Chapter 4
Conservation laws for systems of particles
In this chapter, we shall introduce (but not in this order) the following general concepts:
1. The linear impulse of a force
2. The angular impulse of a force
3. The power transmitted by a force
4. The work done by a force
5. The potential energy of a force.
6. The linear momentum of a particle (or system of particles)
7. The angular momentum of a particle, or system of particles.
8. The kinetic energy of a particle, or system of particles
9. The linear impulse momentum relations for a particle, and conservation of linear momentum
10. The principle of conservation of angular momentum for a particle
11. The principle of conservation of energy for a particle or system of particles.
We will also illustrate how these concepts can be used in engineering calculations. As you will see, to
applying these principles to engineering calculations you will need two things: (i) a thorough
understanding of the principles themselves; and (ii) Physical insight into how engineering systems
behave, so you can see how to apply the theory to practice. The first is easy. The second is hard, and
people who can do this best make the best engineers.
4.1 Work, Power, Potential Energy and Kinetic Energy relations
The concepts of work, power and energy are among the most powerful ideas in the physical sciences.
Their most important application is in the field of thermodynamics, which describes the exchange of
energy between interacting systems. In addition, concepts of energy carry over to relativistic systems and
quantum mechanics, where the classical versions of Newton’s laws themselves no longer apply.
In this section, we develop the basic definitions of mechanical work and energy, and show how they can
be used to analyze motion of dynamical systems. Future courses will expand on these concepts further.
4.1.1 Definition of the power and work done by a force
k
F(t)
Suppose that a force F acts on a particle that moves with speed v.
By definition:
The Power developed by the force, (or the rate of work done by the
force) is P  F  v . If both force and velocity are expressed in
Cartesian components, then
P  Fx vx  Fy v y  Fz vz
Work has units of Nm/s, or `Watts’ in SI units.
i
O
P
j
The work done by the force during a time interval t0  t  t1 is
F(t)
t1
t0
The work done by the force can also be calculated by integrating the
force vector along the path traveled by the force, as
W
r1
P
r0 k
W   F  vdt
i
O
j
r1
 F  dr
r0
where r0 , r1 are the initial and final positions of the force.
Work has units of Nm in SI units, or `Joules’
A moving force can do work on a particle, or on any moving object. For example, if a force acts to
stretch a spring, it is said to do work on the spring.
4.1.2 Definition of the power and work done by a concentrated moment, couple or torque.
‘Concentrated moment, ‘Couple’ and `Torque’ are different names for a
‘generalized force’ that causes rotational motion without causing
translational motion. These concepts are not often used to analyze
motion of particles, where rotational motion is ignored – the only
application might be to analyze rotational motion of a massless frame
connected to one or more particles. For completeness, however, the
power-work relations for moments are listed in this section and applied
to some simple problems. To do this, we need briefly to discuss how
rotational motion is described. This topic will be developed further in
Chapter 6, where we discuss motion of rigid bodies.
Axis of
n
Definition of an angular velocity vector Visualize a
rotation
spinning object, like the cube shown in the figure. The box
rotates about an axis – in the example, the axis is the line
connecting two cube diagonals. In addition, the object
turns through some number of revolutions every minute.
We would specify the angular velocity of the shaft as a
vector ω , with the following properties:
1. The direction of the vector is parallel to the axis of
the shaft (the axis of rotation). This direction would
be specified by a unit vector n parallel to the shaft
2. There are, of course, two possible directions for n. By convention, we always choose a direction
such that, when viewed in a direction parallel to n (so the vector points away from you) the shaft
appears to rotate clockwise. Or conversely, if n points towards you, the shaft appears to rotate
counterclockwise. (This is the `right hand screw convention’)
Viewed along n
Viewed in direction opposite to n
3. The magnitude of the vector is the angular speed d / dt of the object, in radians per second. If
you know the revs per minute n turned by the shaft, the number of radians per sec follows as
d / dt  120 n . The magnitude of the angular velocity is often denoted by   d / dt
The angular velocity vector is then ω 
d
n  n .
dt
Since angular velocity is a vector, it has components ω   x i   y j   z k in a fixed Cartesian basis.
Rate of work done by a torque or moment: If a moment M  M x i  M y j  M z k acts on an object that
rotates with angular velocity ω , the rate of work done on the object by M is
P  M  ω  M x x  M y  y  M z  z
4.1.3 Simple examples of power and work calculations
FD
Example 1: An aircraft with mass 45000 kg flying at 200 knots (102m/s)
climbs at 1000ft/min. Calculate the rate of work done on the aircraft by gravity.
FT
The gravitational force is  mgj , and the velocity vector of the aircraft is
v  vx i  v y j . The rate of work done on the aircraft is therefore
j
mg
i
P  F  v  mgv y
Substituting numbers gives
Example 2: Calculate a formula for the work required to stretch a spring with
stiffness k and unstretched length L0 from length l0 to length l1 .
j x
F
i
The figure shows a spring that held fixed at A and is stretched in the
A
B
horizontal direction by a force F  Fx i acting at B. At some instant the
spring has length l0  x  l1 . The spring force law states that the force acting on the spring at B is related
to the length of the spring x by
F  Fx i  k ( x  L0 )i
The position vector of the force is r  xi , and therefore the work done is
W
r1
l1
r0
l0
 F  dr   k ( x  L0 )i  dxi  2 (l1  L0 )
k
2
 (l0  L0 ) 2

Example 3: Calculate the work done by gravity on a satellite that is launched
from the surface of the earth to an altitude of 250km (a typical low earth orbit).
j
h
Assumptions
1. The earth’s radius is 6378.145km
2. The mass of a typical satellite is 4135kg - see , e.g.
http://www.astronautix.com/craft/hs601.htm
i
R
3. The Gravitational parameter   GM  3.986012  105 km3s 1 (G=
gravitational constant; M=mass of earth)
4. We will assume that the satellite is launched along a straight line path parallel to the i direction,
starting the earths surface and extending to the altitude of the orbit. It turns out that the work
done is independent of the path, but this is not obvious without more elaborate and sophisticated
calculations.
Calculation:
1. The gravitational force on the satellite is
F
GMm
x2
i
2. The work done follows as
Rh

F
R
GMm
x
2
1
 1
i  dxi  GMm 
 
R

h
R


3. Substituting numbers gives 9.7  106 J (be careful with units – if you work with kilometers the
work done is in N-km instead of SI units Nm)
Example 4: A Ferrari Testarossa skids to a stop over a distance of 250ft. Calculate the total work done
on the car by the friction forces acting on its wheels.
L
Assumptions:
j
1. A Ferrari Testarossa has mass 1506kg (see
http://www.ultimatecarpage.com/car/1889/FerrariTestarossa.html)
i TF NF mg
x
2. The coefficient of friction between wheels and road is of
order 0.8
3. We assume the brakes are locked so all wheels skid, and air resistance is neglected
TR
Calculation The figure shows a free body diagram. The equation of motion for the car is
(TF  TR )i  ( N R  N F  mg ) j  max i
1. The vertical component of the equation of motion yields N R  N F  mg
NR
TF   N F  TR  TF    N R  N F    mg
3. The position vectors of the car’s front and rear wheels are rF  xi
rR  ( x  L)i . The work
done follows as. We suppose that the rear wheel starts at some point x0i when the brakes are
applied and skids a total distance d.
2. The friction law shows that TR   N R
W
x0  d

x0
FR  drR 
x0  d  L

x0  L
FF  drF 
x0  d

x0
TR i  (dxi ) 
x0 d  L

TF i  (dxi )  (TR  TF )d
x0  L
4. The work done follows as W    mg . Substituting numbers gives W  9  105 J .
Example 5: The figure shows a box that is pushed up a slope by a force P.
The box moves with speed v. Find a formula for the rate of work done by
each of the forces acting on the box.
The figure shows a free body diagram. The force vectors are
1. Applied force P cos  i  P sin  j
2. Friction T cos  i  T sin  j
3. Normal reaction  N sin  i  T cos  j
4. Weight  mgj
The velocity vector is v cos  i  v sin  j
P

j
i
T
P
j
N
mg
i
Evaluating the dot products F  v for each formula, and recalling that cos 2   sin 2   1 gives
1. Applied force Pv
2. Friction Tv
3. Normal reaction 0
4. Weight  mg sin 
Example 6: The table lists the experimentally measured force-v-draw data for a longbow. Calculate the total work done to draw the bow.
In this case we don’t have a function that specifies the force as a function of position;
instead, we have a table of numerical values. We have to approximate the integral
W
r1
 F  dr
r0
numerically. To understand how to do this, remember that integrating a function can be
visualized as computing the area under a curve of the function, as illustrated in the
figure.
Force
N
0
40
90
140
180
220
270
Draw
(cm)
0
10
20
30
40
50
60
F
We can estimate the integral by dividing the area into a series of
trapezoids, as shown. Recall that the area of a trapezoid is (base x
average height), so the total area of the function is
 f  f3 
 f  f4 
 f  f5 
 f f 
W  x1  1 2   x2  2
 x3  3
 x4  4



 2 
 2 
 2 
 2 
 40  0 
 20  10 
 0.1
  0.1
  ....
 2 
 2 
f1
f2
f3
f4
x
 x1  x2
 x3  x4
You could easily do this calculation by hand – but for lazy people like me MATLAB has a convenient
function called `trapz’ that does this calculation automatically. Here’s how to use it
>> draw = [0,10,20,30,40,50,60]*0.01;
>> force = [0,40,90,140,180,220,270];
>> trapz(draw,force)
ans =
80.5000
>>
So the solution is 80.5J
f5
4.1.4 Definition of the potential energy of a conservative force
F(t)
Preamble: Textbooks nearly always define the `potential energy of a
P
force.’ Strictly speaking, we cannot define a potential energy of a single r0 k
force – instead, we need to define the potential energy of a pair of forces. A
force can’t exist by itself – there must always be an equal and opposite
j
O
reaction force acting on a second body. In all of the discussion to be
i
r1
presented in this section, we implicitly assume that the reaction force is
acting on a second body, which is fixed at the origin. This simplifies calculations, and makes the
discussion presented here look like those given in textbooks, but you should remember that the potential
energy of a force pair is always a function of the relative positions of the two forces.
With that proviso, consider a force F acting on a particle at some position r in space. Recall that the work
done by a force that moves from position vector r0 to position vector r1 is
W
r1
 F  dr
r0
In general, the work done by the force depends on the path between r0 to r1 . For some special forces,
however, the work done is independent of the path. Such forces are said to be conservative.
For a force to be conservative:
The force must be a function only of its position – i.e. it can’t depend on the velocity of the
force, for example.
The force vector must satisfy curl(F )  0
Examples of conservative forces include gravity, electrostatic forces, and the forces exerted by a spring.
Examples of non-conservative (or should that be liberal?) forces include friction, air resistance, and
aerodynamic lift forces.
The potential energy of a conservative force is defined as the negative of the work done by the force in
moving from some arbitrary initial position r0 to a new position r , i.e.
r
V (r )    F  dr  constant
r0
The constant is arbitrary, and the negative sign is introduced by convention (it makes sure that systems try
to minimize their potential energy). If there is a point where the force is zero, it is usual to put r0 at this
point, and take the constant to be zero.
Note that
1. The potential energy is a scalar valued function
2. The potential energy is a function only of the position of the force. If we choose to describe
position in terms of Cartesian components r  xi  yj  zk , then V (r )  V ( x, y, z ) .
3. The relationship between potential energy and force can also be expressed in differential form
(which is often more useful for actual calculations) as
F   grad(V )
If we choose to work with Cartesian components, then
 V
V
V 
Fx i  Fy j  Fz k   
i
j
k
y
z 
 x
Occasionally, you might have to calculate a potential energy function by integrating forces – for example,
if you are interested in running a molecular dynamic simulation of a collection of atoms in a material, you
will need to describe the interatomic forces in some convenient way. The interatomic forces can be
estimated by doing quantum-mechanical calculations, and the results can be approximated by a suitable
potential energy function. Here are a few examples showing how you can integrate forces to calculate
potential energy
Example 1: Potential energy of forces exerted by a spring. A free body diagram showing the forces
exerted by a spring connecting two objects is shown in the figure.
j
x
1. The force exerted by a spring is
F   k ( x  L0 )i
i
2. The position vector of the force is r  xi
3. The potential energy follows as
r
L
r0
L0
1
V (r )    F  dr  constant =  k ( x  L0 )i  dxi  k ( L  L0 ) 2
2
F
where we have taken the constant to be zero.
Example 2: Potential energy of electrostatic forces exerted by charged particles.
The figure shows two charged particles a distance x apart. To
calculate the potential energy of the force acting on particle 2, we
place particle 1 at the origin, and note that the force acting on
particle 2 is
QQ
F   1 22 i
4 x
+Q1
+Q2
j
2 F
i
1
x
where Q1 and Q2 are the charges on the two particles, and  is a fundamental physical constant known
as the Permittivity of the medium surrounding the particles. Since the force is zero when the particles are
infinitely far apart, we take r0 at infinity. The potential energy follows as
x
Q1Q2
QQ
i  dxi  1 2
2
4 x
4 x

