consider
each particle separately, or we can add the momentum and
t2
impulse for the entire system of particles:
 mv1    Fdt   mv2
t1
Note: forces of action & reaction exerted by the particles on
each other have impulses that cancel out. Only the impulses of
external forces need to be considered.
34
Principle of Impulse and Momentum
t2
For entire system of particles:
 mv    Fdt   mv
1
2
t1
If sum of external forces = 0
mv  mv
1
2
Total momentum is conserved
 mv   mv
1
2
0  mAv A  mB vB
vA
mB

vB
mA
Boats move in opposite directions
35
Impulsive Motion
Impulsive force: a force that acts on a particle during a very short
time interval, but is large enough to cause a significant change in
momentum



When impulsive forces act on a tennis ball: mv1   F t  mv2
Non-impulsive forces: forces for which F t
therefore can be neglected (such as weight)
is small, and
36
Sample Problem 13.10
A car weighing 1814.4 kg is driven down a 5° incline at a
speed of 96.54 km/h when the brakes are applied causing a
constant total breaking force of 6672 N.
Determine the time required for the car to come to a stop.
Use the principle of impulse and momentum.
37
Sample Problem 13.10
t2
Principle of impulse
mv1   Fdt  mv2
and momentum:
t
1
Taking components parallel to the incline:
mv1   mg sin 5  t  Ft  0
1814.4  26.817   1814.4  9.81sin 5  t  6672t  0
t  9.5 s
38
Sample Problem 13.11
36.6 m/s
24.4 m/s
A 113.4 g baseball is pitched with a velocity of 24.4 m/s. After
the ball is hit by the bat, it has a velocity of 36.6 m/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. Use the principle of impulse and momentum
39
Sample Problem 13.11
36.6 m/s
Principle of impulse and momentum:
mv1  F t  mv2
x-direction:
mv1  Fx t  mv2 cos 40
0.1134  24.4   Fx  0.015   0.1134  36.6  cos 40
24.4 m/s
Fx  396.4 N
y-direction:
0  Fy t  mv2 sin 40
0  Fy  0.015  0.1134  36.6  sin 40
Fy  177.9 N
177.9 N
F   396.4  i  177.9  j
F  434.5 N
434.5 N
24.2° 396.4 N
40
Sample Problem 13.12
Principle of impulse and
momentum:
t2
mv1   Fdt  mv2
t1
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, and (c) the
fraction of the initial energy lost in the impact. Use the principle
of impulse and momentum.
41
Sample Problem 13.12
Apply the principle of impulse and momentum to the package-cart system:
mv  F t  mv
1
x-direction:
2




m pv1   Imp12  m p  mc v2
m p v1 cos30  0   m p  mc  v2
10  3 cos30  10  25 v2
v2  0.742 m/s
42
Sample Problem 13.12
Apply the same principle to the package alone to determine the impulse
exerted on it from the change in its momentum.
mpv1   F t  mpv2
x direction:
m p v1 cos30  Fx t  m p v2
10  3 cos30  Fx t  10  v2
Fx t  18.56 N  s
y components: m p v1 sin 30  Fy t  0
 10  3 sin 30  Fy t  0
 Imp
12
Fy t  15 N  s
 F t   18.56  i  15  j
F t  23.9 N  s
43
Sample Problem 13.12
To determine the fraction of energy lost, calculate initial and
final energies:
Initial energy:
T1  m v 
Final energy:
T2 
1
2
1
2
2
p 1
m
p
1
2
10 3
 mc  v 
2
2
1
2
2
 45 J
10  25  0.742 
2
 9.63 J
T1  T2 45  9.63

 0.786
T1
45
44
Impact
• 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.
• 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.
• Oblique Impact: Impact for which one or both of
the bodies move along a line other than the line
of impact.
45
Direct Central Impact
• Bodies moving in the same straight line:
v A  vB
• Upon impact the bodies undergo a
period of deformation, at the 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.
• Wish to determine the final velocities of the
two bodies. The total momentum of the
two body system is preserved,
mAv A  mB vB  mAvA  mB vB
• A second relation between the final
velocities is required.
46
Direct Central Impact
• Period of deformation: m Av A   Pdt  m Au
e  coefficient of restitution

