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Chapter 4
Conservation laws for systems of particles
In this chapter, we shall introduce (but not in this order) the following general concepts:
- The linear impulse of a force
- The angular impulse of a force
- The power transmitted by a force
- The work done by a force
- The potential energy of a force.
- The linear momentum of a particle (or system of particles)
- The angular momentum of a particle, or system of particles.
- The kinetic energy of a particle, or system of particles
- The linear impulse momentum relations for a particle, and conservation of linear momentum
- The principle of conservation of angular momentum for a particle
- 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
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
x x y y z z
P F v F v F v
Work has units of Nm/s, or `Watts’ in SI units.
i
j
k
O
P
F (t)
The work done by the force during a time interval 0 1
t t t is
1
0
t
t
W dt
F v
The work done by the force can also be calculated by integrating the
force vector along the path traveled by the force, as
1
0
W d
r
r
F r
where 0 1
r , r 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.
Definition of an angular velocity vector Visualize a
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:
- 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
- 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’)
i
j
k
O
P
F (t)
r
0
r
1
n
Axis of
rotation
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).
Assumptions
- The earth’s radius is 6378.145km
- The mass of a typical satellite is 4135kg - see , e.g.
http://www.astronautix.com/craft/hs601.htm
- The Gravitational parameter
5 3 1 GM 3.986012 10 km s
( G=
gravitational constant; M =mass of earth)
- 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:
- The gravitational force on the satellite is
2
GMm
x
F i
- The work done follows as
2
R h
R
GMm
dx GMm
R h R x
F i i
- Substituting numbers gives
6
9.7 10 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.
Assumptions:
- A Ferrari Testarossa has mass 1506kg (see
http://www.ultimatecarpage.com/car/1889/Ferrari-
Testarossa.html)
- The coefficient of friction between wheels and road is of
order 0.
- We assume the brakes are locked so all wheels skid, and air resistance is neglected
Calculation The figure shows a free body diagram. The equation of motion for the car is
F R R F x
T T i N N mg j ma i
- The vertical component of the equation of motion yields R F
N N mg
2. The friction law shows that
R R F F R F R F
T N T N T T N N mg
- The position vectors of the car’s front and rear wheels are ( ) F R
r x i r x L i. The work
done follows as. We suppose that the rear wheel starts at some point 0
x i when the brakes are
applied and skids a total distance d.
0 0 0 0
0 0 0 0
x d x d L x d x d L
R R F F R F R F
x x L x x L
W d d T dx T dx T T d
F r F r i i i i
- The work done follows as W mg. Substituting numbers gives
5
W 9 10 J.
R
i
h
j
N R
N F
T F
T R
i mg
x
L
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
- Applied force P cos i P sin j
- Friction T cos i T sin j
- Normal reaction N sin i T cos j
- Weight mg j
The velocity vector is v cos i v sin j
Evaluating the dot products F v for each formula, and recalling that
2 2
cos sin 1 gives
- Applied force Pv
- Friction Tv
- Normal reaction 0
- Weight mg sin
Example 6: The table lists the experimentally measured force-v-draw data for a long-
bow. 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
1
0
W d
r
r
F r
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.
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
1 2 2 3 3 4 4 5
1 2 3 4
2 2 2 2
f f f f f f f f
W x x x x
^ ^ ^
^
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 =
So the solution is 80.5J
Force
N
Draw
(cm)
P
i
j
P
i
j (^) N
T
mg
x
1
x
2
x
3
x
4
f
1
f
2
f
3
f
4
f
5
x
F
x y z
V V V
F F F
x y z
i j k i j k
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.
- The force exerted by a spring is
0
F k x ( L ) i
- The position vector of the force is r x i
- The potential energy follows as
0 0
2
0 0
( ) constant = ( ) ( )
L
L
V d k x L dx k L L
r
r
r F r i i
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
1 2
2
Q Q
x
F i
where 1
Q and
2
Q 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 0
r at infinity. The potential energy follows as
1 2 1 2
2
x
Q Q Q Q
V dx
x x
r i i
x
+Q
1
+Q
2
i
j
F
F
i
j x
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
Gravity acting on a
particle near earths
surface
F mg j V mgy
F
m
j
i
y
Gravitational force
exerted on mass m by
mass M at the origin
3
GMm
r
F r
GMm
V
r
r
F
r
m
Force exerted by a
spring with stiffness k
and unstretched length
0
L
0
k r ( L )
r
r
F (^)
2
0
V k r L
F
i
j r
Force acting between
two charged particles
1 2
3
Q Q
r
F r
1 2
Q Q
V
r
r
+Q
1
+Q
2
i
j
F
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.
13 7
E a a
a r r r
F
12 6
a a
E
r r
r
i
j
F
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
- At some time 0
t the particle has some initial position 0
r , velocity 0
v and kinetic energy 0
T
- At some later time t the particle has a new position r , velocity v and kinetic energy T.