V (r )   
Table of potential energy relations
In practice, however, we rarely need to do the integrals to calculate the potential energy of a force,
because there are very few different kinds of force. For most engineering calculations the potential
energy formulas listed in the table below are sufficient.
Type of force
Force vector
Potential energy
F
m
Gravity acting on a
particle near earths
surface
F  mgj
j
V  mgy
y
i
F
Gravitational force
exerted on mass m by
mass M at the origin
F
GMm
r
3
V 
r
m
r
GMm
r
r
Force exerted by a
spring with stiffness k
and unstretched length
L0
j
r
F  k ( r  L0 )
r
r
F
1
2
V  k  r  L0 
2
i
2
Force acting between
two charged particles
F
Q1Q2
r
4 r 3
V
Q1Q2
4 r
j
+Q1
i
r
1
Force exerted by one
molecule of a noble gas
(e.g. He, Ar, etc) on
another (Lennard Jones
potential). a is the
equilibrium spacing
between molecules, and
E is the energy of the
bond.
E a
F  12  2  
a r

13
a
 
r
7
6
 a 12
a 
 E    2   
 r  
 r  r 
F
+Q2
2
F
j
i
1
r
4.1.5 Potential energy of concentrated moments exerted by a torsional spring
A potential energy cannot usually be defined for
concentrated moments, because rotational motion is itself
path dependent (the orientation of an object that is given two
successive rotations depends on the order in which the
rotations are applied). A potential energy can, however, be
defined for the moments exerted by a torsional spring.
A solid rod is a good example of a torsional spring. You
could take hold of the ends of the rod and twist them,
causing one end to rotate relative to the other. To do this, you would apply a moment or a couple to each
end of the rod, with direction parallel to the axis of the rod. The angle of twist increases with the
moment. Various torsion spring designs used in practice are shown in the picture – the image is from
http://www.mollificio.lombardo.molle.com/springs/torsion_springs.html
More generally, a torsional spring resists rotation, by exerting equal and opposite moments on objects
connected to its ends. For a linear spring the moment is proportional to the angle of rotation applied to
the spring.
The figure shows a formal free body diagram for two objects
connected by a torsional spring. If object A is held fixed, and object B
is rotated through an angle  about an axis parallel to a unit vector n,
then the spring exerts a moment
M   n
on object B where  is the torsional stiffness of the spring. Torsional
stiffness has units of Nm/radian.
The potential energy of the moments exerted by the spring can be
determined by computing the work done to twist the spring through an
angle  .
1. The work done by a moment M due to twisting through a very
small angle d about an axis parallel to a vector n is
dW  M  d n
2. The potential energy is the negative of the total work done by
M, i.e.







j
n
B
A
-M
i
M
B
M
A
-M
1
V    M  d n      n   d n    d   2
2
4.1.6 Definition of the Kinetic Energy of a particle
v
Consider a particle with mass m which moves with velocity
v  vx i  v y j  vz k . By definition, its kinetic energy is

1
2 1
T  m v  m vx2  v 2y  vz2
2
2
r0 k

i
P
O
j
r1
4.1.7 Power-Work-kinetic energy relations for a single particle
Consider a particle with mass m that moves under the action of a force F. Suppose that
1. At some time t0 the particle has some initial position r0 , velocity v 0 and kinetic energy T0
2. At some later time t the particle has a new position r, velocity v and kinetic energy T .
3. Let P  F  v denote the rate of work done by the force
t
r1
t0
r0
4. Let W   Pdt   F  dr be the total work done by the force
The Power-kinetic energy relation for the particle states that the rate of work done by F is equal to the rate
of change of kinetic energy of the particle, i.e.
dT
P
dt
This is just another way of writing Newton’s law for the particle: to see this, note that we can take the dot
product of both sides of F=ma with the particle velocity
dv
d 1

F  v  ma  v  m  v   m( v  v) 
dt
dt  2

To see the last step, do the derivative using the Chain rule and note that a  b  b  a .
The Work-kinetic energy relation for a particle says that the total work done by the force F on the particle
is equal to the change in the kinetic energy of the particle.
W  T  T0
This follows by integrating the power-kinetic energy relation with respect to time.
4.1.8 Examples of simple calculations using work-power-kinetic energy relations
There are two main applications of the work-power-kinetic energy relations. You can use them to
calculate the distance over which a force must act in order to produce a given change in velocity. You
can also use them to estimate the energy required to make a particle move in a particular way, or the
amount of energy that can be extracted from a collection of moving particles (e.g. using a wind turbine)
Example 1: Estimate the minimum distance required for a 14 wheeler that travels at the RI speed-limit to
brake to a standstill. Is the distance to stop any different for a Toyota Echo?
This problem can be solved by noting that, since we know the initial and final speed of the vehicle, we
can calculate the change in kinetic energy as the vehicle stops. The change in kinetic energy must equal
the work done by the forces acting on the vehicle – which depends on the distance slid. Here are the
details of the calculation.
Assumptions:
1. We assume that all the wheels are locked and skid over
the ground (this will stop the vehicle in the shortest
possible distance)
2. The contacts are assumed to have friction coefficient
  0.8
3. The vehicle is idealized as a particle.
4. Air resistance will be neglected.
TA
TB
TD
TC
j NA NB NC
mg
i
ND
Calculation:
1. The figure shows a free body diagram.
2. The equation of motion for the vehicle is
 TA  TB  TC  TD  i   N A  N B  NC  N D  mg  j  ma x i
The vertical component of the equation shows that N A  N B  NC  N D  mg .
3. The friction force follows as TA  TB  TC  TD     N A  N B  NC  N D    mg
4. If the vehicle skids for a distance d, the total work done by the forces acting on the vehicle is
d
W   (TA  TB  TC  TD )i  dxi  (TA  TB  TC  TD )d    mgd
0
5. The work-energy relation states that the total work done on the particle is equal to its change in
kinetic energy. When the brakes are applied the vehicle is traveling at the speed limit, with speed
V; at the end of the skid its speed is zero. The change in kinetic energy is therefore
T  0  mV 2 / 2 . The work-energy relation shows that
 mgd   mV 2 / 2  d  V 2 /  g
Substituting numbers gives
This simple calculation suggests that the braking distance for a vehicle depends only on its speed and the
friction coefficient between wheels and tires. This is unlikely to vary much from one vehicle to another.
In practice there may be more variation between vehicles than this estimate suggests, partly because
factors like air resistance and aerodynamic lift forces will influence the results, and also because vehicles
usually don’t skid during an emergency stop (if they do, the driver loses control) – the nature of the
braking system therefore also may change the prediction.
Example 3: Compare the power consumption of a Ford Excursion to that of a Chevy Cobalt during stopstart driving in a traffic jam.
During stop-start driving, the vehicle must be repeatedly accelerated to some (low) velocity; and then
braked to a stop. Power is expended to accelerate the vehicle; this power is dissipated as heat in the brakes
during braking. To calculate the energy consumption, we must estimate the energy required to accelerate
the vehicle to its maximum speed, and estimate the frequency of this event.
Calculation/Assumptions:
1. We assume that the speed in a traffic jam is low enough that air resistance can be neglected.
2. The energy to accelerate to speed V is mV 2 / 2 .
3. We assume that the vehicle accelerates and brakes with constant acceleration – if so, its average
speed is V/2.
4. If the vehicle travels a distance d between stops, the time between two stops is 2d/V.
5. The average power is therefore mV / 4d .
6. Taking V=15mph (7m/s) and d=200ft (61m) are reasonable values – the power is therefore
0.03m, with m in kg. A Ford Excursion weighs 9200 lb (4170 kg), requiring 125 Watts (about
that of a light bulb) to keep moving. A Chevy Cobalt weighs 2681lb (1216kg) and requires only
36 Watts – a very substantial energy saving.
Reducing vehicle weight is the most effective way of improving fuel efficiency during slow driving, and
also reduces manufacturing costs and material requirements. Another, more costly, approach is to use a
system that can recover the energy during braking – this is the main reason that hybrid vehicles like the
Prius have better fuel economy than conventional vehicles.
Example 4: Estimate the power that can be generated by a wind turbine.
The figure shows a wind turbine. The turbine blades deflect the air flowing past
them: this changes the air speed and so exerts a force on the blades. If the blades
move, the force exerted by the air on the blades does work – this work is the power
generated by the turbine. The rate of work done by the air on the blades must equal
the change in kinetic energy of the air as it flows past the blades. Consequently,
we can estimate the power generated by the turbine by calculating the change in
kinetic energy of the air flowing through it.
To do this properly needs a very sophisticated analysis of the air flow around the
turbine. However, we can get a rather crude estimate of the power by assuming
that the turbine is able to extract all the energy from the air that flows through the
circular area swept by the blades.
Calculation: Let V denote the wind speed, and let  denote the density of the air.
1. In a time t, a cylindrical region of air with radius R and height Vt passes through the fan.
2. The cylindrical region has mass  R 2 Vt