 Rdt  u  vA
 Pdt v A  u
0  e 1
• Period of restitution:
m Au   Rdt  m AvA
• A similar analysis of particle B yields
• Combining the relations leads to the desired
second relation between the final velocities.
• Perfectly plastic impact, e = 0: vB  vA  v
e
vB  u
u  vB
vB  vA  ev A  v B 
m Av A  mB v B  m A  mB v
vB  vA  v A  v B
• Perfectly elastic impact, e = 1:
Total energy and total momentum conserved.
e depends on materials, impact velocity, shape & size of colliding surfaces.
47
Coefficient of Restitution
vB  vA  ev A  v B 
In general, e has a value between 0 and 1. The two limiting
conditions are:
• Elastic impact (e = 1): In a perfectly elastic collision, no
energy is lost and the relative separation velocity equals the
relative approach velocity of the particles. In practical
situations, this condition cannot be achieved.
• Plastic impact (e = 0): In a plastic impact, the
relative separation velocity is zero. The particles
stick together and move with a common velocity
after the impact.
Some typical values of e are:
Steel on steel: 0.5 – 0.8
Lead on lead: 0.12 – 0.18
Wood on wood: 0.4 – 0.6
Glass on glass: 0.93 – 0.95
48
Coefficient of Restitution
vB  vA  ev A  v B 
The quality of a tennis ball is
measured by the height of its
bounce. This can be quantified
by the coefficient of restitution.
If the height from which the ball is dropped and the height of
its resulting bounce are known, we can determine the
coefficient of restitution. How?
During a collision, some of the initial kinetic energy will be lost
in the form of heat, sound, or due to localized deformation.
49
Problems Involving Energy and Momentum
• Three methods for the analysis of kinetics problems:
- Direct application of Newton’s second law
- Method of work and energy
- Method of impulse and momentum
• Select the method best suited for the problem or part of a
problem under consideration.
50
Sample Problem 13.13
A 20-Mg railroad car moving at a speed of 0.5 m/s to the
right collides with a 35-Mg car which is at rest. If after
collision the 35-Mg car is observed to move to the right at a
speed of 0.3 m/s, determine the coefficient of restitution.
51
mAv A  mB vB  mAvA  mB vB
 20000 0.5   20000 vA  35000 0.3
vA  0.025 m / s
vB  vA 0.3  (0.025)
e

 0.65
v A  vB
0.5
52
Sample Problem 13.17
A 30 kg block is dropped from a height of 2 m onto the the
10 kg pan of a spring scale. Assuming the impact to be
perfectly plastic, determine the maximum deflection of the
pan. The constant of the spring is k = 20 kN/m.
53
54
Sample Problem 13.17
• Apply principle of conservation of energy to
determine velocity of the block at instant of impact.
T1  0
V1  WA y  309.812  588 J
T2  12 m A v A 22  12 30 v A 22
V2  0
T1  V1  T2  V2
0  588 J  12 30 v A 22  0
v A 2  6.26 m s
• Determine velocity after impact from conservation of
momentum.
mA v A 2  mB vB 2  mA  mB v3
306.26  0  30  10v3
v3  4.70 m s
55
Sample Problem 13.17
• Apply principle of conservation of energy to
determine maximum deflection of spring.
T3  12 m A  mB v32  12 30  10 4.7 2  442 J
V3  Vg  Ve
0
1 kx 2
2 3

1
2
20  10 4.91 10 
3
3 2
 0.241 J
T4  0
Initial spring deflection due to
pan weight:
x3 
WB 109.81
3


4
.
91

10
m
3
k
20  10
V4  Vg  Ve  WA  WB  h   12 kx42


 392x4  4.91  103   12 20  103 x42
 392 x4  x3   12 20  103 x42
T3  V3  T4  V4

 

442  0.241  0  392 x4  4.91  103  12 20  103 x42
x4  0.230 m
h  x4  x3  0.230 m  4.91  103 m
h  0.225 m
56