1
 R 2 Vt V 2
2
dT  2 3
4. The rate of flow of kinetic energy through the fan is therefore
 R V
dt 2
5. If all this energy could be used to do work on the fan blades the power generated would be
dT  2 3
P
 R V
dt 2
3. The kinetic energy of the cylindrical region of air is T 
Representative numbers are (i) Air density 1.2 kg / m3 ; (ii) air speed 25mph (11 m/sec); (iii) Radius 30m
This gives 1.8MW. For comparison, a nuclear power plant generates about 500-1000 MW.
A more sophisticated calculation (which will be covered in EN810) shows that in practice the maximum
possible amount of energy that can be extracted from the air is about 60% of this estimate. On average, a
typical household uses about a kW of energy; so a single turbine could provide enough power for about 510 houses.
4.1.9 Energy relations for a conservative system of particles.
The figure shows a `system of particles’ – this is just a
collection of objects that we might be interested in, which can
be idealized as particles. Each particle in the system can
experience forces applied by:
Other particles in the system (e.g. due to gravity, electric
charges on the particles, or because the particles are physically
connected through springs, or because the particles collide).
We call these internal forces acting in the system. We will
denote the internal force exerted by the ith particle on the jth
particle by R ij . Note that, because every action has an equal
and opposite reaction, the force exerted on the jth particle by
F1ext
m4
F3ext
m1
R13 R31
R12
m
R32 3
R23
R21
m2
F2ext
the ith particle must be equal and opposite, to R ij , i.e. R ij  R ji .
Forces exerted on the particles by the outside world (e.g. by externally applied gravitational or
electromagnetic fields, or because the particles are connected to the outside world through mechanical
linkages or springs). We call these external forces acting on the system, and we will denote the external
force on the i th particle by Fiext (t )
We define the total external work done on the system during a time interval t0  t  t0  t as the sum of
the work done by the external forces.
Wext 
t0 t
 
forces
Fiext (t )  v(t )dt
t0
The total work done can also include a contribution from external moments acting on the system.
The system of particles is conservative if all the internal forces in the system are conservative. This
means that the particles must interact through conservative forces such as gravity, springs, electrostatic
forces, and so on. The particles can also be connected by rigid links, or touch one another, but contacts
between particles must be frictionless.
If this is the case, we can define the total potential energy of the system as the sum of potential energies
of all the internal forces.
We also define the total kinetic energy T total of the system as the sum of kinetic energies of all the
particles.
The work-energy relation for the system of particles can then be stated as follows. Suppose that
1. At some time t0 the system has and kinetic energy T0total
2. At some later time t the system has kinetic energy T total .
3. Let V0total denote the potential energy of the force at time t0
4. Let V total denote the potential energy of the force at time t
5. Let Wext denote the total work done on the system between t0  t  t0  t
Work Energy Relation: This law states that the external work done on the system is equal to the change
in total kinetic and potential energy of the system.
Wext  T  V   T0  V0 
We won’t attempt to prove this result - the proof is conceptually very straightforward: it simply involves
summing the work-energy relation for all the particles in the system; and we’ve already seen that the
work-energy relations are simply a different way of writing Newton’s laws. But when written out the
sums make the proof look scary and difficult to follow so we’ll spare you the gory details. If you are
interested, ask us (or better still see if you can do it for yourself!)
Energy conservation law For the special case where no external forces act on the system, the total energy
of the system is constant
Wext  0  T  V  T0  V0 
It is worth making one final remark before we turn to applications of these law. We often invoke the
principle of conservation of energy when analyzing the motion of an object that is subjected to the earth’s
gravitational field. For example, the first problem we solve in the next section involves the motion of a
projectile launched from the earth’s surface. We usually glibly say that `the sum of the potential and
kinetic energies of the particle are constant’ – and if you’ve done physics courses you’ve probably used
this kind of thinking. It is not really correct, although it leads to a more or less correct solution.
Properly, we should consider the earth and the projectile together as a conservative system. This means
we must include the kinetic energy of the earth in the calculation, which changes by a small, but finite,
amount due to gravitational interaction with the projectile. Fortunately, the principle of conservation of
linear momentum (to be covered later) can be used to show that the change in kinetic energy of the earth
is negligibly small compared to that of the particle.
4.1.10 Examples of calculations using kinetic and potential energy in conservative systems
The kinetic-potential energy relations can be used to quickly calculate relationships between the velocity
and position of an object. Several examples are provided below.
Example 1: (Boring FE exam question) A projectile with mass m is
launched from the ground with velocity V0 at angle  . Calculate an
expression for the maximum height reached by the projectile.
V0
j

h
If air resistance can be neglected, we can regard the earth and the
i
projectile together as a conservative system. We neglect the change in
the earth’s kinetic energy. In addition, since the gravitational force
acting on the particle is vertical, the particle’s horizontal component of velocity must be constant.
Calculation:
1. Just after launch, the velocity of the particle is v  V0 cos  i  V0 sin  j
2. The kinetic energy of the particle just after launch is mV02 / 2 . Its potential energy is zero.
3. At the peak of the trajectory the vertical velocity is zero. Since the horizontal velocity remains
constant, the velocity vector at the peak of the trajectory is v  V0 cos  i . The kinetic energy at
this point is therefore mV02 cos 2  / 2
4. Energy is conserved, so
mV02 / 2  mgh  mV02 cos 2  / 2
 h  V02 (1  cos 2  ) / 2  V02 sin 2 
Example 2: You are asked to design the packaging for a sensitive instrument.
The packaging will be made from an elastic foam, which behaves like a spring.
The specifications restrict the maximum acceleration of the instrument to 15g.
Estimate the thickness of the packaging that you must use.
This problem can be solved by noting that (i) the max acceleration occurs when
d
the packaging (spring) is fully compressed and so exerts the maximum force on
the instrument; (ii) The velocity of the instrument must be zero at this instant,
(because the height is a minimum, and the velocity is the derivative of the height); and (iii) The system is
conservative, and has zero kinetic energy when the package is dropped, and zero kinetic energy when the
spring is fully compressed.
Assumptions:
1. The package is dropped from a height of 1.5m
2. The effects of air resistance during the fall are neglected
3. The foam is idealized as a linear spring, which can be fully
compressed.
d
Calculations: Let h denote the drop height; let d denote the foam thickness.
1. The potential energy of the system just before the package is dropped
is mgh
2. The potential energy of the system at the instant when the foam is
1
compressed to its maximum extent is kd 2
2
3. The total energy of the system is constant, so
1 2
kd  mgh
2
4. The figure shows a free body diagram for the instrument at the instant
of maximum foam compression. The resultant force acting on the instrument
is (kd  mg ) j , so its acceleration follows as a  (kd / m  g ) j . The
acceleration must not exceed 15g, so
kd / m  16 g
5. Dividing (3) by (4) shows that
d h/8
The thickness of the protective foam must therefore exceed 18.8cm.
h
mg
kd
Example 3: The Charpy Impact Test is a way to measure the work of fracture of
a material (i.e. the work per unit area required to separate a material into two
pieces). An example (from www.qualitest-inc.com/qpi.htm) is shown in the
picture. You can see one in Prince Lab if you are curious.
It consists of a pendulum, which swings down from a prescribed initial angle to
strike a specimen. The pendulum fractures the specimen, and then continues to
swing to a new, smaller angle on the other side of the vertical. The scale on the
pendulum allows the initial and final angles to be measured. The goal of this
example is to deduce a relationship between the angles and the work of fracture
of the specimen.
The figure shows the pendulum before and after it hits
the specimen.
1. The potential energy of the mass before it is
released is V1   mgl cos 1 . Its kinetic energy
is zero.
2. The potential energy of the mass when it comes
to rest after striking the specimen is
V2   mgl cos  2 . The kinetic energy is again
zero.
l
The work of fracture is equal to the change in potential energy WF  V1  V2  mgl  cos  2  cos 1 
m

m 
Example 4: Estimate the maximum distance that a long-bow can fire an arrow.
We can do this calculation by idealizing the bow as a spring, and estimating the maximum force that a
person could apply to draw the bow. The energy stored in the bow can then be estimated, and energy
conservation can be used to estimate the resulting velocity of the arrow.
Assumptions
1. The long-bow will be idealized as a linear spring
2. The maximum draw force is likely to be around 60lbf (270N)
3. The draw length is about 2ft (0.6m)
4. Arrows come with various masses – typical range is between 250-600 grains (16-38 grams)
5. We will neglect the mass of the bow (this is not a very realistic assumption)
Calculation: The calculation needs two steps: (i) we start by calculating the velocity of the arrow just
after it is fired. This will be done using the energy conservation law; and (ii) we then calculate the
distance traveled by the arrow using the projectile trajectory equations derived in the preceding chapter.
1. Just before the arrow is released, the spring is stretched to its maximum length, and the arrow is
1
stationary. The total energy of the system is T0  V0  kL2 , where L is the draw length and k is
2
the stiffness of the bow.
2. We can estimate values for the spring stiffness using the draw force: we have that FD  kL , so
1
k  FD / L . Thus T0  V0  LFD .
2
3. Just after the arrow is fired, the spring returns to its un-stretched length, and the arrow has
1
velocity V. The total energy of the system is T1  V1  mV 2 , where m is the mass of the arrow
2
1
1
4. The system is conservative, therefore T0  V0  T1  V1  LFD  mV 2  V  LFD / m
2
2
5. We suppose that the arrow is launched from the origin at an angle  to the horizontal. The
horizontal and vertical components of velocity are Vx  V cos  V y  V sin  . The position
vector of the arrow can be calculated using the method outlined in Section 3.2.2 – the result is
1


r  Vt cos   i   Vt sin   gt 2  j
2


We can calculate the distance traveled by noting that its position vector when it lands is di. This
gives
1
Vt cos  i   Vt sin   gt 2  j  di
2


where t is the time of flight. The i and j components of this equation can be solved for t and d,
with the result
2V sin 
2V sin 
t
d
g
g cos
The arrow travels furthest when fired at an angle that maximizes sin  / cos - i.e. 45 degrees.
The distance follows as
d
2V 2 2 LFD

g
mg
6. Substituting numbers gives 2064m for a 250 grain arrow – over a mile! Of course air resistance
will reduce this value, and in practice the kinetic energy associated with the motion of the bow
and bowstring (neglected here) will reduce the distance.
Example 5: Find a formula for the escape velocity of a space vehicle as a
function of altitude above the earths surface.
The term ‘Escape velocity’ means that the space vehicle has a large enough
velocity to completely escape the earth’s gravitational field – i.e. the space
vehicle will never stop after being launched.
Assumptions
1. The space vehicle is initially in orbit at an altitude h above the
earth’s surface
2. The earth’s radius is 6378.145km
3. While in orbit, a rocket is burned on the vehicle to increase its speed
to v (the escape velocity), placing it on a hyperbolic trajectory that
will eventually escape the earth’s gravitational field.
h
R
4. The Gravitational parameter   GM  3.986012  105 km3s 1
(G= gravitational constant; M=mass of earth)
5.
Calculation
1. Just after the rocket is burned, the potential energy of the system is V  GMm / ( R  h) , while
its kinetic energy is T  mv 2 / 2
2. When it escapes the earth’s gravitational field (at an infinite height above the earth’s surface) the
potential energy is zero. At the critical escape velocity, the velocity of the spacecraft at this point
drops to zero. The total energy at escape is therefore zero.
3. This is a conservative system, so
T  V  mv 2 / 2  GMm / ( R  h)  0
 v  2GM / ( R  h)
4. A typical low earth orbit has altitude of 250km.
10.9km/sec.
For this altitude the escape velocity is
4.1.11 Force, Torque and Power curves for actuators and motors
Concepts of power and work are particularly useful to calculate the behavior of a system that is driven by
a motor. Calculations like this are described in more detail in Section 4.12. In this section, we discuss
the relationships between force, torque, speed and power for motors.
There are many different types of motor, and each type has its own unique characteristics. They all have
the following features in common, however:
1. The motors convert some non-mechanical form of energy into mechanical work. For example, an
electric motor converts electrical energy; an internal combustion engine, your muscles, and other
molecular motors convert chemical energy, and so on. You may be interested in an extensive
discussion of muscles at http://www.unmc.edu/physiology/Mann/mann14.html
2. A linear motor or linear actuator applies a force to an object (or more accurately, it applies
roughly equal and opposite forces on two objects connected to its ends). Your muscles are good
examples of linear actuators. You can buy electrical or hydraulically powered actuators as well.
L
Force
Actuator
-F
Power
Electrical
-F
A
FB
Muscle
Muscle
B
FA
Electrical
Contraction rate
_ dL/dt
Contraction rate
_ dL/dt
3. The forces applied by an actuator depend on (i) the amount of power supplied to the actuator –
electrical, chemical, etc; and (ii) the rate of contraction, or stretching, of the actuator. The typical
variation of force with stretch rate is shown in the figure. The figure assumes that the actuator
pulls on the objects attached to its ends (like a muscle); but some actuators will push rather than
pull; and some can do both. If the actuator extends slowly and is not rotating it behaves like a
two-force member, and the forces exerted by its two ends are equal and opposite.
motor
-M
-M
B
A
MA
MB
Power
Moment
Electric motor
Internal
Combustion
engine
Rotation rate
Electrical
Internal
Combustion
engine
Rotation rate
4. A rotary motor or rotary actuator applies a moment or torque to an object (or, again, more
precisely
it
applies
equal and opposite moments to two objects). Electric motors, internal combustion engines,
bacterial motors are good examples of rotary motors.
5. The torques exerted by a rotary actuator depend on (i) the amount of power supplied to the
actuator – electrical, chemical, etc; and (ii) the relative velocity of the two objects on which the
forces act. The typical variation of force with external power applied and with relative velocity is
shown in the figure. An electric motor applies a large torque when it is stationary, and its torque
decreases with rotational speed. An internal combustion engine applies no torque when it is
stationary. It’s torque is greatest at some intermediate speed, and decreases at high speed.
The power curve describes the variation of the rate of work done by an actuator or motor with its
extension or contraction rate or rotational speed. Note that
1. The rate of work done (or power expended) by a linear actuator that applies forces FA , FB to the
objects attached to its two ends can be calculated as P  FA  v A  FB  v B , where v A and v B are
the velocities of the two ends.
2. The rate of work (or power expended) by a rotational motor that applies moments M A , M B to
the objects attached to its two ends is P  M A  ω A  M B  ω B .
Typical power curves for motors and actuators are sketched in the figures.
The efficiency of a motor or actuator is the ratio of the rate of work done by the forces or moments
exerted by the motor to the electrical or chemical power. The efficiency is always less than 1 because
some fraction of the power supplied to the motor is dissipated as heat. An electric motor has a high
efficiency; heat engines such as an internal combustion engine have a much lower efficiency, because
they operate by raising the temperature of the air inside the cylinders to increase its pressure. The heat
required to increase the temperature can never be completely converted into useful work (EN72 will
discuss the reasons for this in more detail). The efficiency of a motor always varies with its speed – there
is a special operating speed that maximizes its efficiency. For an internal combustion engine, the speed
corresponding to maximum efficiency is usually quite low – 1500rpm or so. So, to minimize fuel
consumption during your driving, you need to operate the engine at this speed for as much of your drive
as possible.
There is not enough time in this course to be able to discuss the characteristics of actuators and motors in
great detail. Instead, we focus on one specific example, which helps to illustrate the general
characteristics in more detail.
Torque Curves for a brushed electric motor. There are several different types of electric motor – here,
we will focus on the simplest and cheapest – the so called `Brushed DC electric motor.’ The term
`Brushed’ refers to the fact that the motor is driven by a coil of wire wrapped around its rotating shaft,
and `brushes’ are used to make an electrical contact between the power supply and the rotating shaft.
`DC’ means that the motor is driven by direct, rather than alternating current.
Here is a very brief summary of the basic principles of this type of motor. The underlying theory will be
discussed in more detail in EN510
An electric motor applies a moment, or torque to an object
that is coupled to its output shaft. The moment is developed
motor
by electromagnetic forces acting between permanent
magnets in the housing and the electric current flowing
MB
through the winding of the motor. In the following
discussion, we shall assume that the body of the motor is
stationary ω A  0 , and its output shaft rotates with angular
velocity ω B  n where n is a unit vector parallel to the
n
MB

shaft and  is its angular speed. In addition, we assume
that the output shaft exerts a moment M B  Tn .
-
Power to drive the motor is supplied by connecting an electrical power source (e.g. a battery) to
its terminals. The power supply generally applies a fixed voltage to the terminals, which then
causes current to flow through the winding. The current is proportional to the voltage, and
decreases in proportion to the angular (rotational) speed of the output shaft of the motor.
The moment applied by the output shaft is proportional to this electric current.
The electric current I flowing through the winding is related to the voltage V and angular speed of the
motor  (in radians per second) by
I  (V   ) / R
where R is the electrical resistance of the winding, and  is a constant that depends on the arrangement
and type of magnets used in the motor, as well as the geometry of the wire coil.
The magnitude of moment exerted by output shaft of the motor T is related to the electric current and the
speed of the motor by
  I  T0   0 I  T0 / 
T 
I  T0 / 
0
where T0 and  0 are constants that account for losses such as friction in the bearings, eddy currents, and
air resistance.
The torque-current and the current-voltage-speed relations can be combined into a single formula relating
torque to voltage and motor speed
2
  V

 
 V / R  T0

 T0   
  0 

T   R
  R

V / R  T0

0

This relationship is sketched in the figure – this is called the
torque curve for the motor. The two most important points on
Moment
the curve are
Stall torque Ts
1. The `stall torque’ Ts  (  V / R )  T0
2. The `No load’ speed nl  (  V  T0 R ) / ( 0 R   2 )
The torque curve can be expressed in
as


T  Ts  1 

nl

terms of these quantities
No load speed
nl
Rotation rate



Motor manufacturers generally provide values of the stall torque and no load speed for a motor. Some
manufacturers provide enough data so that you can usually calculate the values of R,  , T0 and  0 for
their product. For example, a very detailed set of motor specs are shown in the table below. The table is
from http://www.motortech.com/dcmotorCIR_MD.htm Each row of the table refers to a different motor.
Here
1. The `input voltage’ is the recommended voltage V to apply to the motor. All other data in the
table assume that the motor is used with this voltage. If you wish, you can run a motor with a
lower voltage (this will make it run more slowly) and if you are feeling brave you can increase
the voltage slightly – but this might cause it to burn out, invalidate your warranty and expose you
to a potential lawsuit from your customers…
2. The `No load speed’ nl is the speed of the motor when it is spinning freely with T=0
3. The `No load current’ I nl is the corresponding current
4. The `Stall torque’ Ts is the value of T when the motor is not spinning   0
5. The ‘Stall current’ I s is the corresponding current
6. The `rated torque’ and `rated current’ are limits to the steady running of the motor – the torque
and current should not exceed these values for an extended period of time.
The values in the table can be substituted into the formulas relating current, voltage, torque and motor
speed, which yield four equations for the unknown values of R,  , T0 and  0
I nl  (V  nl ) / R
Is  V / R
0   I nl  T0   0nl
Solving these equations gives
R  V / Is
T0 

V ( I s  I nl )
 Ts
nl
Ts   I s  T0   0nl
V ( I s  I nl )
I snl
0 
Ts V ( I s  I nl ) 2

2
nl
I snl
HEALTH WARNING: When you use these formulas it is critical to use a consistent set of units. SI
units are very strongly recommended! Volts and Amperes are SI units of electrical potential and electric
current. But the torques must be converted into Nm. The conversion factor is 1 ‘oz-in’ is 0.0070612 Nm.
Not all motor manufacturers provide such detailed specifications – it is more common to be given (i) the
stall torque; (ii) the no load speed; and (iii) the stall current. If this is the case you have to assume I nl  0
and take  0  0 . If you aren’t given the stall current you have to take T0  0 as well.
Power Curves for a brushed electric motor driven from a
Power
constant voltage power supply: The rate of work done by the
Max Power Ts nl /2
motor is P  M B  ω B  T  . The power can be calculated in
terms of the angular speed of the motor as
No load speed nl

 
P  Ts  1 

 nl 
The power curve is sketched in the figure. The most
Rotation rate
important features are (i) The motor develops no power at
both zero speed and the no load speed; and (ii) the motor develops its maximum power Pmax  Tsnl / 2
at speed nl / 2 .
Efficiency of a brushed electric motor: The electrical power supplied to a motor can be calculated from
the current I and voltage V as Pin  VI . The efficiency can therefore be calculated as
Ts (1   / nl )  T0   0  R


VI
V
(V   )
The second expression can be used to calculate the speed that maximizes efficiency.

4.1.12 Power transmission in machines
A machine is a system that converts one form of motion into another. A lever is a simple example – if
point A on the lever is made to move, then point B will move either faster or slower than A, depending on
the lengths of the two lever arms. Other, more practical examples include
1. Your skeleton; used to help convert muscle motion into
j
FB
various more useful forms…
2. The transmission of a vehicle – used to convert rotational F
A
B
motion induced by the engine’s crank-shaft to translational
motion of the vehicle

3. A gearbox – converts rotational motion at one angular speed to
i
rotational motion at another.
A
4. A system of pulleys.
Concepts of work and energy are extremely useful to analyze transmission of forces and moments
through a machine. The goal of these calculations is to quickly determine relationships between external
forces acting on the machine, rather than to analyze the motion of the system itself.
To do this, we usually make two assumptions:
1. The machine is a conservative system. This is true as long as friction in the machine can be
neglected.
2. The kinetic and potential energy of the mechanism within the machine can be neglected. This is
true if either (i) the machine has negligible mass and is very stiff (i.e. it does not deform); or (ii)
the machine moves at steady speed or is stationary.
If this is the case, the total rate of work done by external forces acting on the machine is zero. This
principle can be used to relate external forces acting on the machine, without having to work through the
very lengthy and tedious process of computing all the internal forces.
The procedure to do this is always:
1. Derive a relationship between the velocities of the points where external forces act – this is
always a geometric relationship.
2. Write down the total rate of external work done on the system
3. Use (1) and (2) to relate the external forces to each other.
The procedure is illustrated using a number of examples below.
Example 1: As a first example, we derive a formula relating the forces
acting on the two ends of a lever. Assume that the forces act
perpendicular to the lever as shown in the figure.
j
FB
FA
B
Calculation

1. Suppose that the lever rotates about the pivot at some angular
rate d / dt   .
i
2. The
ends
of
the
lever
have
velocities A
v A   d A (sin  i  cos j)
v B   d B (  sin  i  cos  j)
(see
the circular motion example in the preceding chapter if you can’t remember how to show this.
3. The force vectors can be expressed as FA  FA (sin  i  cos  j) FB  FB (sin  i  cos  j)
4. The reaction forces at the pivot are stationary and so do no work.
5. The total rate of work done on the lever is therefore
FA  v A  FB  v B
 FA (sin  i  cos  j)   d A (sin  i  cos  j)  FB (sin  i  cos  j)   d B (  sin  i  cos  j)

  FA d A  FB d B   sin 2   cos 2 

6. The rate of external work is zero, so FA d A  FB d B
Example 2: Here we revisit an EN3 statics problem. The rabbit uses a
pulley system to raise carrot with weight W. Calculate the force applied by
the rabbit on the cable.
1. The velocity of the end of the cable held by the rabbit is
dl
, with
dt
direction parallel to the cable
dh
2. The velocity of the carrot is
; its direction is vertical
dt
3. The total length of the cable (aside from small, constant lengths going around the pulleys) is
l  2h . Since the length is constant, it follows that
dl
dh
2 0
dt
dt
4. The external forces acting on the pulley system are (i) the weight of the carrot W, acting
vertically; (ii) the unknown force F applied by the rabbit, acting parallel to the cable; and (iii) the
dl
reaction force acting on the topmost pulley. The rate of work done by the rabbit is F , the rate
dt
dh
. The reaction force is
dt
stationary, and so does no work. The total rate of work done is therefore
dh
dl
W
F 0
dt
dt
5. Using (3) and (4) we see that
dh
dl
dh
dh
W
 F W
 2F
0
dt
dt
dt
dt
 F W / 2
of work done by the gravitational force acting on the carrot is W
Example 3 Calculate the torque that must be applied to a screw with pitch d in
order to raise a weight W.
Here, we regard the screw as a machine subjected to external forces.
1. When the screw turns through one complete turn ( 2 radians) it raises
the weight by a distance d
2. The work done by the torque is 2 T ; the work done by the weight is –
Wd. The reaction forces acting on the thread of the screw do no work
3. The total work is zero, therefore 2 T  Wd  0  T  W / 2 .
W
d
T
Example 4: A gearbox is used for two purposes: (i) it can amplify or attenuate torques, or moments; and
(ii) it can change the speed of a rotating shaft. The
j
Input
i
characteristics of a gearbox are specified by its gear ratio
i
shaft k
specifies the ratio of the rotational speed of the output
Output
shaft to that of the input shaft r  o / i .
shaft
Mi

0
The sketch shows an example. A power source (a
rotational motor) drives the input shaft; while the output
M0
shaft is used to drive a load. The rotational motion of the
input and output shafts can be described by angular
velocity vectors ωi  i i ωo  o i (note that the shafts don’t necessarily have to rotate in the same
direction – if they don’t then one of o or i will be negative).
A quick power calculation can be used to relate the moments M i  Ti i M o  To i acting on the input
and output shafts.
1. The gear ratio relates the input and output angular velocities as ωi  i i ωo  ri i
2. The rate of work done on the gearbox by the moment acting on the input shaft is Ti i  i i  Tii .
The work done by the moment acting on the output shaft is To i  o i  Too  To ri . A reaction
moment must act on the base of the gearbox, but since the gearbox is stationary this reaction
moment does no work.
3. The rate of work done by all the external forces on the gearbox is zero, so
Tii  To ri  0  T0  Ti / r
Example 5: Find a formula for the engine power required for a
vehicle with mass m to climb a slope with angle  at speed v.
Estimate values for engine power required for various vehicles to

climb a slopes of 3%, 4% and 5% (these are typical standard grades
for flat, rolling and mountain terrain freeways, see, e.g. the Roadway
Engineering Highway
Design
Manual for
Oregon
at
http://www.oregon.gov/ODOT/HWY/ENGSRVICES/hwy_manual.shtml#2003_English_Manual
Also, estimate (a) the torque exerted by the engine crankshaft on the transmission; and (b) the force acting
on the pistons of the engine.
Assumptions:
1. The only forces acting on the vehicle are aerodynamic drag, gravity, and reaction forces at the
driving wheels.
2. Aerodynamic drag will be assumed to be in the high Reynolds number regime, with a drag
coefficient of order
3. The crankshaft rotates at approximately 2000rpm.
4. As a representative examples of engines, we will consider (i) a large 6 cylinder 4 stroke diesel
engine, with stroke (distance moved by the pistons) of 150mm; and (ii) a small 4 cylinder
gasoline powered engine with stroke 94mm
(see http://www.automotive.com/2009/12/chevrolet/cobalt/specifications/index.html)
Calculation. The figure shows a free body diagram for the
truck.
It shows (i) gravity; (ii) Aerodynamic drag; (iii)
reaction forces acting on the wheels.
1. The velocity vector of the vehicle is v(cos  i  sin  j)
2. The rate of work done by the gravitational and
aerodynamic forces is
P   FD (cos  i  sin  j)  v(cos  i  sin  j)  mgj  v(cos  i  sin  j)
j
FD
i
NA NB mg
T
NC
ND
  FD v  mgv sin 
1
where FD  C D A v 2 , with CD the drag coefficient, A the frontal area of the vehicle, and 
2
the air density.
3. The reaction forces acting on the wheels do no work, because they act on stationary points (the
wheel is not moving where it touches the ground).
4. The kinetic energy of the vehicle is constant, so the total rate of work done on the vehicle must be
zero. The moving parts of the engine (specifically, the pressure acting on the pistons) do work on
the vehicle. Denoting the power of the engine by PE , we have that
PE  FD v  mgv sin   0
 PE  FD v  mgv sin 
5. The torque exerted by the crankshaft can be estimated by noting that the rate of work done by the
engine is related to the torque T and the rotational speed  of the crankshaft by PE  T  . The
torque follows as T   FD v  mgv sin   / 
6. The force F on the pistons can be estimated by noting that the rate of work done by a piston is
related to the speed of the piston v by P  Fv . In a four-stroke engine, each piston does work
25% of the time (during the firing stroke). If the engine has n cylinders, the total engine power is
therefore PE  nFv / 4 . Finally, the piston speed can be estimated by noting that the piston travels
the distance of the stroke during half a revolution, and so has average speed v  d  /  . This
shows that PE  nFd  / 4  F  4 PE / (nd  ) .
The table below calculates values for two examples of vehicles
Vehicle
Grade
Weight
(kg)
Frontal
area
Drag
Coeft
Speed
m/s
m2
18
wheeler
Chevy
cobalt
3%
4%
5%
3%
4%
5%
Engine Stroke Engine
speed
m
power
Rad/sec
Watts
36000
10
0.1
29
12600
0.15
1250
2.5
.1
29
12600
0.095
320000
420000
520000
14000
17000
21000
Torque
Nm
Piston
force
N
25
34
42
1.1
1.4
1.7
355
470
580
36
46
55
4.2 Linear impulse-momentum relations
4.2.1 Definition of the linear impulse of a force
k
In most dynamic problems, particles are subjected to forces that vary with
time. We can write this mathematically by saying that the force is a vector
valued function of time F (t ) . If we express the force as components in a
fixed basis i, j, k , then
F (t )  Fx (t )i  Fy (t ) j  Fz (t )k
i
F(t)
O
P
j
where each component of the force is a function of time.
The Linear Impulse exerted by a force during a time interval t0  t  t0  t is defined as

t0 t

F(t )dt
t0
The linear impulse is a vector, and can be expressed as components in a basis
   x i   y j  z k
x 
t0 t

t0
Fx (t )dt
y 
t0 t

Fy (t )dt
z 
t 0 t
t0

Fz (t )dt
t0
If you know the force as a function of time, you can calculate its impulse using simple calculus. For
example:
1. For a constant force, with vector value F0 , the impulse is   F0 t
2. For a harmonic force of the form F (t )  F0 sin t the impulse is
  F0  cos( (t0  t ))  cos t0  / 
It is rather rare in practice to have to calculate the impulse of a force from its time variation.
4.2.2 Definition of the linear momentum of a particle
The linear momentum of a particle is simply the product of its mass and velocity
p  mv
The linear momentum is a vector – its direction is parallel to the velocity of the particle.
4.2.3 Impulse-momentum relations for a single particle
Consider a particle that is subjected to a force F (t ) for a time interval t0  t  t0  t .
Let  denote the impulse exerted by F on the particle
Let p  mv (t0  t )  mv (t0 ) denote the change in linear momentum during the time interval
t .
The momentum conservation equation can be expressed either in differential or integral form.
1. In differential form
F (t ) 
dp
dt
k
This is clearly just a different way of writing Newton’s law
F=ma.
2. In integral form
P
r
O
x
  p
This is the integral of Newton’s law of motion with respect
to time.
i
j
z
y
The impulse-momentum relations for a single particle are useful if you need to calculate the change in
velocity of an object that is subjected to a prescribed force.
4.2.4 Examples using impulse-momentum relations for a single particle
Example 1: Estimate the time that it takes for a Ferrari Testarossa
traveling at the RI speed limit to make an emergency stop. (Like many
textbook problems this one is totally unrealistic – nobody in a Ferrari is
going to travel at the speed limit!)
We can do this calculation using the impulse-momentum relation for a
single particle. Assume that the car has mass m, and travels with speed
V before the brakes are applied. Let t denote the time required to stop.
1. Start by estimating the force acting on the car during an emergency stop. The figure shows a free
body diagram for the car as it brakes to a standstill.
j
We assume that the driver brakes hard enough to lock
the wheels, so that the car skids over the road. The
i
horizontal friction forces must oppose the sliding, as
shown in the picture. F=ma for the car follows as
TF
TR
TR  TF  i  ( N F  N R  mg ) j  max i
NF
NR
The vertical component of the equation of motion shows that N F  N R  mg . The friction law
then shows that TF  TR    N F  N R    mg
2. The force acting on the car is constant, so the impulse that brings the car to a halt is
  TR  TF  ti  ( N F  N R  mg )tj   mgi
3. The linear momentum of the car before the brakes are applied is p 0   mVi .
The linear
momentum after the car stops is zero. Therefore, p  mVi .
4. The linear impulse-momentum relation shows that
  p   mg ti  mVi
 t  V / (  g )
5. We can take the friction coefficient as   0.8 , and 65mph is 29m/s. We take the gravitational
acceleration g  9.81 . The time follows as t  3.7s . Note that a testarossa can’t stop any
faster than a Honda Civic, despite the price difference…
Example 2: The figure shows a pendulum that swings with frequency f swings per
second. Find an expression for the average reaction force at the pivot of a pendulum
during one cycle of oscillation. (This seems a totally academic problem – and indeed it
is – but the important point is that certain quantities, such as average forces, can often
be computed without solving the full equations of motion. These quantities can be
extremely useful for checking computer simulations)
Calculation: Consider the pendulum at the extreme limits of its swing, at the start and
end of one cycle of oscillation. Note that
1. The velocity of the particle is zero both at the start and end of the cycle. Its
change in momentum is therefore zero, and the total impulse exerted on the
particle must be zero.
2. The particle is subjected to two forces (i) gravity; and (ii) the reaction force
exerted on the particle. During a swing of the pendulum these forces exert
impulses g , R , respectively.
R
mg
3. The impulse exerted by the gravitational force is g   mgTj where
T  1/ f is the time for one complete swing of the pendulum.
4. Since the total impulse is zero, it follows that g  R  0  R  mgj / f
The MATLAB script below shows how this result can be used to check the program developed in Section
???? to model the swing of a pendulum.
function pendulum_reactions
g = 9.81;
l = 1;
m = 1;
theta = pi*45/180;
time = 10;
Abstol = 10^(-03)*ones(4,1);
options = odeset('RelTol',0.000001,'AbsTol',Abstol,'Events',@event);
y0 = [l*sin(theta),l*cos(theta),0,0]; % Initial conditions
[t,y,tevent,yevent,ievent] = ode45(@eom,[0,time],y0,options);%,options);
plot(t,y(:,1:2));
% Calculate and plot the reaction forces as a function of time
for i=1:length(t)
[Reaction(i,1),Reaction(i,2)] = reactions(y(i,:));
end
figure
plot(t,Reaction(:,1:2));
% Integrate the reaction forces with respect to time
Ix = trapz(t,Reaction(:,1))
Iy = trapz(t,Reaction(:,2))
AverageRy = trapz(t,Reaction(:,2))/time(length(t))
function dydt = eom(t,y)
% The vector y contains [x,y,vx,vy]
M = eye(5); % This sets up a 5x5 matrix with 1s on all the diags
M(3,5) = (y(1)/m)/sqrt(y(1)^2+y(2)^2);
M(4,5) = (y(2)/m)/sqrt(y(1)^2+y(2)^2);
M(5,3) = y(1);
M(5,4) = y(2);
M(5,5) = 0;
f = [y(3);y(4);0;g;-y(3)^2-y(4)^2];
sol = M\f; % This solves the matrix equation M*[dydt,R]=f for [dydt,R]
dydt = sol(1:4); % only need to return time derivatives dy/dt
end
function [Rx,Ry] = reactions(y)
% This function calculates the reaction forces Rx, Ry, given the vector
% y containing [x,y,vx,vy]
M = eye(5); % This sets up a 5x5 matrix with 1s on all the diags
M(3,5) = (y(1)/m)/sqrt(y(1)^2+y(2)^2);
M(4,5) = (y(2)/m)/sqrt(y(1)^2+y(2)^2);
M(5,3) = y(1);
M(5,4) = y(2);
M(5,5) = 0;
f = [y(3);y(4);0;g;-y(3)^2-y(4)^2];
sol = M\f; % This solves the matrix equation M*[dydt,R]=f for [dydt,R]
Rx = -sol(5)*y(1)/sqrt(y(1)^2+y(2)^2);
Ry = -sol(5)*y(2)/sqrt(y(1)^2+y(2)^2);
end
function [eventvalue,stopthecalc,eventdirection] = event(t,y)
% Function to stop the calculation after one swing
% Note that v_x=0 at the end of the swing and crosses zero from above
eventvalue = y(3);
stopthecalc = 1; % This stops the calculation
eventdirection = -1;
end
end
4.2.5 Impulse-momentum relation for a system of particles
Suppose we are interested in calculating the motion of several
particles, sketched in the figure.
Total external impulse on a system of particles: Each
particle in the system can experience forces applied by:
Other particles in the system (e.g. due to gravity,
electric charges on the particles, or because the
particles are physically connected through springs, or
because the particles collide). We call these internal
forces acting in the system. We will denote the
internal force exerted by the ith particle on the jth
particle by R ij . Note that, because every action has
F1ext
m4
F3ext
m1
R13 R31
R12
m
R32 3
R23
R21
m2
F2ext
an equal and opposite reaction, the force exerted on the jth particle by the ith particle must be
equal and opposite, to R ij , i.e. R ij  R ji .
Forces exerted on the particles by the outside world (e.g. by externally applied gravitational or
electromagnetic fields, or because the particles are connected to the outside world through
mechanical linkages or springs). We call these external forces acting on the system, and we will
denote the external force on the i th particle by Fiext (t )
We define the total impulse exerted on the system during a time interval t0  t  t0  t as the sum of all
the impulses on all the particles. It’s easy to see that the total impulse due to the internal forces is zero –
because the ith and jth particles must exert equal and opposite impulses on one another, and when you
add them up they cancel out. So the total impulse on the system is simply
tot 
t0 t
 
particles
Fiext (t )dt
t0
Total linear momentum of a system of particles: The total linear momentum of a system of particles is
simply the sum of the momenta of all the particles, i.e.
ptot 

mi vi
particles
The impulse-momentum equation can be expressed either in differential or integral form, just as for a
single particle:
1. In differential form

Fiext (t ) 
particles
dptot
dt
This is clearly just a different way of writing Newton’s law F=ma.
2. In integral form
tot  ptot
This is the integral of Newton’s law of motion with respect to time.
Conservation of momentum: If no external forces act on a system of particles, their total linear
momentum is conserved, i.e.  p  0
ptot (t0  t )  ptot (t0 )
4.2.6 Examples of applications of momentum conservation for systems of particles
The impulse-momentum equations for systems of particles are particularly useful for (i) analyzing the
recoil of a gun; and (ii) analyzing rocket and jet propulsion systems. In both these applications, the
internal forces acting between the gun on the projectile, or the motor and propellant, are much larger than
any external forces, so the total momentum of the system is conserved.
j
Example 1: Estimate the recoil velocity of a rifle
(youtube abounds with recoil demonstrations – see. e.g.
http://www.youtube.com/watch?v=F4juEIK_zRM for
samples. Be warned, however – a lot of the videos are
tasteless and/or sexist… )
The recoil velocity can be estimated by noting that the
V
mg
i
v
mb
total momentum of bullet and rifle together must be conserved. If we can estimate the mass of rifle and
bullet, and the bullet’s velocity, the recoil velocity can be computed from the momentum conservation
equation.
Assumptions:
1. The mass of a typical 0.22 (i.e. 0.22” diameter) caliber rifle bullet is about 7  10  4 kg (idealizing
the bullet as a sphere, with density 7860 kg / m 3 )
2. The muzzle velocity of a 0.22 is about 1000 ft/sec (305 m/s)
3. A typical rifle weighs between 5 and 10 lb (2.5-5 kg)
Calculation:
1. Let mb denote the bullet mass, and let m R denote the mass of the rifle.
2. The rifle and bullet are idealized as two particles. Before firing, both are at rest. After firing, the
bullet has velocity v b  v i ; the rifle has velocity v R  V i .
3. External forces acting on the system can be neglected, so
ptot (t0  t )  ptot (t0 )  mb vb  mR v R  0
 mb vi  mRVi  0  V   mb v / mR
.
4. Substituting numbers gives |V| between 0.04 and 0.08 m/s (about 0.14 ft/sec)
Example 2: Derive a formula that can be used to estimate the mass of a handgun required to keep its
recoil within acceptable limits.
The preceding example shows that the firearm will recoil with a velocity that depends on the ratio of the
mass of the bullet to the firearm. The firearm must be brought to rest by the person holding it.
Assumptions:
1. We will idealize a person’s hand holding the gun as a spring, with stiffness k, fixed at one end.
The ‘end-point stiffness’ of a human hand has been extensively studied – see, e.g. Shadmehr et al
J. Neuroscience, 13 (1) 45 (1993). Typical values of stiffness during quasi-static deflections are
of order 0.2 N/mm. During dynamic loading stiffnesses are likely to be larger than this.
2. We idealize the handgun and bullet as particles, with mass mb and mg , respectively.
Calculation.
1. The preceding problem shows that the firearm will recoil with velocity V   mb v / mg
2. Energy conservation can be used to calculate the recoil distance. We consider the firearm and the
hand holding it a system. At time t=0 it has zero potential energy; and has kinetic energy
1
1
2
mgV 2   mb v  / mg . At the end of the recoil, the gun is at rest, and the spring is fully
2
2
1
compressed – the kinetic energy is zero, and potential energy is kd 2 . Energy conservation
2
2
2
gives kd   mb v  / mg
T
3. The required mass follows as mg   mb v  / kd 2 .
2
4. Substituting numbers gives
Example 4 Rocket propulsion equations. Rocket motors and jet
engines exploit the momentum conservation law in order to produce
motion. They work by expelling mass from a vehicle at very high
velocity, in a direction opposite to the motion of the vehicle. The
momentum of the expelled mass must be balanced by an equal and
opposite change in the momentum of the vehicle; so the velocity of the
vehicle increases.
Rocket, mass M
Motor, mass m0
Analyzing a rocket engine is quite complicated, because the propellant carried by the engine is usually a
very significant fraction of the total mass of the vehicle. Consequently, it is important to account for the
fact that the mass decreases as the propellant is used.
Assumptions:
1. The figure shows a rocket motor attached to a rocket with mass M.
2. The rocket motor contains an initial mass m0 of propellant and expels propellant at rate
dm
  0 (kg/sec) with a velocity v 0   v0 i relative to the rocket.
dt
3. We assume straight line motion, and assume that no external forces act on the rocket or motor.
Calculations:
The figure shows the rocket at an instant of time t, and then a very short
time interval dt later.
1. At time t, the rocket moves at speed v, and the system has
momentum p  ( M  m)vi , where m is the motor’s mass.
2. During the time interval dt a mass dm   dt is expelled from
the motor. The velocity of the expelled mass is v  (v  v0 )i .
3. At time t+dt the mass of the motor has decreased to
m   dt .
4. At time t+dt, the rocket has velocity v  (v  dv)i .
5. The total momentum of the system at time t+dt is
Rocket, mass M
i
v
Motor, mass m
v-v0
Rocket, mass M
 dt
Motor, mass m-  dt
v+dv
therefore
p  ( M  m  0 dt )(v  dv)i  0 dt (v  v0 )i
6. Momentum must be conserved, so
( M  m)vi  ( M  m  0 dt )(v  dv)i  0 dt (v  v0 )i
7. Multiplying this out and simplifying shows that
( M  m)dvi   dtv0i  0
where the term 0 dtdv has been neglected.
8. Finally, we see that
dv
( M  m)   v0
dt
This result is called the `rocket equation.’
Specific Impulse of a rocket motor: The performance of a rocket engine is usually specified by its
`specific impulse.’ Confusingly, two different definitions of specific impulse are commonly used:
I sp  v0
v
I sp  0
g
The first definition quantifies the impulse exerted by the motor per unit mass of propellant; the second is
the impulse per unit weight of propellant. You can usually tell which definition is being used from the
units – the first definition has units of m/s; the second has units of s. In terms of the specific impulse, the
rocket equation is
dv
( M  m)   I sp   gI sp
dt
Integrated form of the rocket equation: If the motor expels propellant at constant rate, the equation of
motion can be integrated. Assume that
1. The rocket is at the origin at time t=0;
2. The rocket has speed v0 at time t=0
3. The motor has mass m0 at time t=0; this means that at time t it has mass m0   t
Then the rocket’s speed can be calculated as a function of time:
( M  m0   t )
dv
  I sp
dt
v

 I sp dt
t

dv  
v0
0
( M  m0  t )
 M  m0 
 v  v0  I sp log 

 M  m0   t 
Similarly, the position follows as
v
x
t


 M  m0 
 M  m0 
dx
 I sp log 

v

dx

 I sp log 
 0
  v0 dt


dt
 M  m0   t 
 M  m0   t 


0
 x  (v0  I sp )t   M  m0   t 
0
 M  m0 
log 


 M  m0   t 
I sp
These calculations assume that no external forces act on the rocket. It is quite straightforward to
generalize them to account for external forces as well – the details are left as an exercise.
Example 5 Application of rocket propulsion equation: Calculate the maximum payload that can be
launched to escape velocity on the Ares I launch vehicle.
‘Escape velocity’ means that after the motor burns out, the space vehicle can escape the earth’s
gravitational field – see example 5 in Section 4.1.6.
Assumptions
1. The specifications for the Ares I are at http://www.braeunig.us/space/specs/ares.htm Relevant
variables are listed in the table below.
Stage I
Stage II
Total mass (kg)
586344
183952
Specific impulse (s)
268.8
452.1
Propellant mass (kg)
504516
163530
2. As an approximation, we will neglect the motion of the rocket during the burn, and will neglect
aerodynamic forces.
3. We will assume that the first stage is jettisoned before burning the second stage.
4. Note that the change in velocity due to burning a stage can be expressed as
 M0 
v  v0  I sp g log 

 M 0  m 
where M 0 is the total mass before the burn, and m is the mass of propellant burned.
5. The earth’s radius is 6378.145km
6. The Gravitational parameter   GM  3.986012  105 km3s 1 (G= gravitational constant;
M=mass of earth)
7. Escape velocity is from the earths surface is ve  2GM / R , where R is the earth’s radius.
Calculation:
1. Let m denote the payload mass; let m1, m2 denote the total masses of stages I and II, let m01, m02
denote the propellant masses of stages I and II; and let I sp1 , I sp 2 denote the specific impulses of
the two stages.
2. The rocket is at rest before burning the first stage; and its total mass is m  m1  m2  m01  m02 .
After burn, the mass is m  m1  m2  m02 . The velocity after burning the first stage is therefore
 m  m1  m2  m01  m02 
v1  I sp1g log 

 m  m1  m2  m02

3. The first stage is then jettisoned – the mass before starting the second burn is m  m2  m02 , and
after the second burn it is m  m2 . The velocity after the second burn is therefore
 m  m1  m2  m01  m02 
 m  m2  m02 
v2  I sp1g log 
  I sp 2 g log 

 m  m2

 m  m1  m2  m02

4. Substituting numbers into the escape velocity formula gives ve  11.18 km/sec. Substituting
numbers for the masses shows that to reach this velocity, the payload mass must satisfy
 m  770296 
 m  183952 
11.18  2.621log 
  4.435log 

m

265780


 m  20422 
where m is in kg.
5. We can use MAPLE to solve for the critical value of m for equality
so the solution is 8300kg – a very small mass compared with that of the launch vehicle, but you could
pack in a large number of people you would like to launch into outer space nonetheless (the entire faculty
of the division of engineering, if you wish).
4.2.7 Analyzing collisions between particles: the restitution coefficient
The momentum conservation equations are particularly helpful if you want to analyze collisions between
two or more objects. If the impact occurs over a very short time, the impulse exerted by the contact
forces acting at the point of collision is huge compared with the impulse exerted by any other forces. If
we consider the two colliding particles as a system, the external impulse exerted on the system can be
taken to be zero, and so the total momentum of the system is conserved.
The momentum conservation equation can be used to relate the velocities of the particles before collision
to those after collision. These relations are not enough to be able to determine the velocities completely,
however – to do this, we also need to be able to quantify the energy lost (or more accurately, dissipated as
heat) during the collision.
In practice we don’t directly specify the energy loss during a collision – instead, the relative velocities are
related by a property of the impact called the coefficient of restitution.
B0
A0
Restitution coefficient for straight line motion
vx
Suppose that two colliding particles A and B move in a straight line parallel
to a unit vector i. Let
v A0  vxA0 i, v B 0  vxB 0i denote the velocities of A and B just before the
collision
v A1  vxA1i, v B1  vxB1i denote the velocities of A and B just after the
collision.
A
vx
B
*
A1
vx
A
The velocities before and after impact are related by
B1
vx
B
v B1  v A1  e  v B 0  v A0 
where e is the restitution coefficient. The negative sign is needed because the particles approach one
another before impact, and separate afterwards.
Restitution coefficient for 3D frictionless collisions
A
For a more general contact, we define
v A0 v B 0 to denote velocities of the particles before collision
v
A1
v
B1
v
to denote velocities of the particles after collision
A0
In addition, we let n be a unit vector normal to the common tangent plane
at the point of contact (if the two colliding particles are spheres or disks
the vector is parallel to the line joining their centers).
B
B0
v
n
A
v
A1
B
B1
v
The velocities before and after impact are related by two vector equations:
 v B1  v A1   n  e  v B0  v A0   n
 v B1  v A1    v B1  v A1   n  n   v B0  v A0    v B0  v A0   n  n
To interpret these equations, note that
1. The first equation states that the component of relative velocity normal to the contact plane is
reduced by a factor e (just as for 1D contacts)
2. The second equation states that the component of relative velocity tangent to the contact plane is
unchanged (this is because the contact is frictionless, and so the reaction forces acting on the
particles during the collision exert no impulse is exerted in a direction tangent to the contact
plane).
The two equations can be combined into a single vector equation relating velocities before and after
impact – this form is more compact, and often more useful, but more difficult to visualize physically
v
B1
 v A1    v B 0  v A0   (1  e)  v B 0  v A0   n  n
It is possible to account for friction at the contact, but friction will always make the colliding particles
rotate, so they cannot be idealized as particles.
Values of restitution coefficient
The restitution coefficient almost always lies in the range 0  e  1 . It can only be less than zero if one
object can penetrate and pass through the other (e.g. a bullet); and can only be greater than 1 if the
collision generates energy somehow (e.g. releasing a preloaded spring, or causing an explosion).
If e=0, the two colliding objects stick together; if e=1 the collision is perfectly elastic, with no energy
loss.
The restitution coefficient is strongly sensitive to the material properties of the two colliding objects, and
also varies weakly with their geometry and the velocity of impact. The two latter effects are usually
ignored.
Collision between a particle and a fixed rigid surface. The collision formulas
can be applied to impact between a rigid fixed surface by taking the surface to
be particle B, and noting that the velocity of particle B is then zero both before
and after impact.
A0
vx
A
*
For straight line motion, v A1  ev A 0
For collision with an angled wall v A1  v A0  (1  e)  v A 0  n  n , where
n is a unit vector perpendicular to the wall.
A1
vx
A
n
A
v
v
A1
A
A0
4.2.8 Examples of collision problems
Example 1 Suppose that a moving car hits a stationary (parked) vehicle from behind. Derive a formula
that will enable an accident investigator to determine the velocity of the moving car from the length of the
skid marks left on the road.
Assumptions:
1. We will assume both cars move in a straight line
2. The moving and stationary cars will be assumed to have masses m1, m2 , respectively
3. We will assume the cars stick together after the collision (i.e. the restitution coefficient is zero)
4. We will assume that only the parked car has brakes applied after the collision
This calculation takes two steps: first, we will use work-energy to relate the distance slid by the cars after
impact to their velocity just after the impact occurs. Then, we will use momentum and the restitution
formula to work out the velocity of the moving car just before impact.
Calculation: Let V denote the velocity of the moving car just before impact; let v1 denote the velocity of
the two (connected) cars just after impact, and let d denote the distance slid.
1. The figure shows a free body diagram for each of the two cars during sliding after the collision.
Newton’s law of motion for each car shows that
 Ri   N1  N 2  m1g  j  m1a x i
 R  T3  T4  i   N3  N 4  m2 g  j  m2a x i
2. The vertical component of the equations of motion give N1  N 2  m1 g ; N3  N 4  m2 g .
3. The parked car has locked wheels and skids over the road; the friction law gives the tangential
forces at the contacts as T3  T4   ( N3  N 4 )   m2 g
4. We can calculate the velocity of the cars just after impact using the work-kinetic energy relation
during skidding. For this purpose, we consider the two connected cars as a single particle. The
work done on the particle by the friction forces is  T3  T4  d    m2 gd . The work done is
equal to the change in kinetic energy of the particle, so
1
  m2 gd  0  (m1  m2 )v12
2
2 m2 gd
v1 
m1  m2
5. Finally, we can use momentum conservation to calculate the velocity just before impact. The
momentum after impact is h1  (m1  m2 )v1i , while before impact h0  m1Vi . Equating the two
gives
V
(m1  m2 )
2  (m1  m2 )m2 gd
v1 
m1
m12
Example 2: Two frictionless spheres with radius R have initial velocity v A0 v B 0 . At some instant of
time, the two particles collide. At the point of collision, the centers of the spheres have position vectors
y A y B . The restitution coefficient for the contact is denoted by e. Find a formula for the velocities of the
spheres after impact. Hence, deduce an expression for the change in kinetic energy during the impact.
This is a straightforward vector algebra exercise. We have two unknown
velocity vectors: v A1 v B1 , and two vector equations – momentum
conservation, and the restitution coefficient formula.
A
v
Calculation
1. Note that a unit vector normal to the tangent plane can be
calculated from the position vectors of the centers at the impact


as n  y A  y B / y A  y B .
B
mB v  mA v  mB v
n
A1
B0
 mA v
A0
A
v
A1
3. The restitution coefficient formula gives
v
B1
B0
v
It doesn’t matter whether you
choose to take n to point from A to B or the other way around –
the formula will work either way.
2. Momentum conservation requires that
B1
A0
B
B1
v
 v A1    v B 0  v A0   (1  e)  v B 0  v A0   n  n
4. We can solve (2) and (3) for v B1 by multiplying (3) by m A and adding the equations, which
gives
 mB  mA  v B1  mB
 v B1  v B 0 
v B 0  mA v A0  mA  v B 0  v A0   (1  e)  v B 0  v A0   n  n 


mA
(1  e)  v B 0  v A0   n  n
mB  mA
5. Similarly, we can solve for v A1 by multiplying (3) by mB and subtracting (3) from (2), with the
result
v A1  v A0 
mB
(1  e)  v B 0  v A0   n  n
mB  mA
6. The change in kinetic energy during the collision can be calculated as
mA A1 A1 mB B1 B1  mA A0 A0 mB B 0 B 0 
v v 
v v 
v v 
v v 
2
2
2
 2

B1
A1
7. Substituting the results of (4) and (5) for v and v and simplifying the result gives
T 
T 

  v B 0  v A0  n  2

 
2  m A  mB  
m A mB e2  1
Note that the energy change is zero if e=1 (perfectly elastic collisions) and is always negative for e<1
(i.e. the kinetic energy after collision is less than that before collision).
Example 3: This is just a boring example to help illustrate the practical application of
the vector formulas in the preceding example. In the figure shown, disk A has a
vertical velocity V at time t=0, while disk B is stationary. The two disks both have
radius R, have the same mass, and the restitution coefficient between them is e.
Gravity can be neglected. Calculate the velocity vector of each disk after collision.
Calculation
1. Before impact, the velocity vectors are v A0  Vj
vB0  0
2. A unit vector parallel to the line joining the two centers is n  (3i  7 j) / 4
(to see this, apply Pythagoras theorem to the triangle shown in the figure).
3. The velocities after impact are
v B1  v B 0 
mA
(1  e)  v B 0  v A0   n  n
mB  mA
j
B
i
V
A
2R
R/2
3R/2
mB
(1  e)  v B 0  v A0   n  n
mB  mA
Substituting the vectors gives
1
1 e
v B1   (1  e)  Vj  (3i  7 j) / 4  (3i  7 j) / 4 
V (3 7i  7 j)
2
32
v A1  v A0 
1
3 7(1  e)
25  7e
v A1  Vj  (1  e)  Vj  (3i  7 j) / 4  (3i  7 j) / 4  
Vi 
Vj
2
32
32
Example 4: How to play pool (or snooker, billiards, or your own
favorite bar game involving balls, a stick, and a table…). The
figure shows a typical problem faced by a pool player – where
should the queue ball hit the eight ball to send it into the pocket?
This is easily solved – the eight ball is stationary before impact, and
after impact has a velocity
1
v B1  (1  e)  v A0  n  n
2
Notice that the velocity is parallel to the unit vector n. This vector
is parallel to a line connecting the centers of the two balls at the
instant of impact. So the correct thing to do is to visualize an imaginary ball just touching the eight ball,
in line with the pocket, and aim the queue ball at the imaginary ball. Easy!
The real secret to being a successful pool player is not potting the
balls – that part is easy. It is controlling where the queue ball goes
after impact.
You may have seen experts make a queue ball
reverse its direction after an impact (appearing to bounce off the
stationary ball); or make the queue ball follow the struck ball after
the impact. According to the simple equations developed here, this
is impossible – but a pool table is more complicated, because the
balls rotate, and are in contact with a table. By giving the queue
ball spin, an expert player can move the queue ball around at will.
To make the ball rebound, it must be struck low down (below the
‘center of percussion’) to give it a reverse spin; to make it follow
the struck ball, it should be struck high up, to make it roll towards
the ball to be potted. Giving the ball a sideways spin can make it rebound in a controllable direction
laterally as well. And it is even possible to make a queue ball travel in a curved path with the right spin.
Never let it be said that you don’t learn useful skills in engineering classes!
Example 5: In this problem we set up a simple MATLAB program that
analyzes motion with impacts. The problem to be solved in illustrated in the
figure. A square box contains a number of rigid, frictionless circular disks.
The disks all have the same mass. At time t=0 the disks are given some
random distribution of velocities. The goal of the program is to compute and
animate the subsequent motion of the disks, accounting for all collisions.
To do this, we need to:
1. Derive the equations of motion for each disk and integrate them;
2. Detect collisions between particles and the wall, and change the
particle velocity to account for the collision;
3. Detect collisions between two particles, and change the velocities of both colliding particles to
account for the collision.
We will model the system by calculating the (x,y) coordinates of each disk. In the following, we will use
 xi , yi   vxi , vyi  to denote the position and velocity of the ith disk.
Equations of motion : No forces act on the disks except during the collisions. Therefore, between
collisions, the equations of motion for the ith disk is
d 2 xi
d 2 yi
i

j0
dt 2
dt 2
As always, we turn this into MATLAB form by writing
 x1   vx1 
 y  v 
 1   y1 
 vx1   0 
d    
v y1  0
dt    
 x2  vx1
   
 y 2   v y1 
   
   
Collisions with the walls: The collisions occur at the instants when the distance of a disk to the wall is
equal to the radius of the disk. Therefore
1. Collision with the wall at x=L occurs when L  xi  R  0
2. Collision with the wall at x=-L xi  R  L  0
3. Collision with the roof at y=L occurs when L  yi  R  0
4. Collision with the floor at y=-L occurs when yi  R  L  0
A collision can be detected by detecting the point when
d  min{L  xi  R, xi  R  L, L  yi  R, yi  R  L}
i
crosses zero from above.
When a collision occurs, the velocities must be corrected as follows to account for the collision:
1. Collision with the walls at x   L : v1xi  evxi0
v1yi  v 0yi
2. Collision with the walls at y   L :
v1yi  ev 0yi
v1xi  vxi0
Collisions between particles The collisions occur at the instants when the distance between the centers
of any two disks is equal to the diameter of the disk.
When a collision occurs between the ith and jth particles, their velocities must be corrected as follows to
account for the collision:
1
1
v1xi  vxi0  (1  e)vn ( xi  x j ) / (2 R )
v1yi  v0yi  (1  e)vn ( yi  y j ) / (2 R )
2
2
1
1
v1xj  vxj0  (1  e)vn ( xi  x j ) / (2 R )
v1yi  v0yi  (1  e)vn ( yi  y j ) / (2 R )
2
2
0
0
0
0
 vxi  vxi   xi  x j    vyi  vyi   yi  y j 
vn 
2R
To see this, note that n   ( xi  x j )i  ( yi  y j ) j / (2 R) is a unit vector joining the two centers; and
let v j  vxj i  v yj j,
v i  vxi i  v yi j denote the velocities of the two particles, and substitute into the
general equations in Example 4.
Here is a MATLAB script that implements these calculations. It will produce the animation shown in the
picture. You can make various modifications to solve similar problems. The only new feature of this
code that has not been discussed before is the ODE solver – it turns out that the standard solver ode45 is
not accurate enough for this problem and misses some collisions. The more accurate solver ode113 is
used instead.
function disk_collisions
% Function to compute and animate motion of colliding disks
% in a box
n_disks = 3;
disk_radius = 1;
box_size = 10.0;
restitution_wall = 1;
restitution_disk = 1;
% These variables are defined in the outer function so they are available
% in all functions
disk_wall_index = 1;
%specifies the disk that is closest to a wall
colliding_pair(1) = 1; %specifies first disk in pair of closest disks
colliding_pair(2) = 1; %specifies second disk in pair of closest disks
tstart = 0;
tstop = 60;
count = 0;
y0 = [1.,1.,0.,0.,-1.,-1.1,1.,1.,-3.,3.,-0.25,0.25];
while(tstart<tstop)
Abstol = 10^(-05)*ones(4*n_disks,1);
options =
odeset('RelTol',0.0000001,'AbsTol',Abstol,'Events',@detect_collision);
[t,y,tevent,yevent,ievent] = ode113(@eom,[tstart,tstop],y0,options);
if (~isempty(tevent))
y0 = collide(tevent,yevent,ievent);
end
tstart = t(length(t));
for i=1:length(t)
% The disks are a single point on an x-y plot
hold off
plot(y(i,1),y(i,2),'ro','MarkerSize',52,'MarkerFaceColor','r');
hold on
for n = 2:n_disks
plot(y(i,4*(n-1)+1),y(i,
4*(n-1)+2),'ro','MarkerSize',52,'MarkerFaceColor','r');
end
axis([-5,5,-5,5]);
axis square
pause(0.025) % This just slows down the animation a bit
end
end
function dydt = eom(t,y)
% Equations of motion for the disks
% y contains [x,y,vx,vy,…] for each disk
for n=1:n_disks
i = 4*(n-1)+1;
dydt(i) = y(i+2);
dydt(i+1) = y(i+3);
dydt(i+2) = 0;
dydt(i+3) = 0.;
end
dydt = transpose(dydt);
end
function [eventvalue,stopthecalc,eventdirection] = detect_collision(t,y)
% Function to detect a collision of a disk with a wall or another disk
%
% Find the shortest distance between a disk and a wall
dmin = box_size;
for n=1:n_disks
i = 4*(n-1)+1;
d1 = box_size/2-(y(i)+disk_radius);
d2 = (y(i)-disk_radius)+box_size/2;
d3 = box_size/2-(y(i+1)+disk_radius);
d4 = (y(i+1)-disk_radius)+box_size/2;
[d,min_wall] = min([d1,d2,d3,d4]);
if (d<dmin)
dmin = d;
disk_wall_index = n;
end
end
% First event is collision of disk with a wall
eventvalue(1) = dmin;
stopthecalc(1) = 1;
eventdirection(1)=-1;
dmin = 10*box_size;
% Find the min dist between any pair of disks
if (n_disks>1)
for n=1:n_disks-1
i = 4*(n-1)+1;
for m = n+1:n_disks
j = 4*(m-1)+1;
d = sqrt( (y(j)-y(i))^2 + (y(j+1)-y(i+1))^2)-2*disk_radius;
if (d<dmin)
dmin = d;
colliding_pair(1)=n;
colliding_pair(2)=m;
end
end
end
% Second event is collision of two disks
eventvalue(2) = dmin;
stopthecalc(2) = 1;
eventdirection(2)=-1;
end
end
function y0 = collide(t,y,ievent)
%
Function to change velocity of disks when a collision occurs
y0 = y;
if (ievent==1)
%collision with a wall
i = 4*(disk_wall_index-1)+1;
d1 = box_size/2-(y(i)+disk_radius);
d2 = (y(i)-disk_radius)+box_size/2;
d3 = box_size/2-(y(i+1)+disk_radius);
d4 = (y(i+1)-disk_radius)+box_size/2;
if (d1<0.0001 || d2<0.0001)
%collision with left or right wall
y0(i+2) = -restitution_wall*y0(i+2);
end
if (d3<0.0001 || d4<0.0001) % collision with top or bottom wall
y0(i+3) = -restitution_wall*y0(i+3);
end
else
i = 4*(colliding_pair(1)-1)+1;
j = 4*(colliding_pair(2)-1)+1;
normal(1) = (y(j)-y(i))/(2*disk_radius);
normal(2)= (y(j+1)-y(i+1))/(2*disk_radius);
vn = (y(j+2)-y(i+2))*normal(1) + (y(j+3)-y(i+3))*normal(2);
y0(i+2) = y0(i+2) + (1+restitution_disk)*vn*normal(1)/2;
y0(i+3) = y0(i+3) + (1+restitution_disk)*vn*normal(2)/2;
y0(j+2) = y0(j+2) - (1+restitution_disk)*vn*normal(1)/2;
y0(j+3) = y0(j+3) - (1+restitution_disk)*vn*normal(2)/2;
end
end
end
4.3 Angular impulse-momentum relations
4.3.1 Definition of the angular impulse of a force
The angular impulse of a force is the time integral of the moment exerted by the force.
To make the concept precise, consider a particle that is subjected to a time varying force F (t ) , with
components in a fixed basis i, j, k , then
F (t )  Fx (t )i  Fy (t ) j  Fz (t )k
Let
r (t )  x(t )i  y (t ) j  z (t )k
k
F(t)
denote the position vector of the particle, and
M (t )  r (t )  F (t )   y (t ) Fz (t )  z (t ) Fy (t )  i   z (t ) Fx (t )  x(t ) Fz (t )  j   x(t ) Fy (t )  y (t ) Fx (t )  k
r(t)
O
x
the moment of the force about the origin.
The Angular Impulse exerted by the force about O during a time
interval t0  t  t0  t is defined as

t0 t

t0
M (t )dt 
t0 t

i
r (t )  F(t )dt
t0
The angular impulse is a vector, and can be expressed as components in a basis
j
z
y
  x i y j z k
t0 t
  y (t ) F (t )  z (t ) F (t )  dt
x 
z
y
t0
t0 t
  z (t ) F (t )  x(t ) F (t )  dt
y 
x
z
t0
z 
t0 t
  x(t ) F (t )  y (t ) F (t )  dt
y
x
t0
If you know the moment as a function of time, you can calculate its angular impulse using simple
calculus. For example for a constant moment, with vector value M 0 , the impulse is   M 0 t
4.3.2 Definition of the angular momentum of a particle
The angular momentum of a particle is simply the cross product of the particle’s position vector with its
linear momentum
h  r  p  r  mv
The angular momentum is a vector – the direction of the vector is perpendicular to its velocity and its
position vectors.
4.3.3 Angular impulse – Angular Momentum relation for a single particle
Consider a particle that is subjected to a force F (t ) for a time interval t0  t  t0  t .
Let r (t ) denote the position vector of the particle
Let M (t )  r (t )  F (t ) denote the moment of F about the origin
Let  denote the angular impulse exerted on the particle
Let η  r (t0  t )  mv (t0  t )  r (t0 )  mv (t0 ) denote the change in angular momentum during
the time interval t .
The momentum conservation equation can be expressed either in differential or integral form.
1. In differential form
M
2. In integral form
dh
dt
k
P
  h
Although it’s not obvious, these are just another way of writing
Newton’s laws of motion. To show this, we’ll derive the
differential form. Start with Newton’s law
i
F  ma  m
dv
dt
Take the cross product of both sides with r
r
O
x
j
z
y
rF  rm
dv
dt
Note that v  mv  0 since the cross product of two parallel vectors is zero. We can add this to the right
hand side, which shows that
rF  rm
dv
dv dr
d
dh
 v  mv  r  m   mv   r  mv  
dt
dt dt
dt
dt
This yields the required relation.
Angular momentum conservation: For the special case where the force is parallel to r, the moment of
the force acting on the particle is zero ( r  F  0 ), and angular momentum is constant
d
 r  mv   0
dt
4.3.4 Examples using Angular Impulse – Angular Momentum relations for a single particle
The angular impulse-angular momentum equations are particularly helpful when you need to solve
problems where particles are subjected to a single force, which acts through a fixed point. They can also
be used to analyze rotational motion of a massless frame containing one or more particles.
Example 1: Orbital motion. A satellite is launched into a
geostationary transfer orbit by the ARIANE V launch facility.
At its perigee (the point where the satellite is closest to the
earth) the satellite has speed 10.2km/sec and altitude 250km.
At apogee (the point where the satellite is furthest from the
earth) the satellite has altitude 35950 km. Calculate the speed
of the satellite at apogee.
j
rp
i
ra
va
va
Assumptions:
1. We assume that the only force acting on the satellite is the gravitational attraction of the earth
2. The earth’s radius is 6378.145km
Calculation:
1. Since the gravitational force on the satellite always acts towards the center of the earth,
angular momentum about the earth’s center is conserved.
2. At both perigee and apogee, the velocity vector of the satellite must be perpendicular to its
position vector. To see this, note that at the point where the satellite is closest and furthest
from the earth, the distance to the earth is a max or min, and so the derivative of the distance
of the satellite from the earth must vanish, i.e.


1 1  dr dr 
r   r   0  r  v  0
2 r  r  dt dt 
where we have used the chain rule to evaluate the time derivative of r  r . Recall that if the
d
d
r

dt
dt
r r  0

dot product of two vectors vanishes, they are mutually perpendicular. We take a coordinate
system with i and j in the plane of the orbit, and k perpendicular to the orbit.
3. We take the satellite orbit to lie in the i , j plane with k perpendicular to the orbit. The
angular momentum at apogee and perigee is then
h p  rp  mv p  rp mv p k
h a  ra  mv a  ra mva k
where ra , rp are the distance of the satellite from the center of the earth at apogee and
perigee, and va , v p are the corresponding speeds.
4. Since angular momentum is conserved it follows that
h p  h a  rp mv p  ra mva  va  rp v p / ra
5. Substituting numbers yields 1.6 km/s
Example 2: More orbital motion. The orbit
for a satellite is normally specified by a set
of angles specifying the inclination of the
orbit, and by quoting the distance of the
satellite from the earths center at apogee and
perigee ra , rp . It is possible to calculate the
speed of the satellite at perigee and apogee
from this information.
Calculation
1. From the previous example, we
know that the distances and
velocities are related by
rp mv p  ra mva
2. The system is conservative, so the total energy of the system is conserved.
3. The potential energies when the satellite is at perigee and apogee are
Vp  
GMm
rp
Va  
GMm
ra
4. The kinetic energies of the satellite at perigee and apogee are
Tp 
1 2
mv p
2
Ta 
1 2
mva
2
5. The kinetic energy of the earth can be assumed to be constant. Energy conservation therefore
shows that
1 2 GMm 1 2 GMm
mv p 
 mva 
2
rp
2
ra
6. The results of (1) and (5) give two equations that can be solved for va , v p in terms of known
parameters. For example, (1) shows that va  v p rp / ra - this can be substituted into (5) to see
that
2
1 2 GMm 1 2 rp GMm
mv p 
 mv p 2 
2
rp
2
ra
ra
 v 2p
ra2  rp2
2
a
r
1 1
 2GM   
 rp ra 


 vp 
Similarly, at apogee
va 
2GMrp
ra (ra  rp )
2GMra
rp (ra  rp )