- Let P F v denote the rate of work done by the force
- Let
1
0 0
t
t
W Pdt d
r
r
F r 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 =m a with the particle velocity
d d
m m m
dt dt
v
F v a v v v v
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.
0
W T T
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:
- We assume that all the wheels are locked and skid over
the ground (this will stop the vehicle in the shortest
possible distance)
- The contacts are assumed to have friction coefficient
- The vehicle is idealized as a particle.
- Air resistance will be neglected.
N A
N B
N C
N D
T D
T C
T T B A
mg
i
j
Calculation:
- The figure shows a free body diagram.
- The equation of motion for the vehicle is
A B C D A B C D x
T T T T i N N N N mg j ma i
The vertical component of the equation shows that A B C D
N N N N mg.
- The friction force follows as (^) A B C D A B C D
T T T T N N N N mg
- If the vehicle skids for a distance d , the total work done by the forces acting on the vehicle is
0
d
A B C D A B C D
W T T T T dx T T T T d mgd
i i
- 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
2
T 0 mV / 2. The work-energy relation shows that
2 2 mgd mV / 2 d V / 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 stop-
start 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:
- We assume that the speed in a traffic jam is low enough that air resistance can be neglected.
- The energy to accelerate to speed V is
2
mV / 2.
- We assume that the vehicle accelerates and brakes with constant acceleration – if so, its average
speed is V /2.
- If the vehicle travels a distance d between stops, the time between two stops is 2 d/V.
- The average power is therefore mV / 4 d.
- Taking V=15 mph (7m/s) and d=200 ft (61m) are reasonable values – the power is therefore
0.03 m , 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.
the i th particle must be equal and opposite, to ij
R , i.e. ij ji
R R.
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 ( )
ext
i
F t
We define the total external work done on the system during a time interval 0 0
t t t t as the sum of
the work done by the external forces.
0
0
t t
ext
ext i
forces t
W t t dt
(^)
F v
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
total
T 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
- At some time 0
t the system has and kinetic energy 0
total
T
- At some later time t the system has kinetic energy
total
T.
- Let 0
total
V denote the potential energy of the force at time 0
t
- Let
total
V denote the potential energy of the force at time t
- Let ext
W denote the total work done on the system between 0 0
t t t 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.
ext 0 0
W T V T V
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
0 0
ext
W T V T V
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 0
V at angle . Calculate an
expression for the maximum height reached by the projectile.
If air resistance can be neglected, we can regard the earth and the
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:
- Just after launch, the velocity of the particle is 0 0
v V cos i V sin j
- The kinetic energy of the particle just after launch is
2
0
mV / 2. Its potential energy is zero.
- 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 0
v V cos i. The kinetic energy at
this point is therefore
2 2
0
mV cos / 2
- Energy is conserved, so
2 2 2
0 0
2 2 2 2
0 0
/ 2 cos / 2
(1 cos ) / 2 sin
mV mgh mV
h V V
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
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.
i
j
V 0
h
d
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
- The long-bow will be idealized as a linear spring
- The maximum draw force is likely to be around 60lbf (270N)
- The draw length is about 2ft (0.6m)
- Arrows come with various masses – typical range is between 250-600 grains (16-38 grams)
- 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.
- Just before the arrow is released, the spring is stretched to its maximum length, and the arrow is
stationary. The total energy of the system is
2
0 0
T V kL , where L is the draw length and k is
the stiffness of the bow.
- We can estimate values for the spring stiffness using the draw force: we have that D
F kL , so
D
k F L. Thus 0 0
D
T V LF.
- Just after the arrow is fired, the spring returns to its un-stretched length, and the arrow has
velocity V. The total energy of the system is
2
1 1
T V mV , where m is the mass of the arrow
- The system is conservative, therefore
2
0 0 1 1
D D
T V T V LF mV V LF m
- We suppose that the arrow is launched from the origin at an angle to the horizontal. The
horizontal and vertical components of velocity are cos sin x y
V V V V . The position
vector of the arrow can be calculated using the method outlined in Section 3.2.2 – the result is
2
cos sin
Vt Vt gt
r i j
We can calculate the distance traveled by noting that its position vector when it lands is d i. This
gives
2
cos sin
Vt Vt gt d
i j i
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
2 sin 2 sin
cos
V V
t d
g g
The arrow travels furthest when fired at an angle that maximizes sin / cos - i.e. 45 degrees.
The distance follows as
2
2 2 D
V LF
d
g mg
- 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
- The space vehicle is initially in orbit at an altitude h above the
earth’s surface
- The earth’s radius is 6378.145km
- 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.
- The Gravitational parameter
5 3 1
GM 3.986012 10 km s
( G= gravitational constant; M =mass of earth)
Calculation
- Just after the rocket is burned, the potential energy of the system is V GMm / ( R h ), while
its kinetic energy is
2
T mv / 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.
- This is a conservative system, so
2 / 2 / ( ) 0
T V mv GMm R h
v GM R h
- A typical low earth orbit has altitude of 250km. For this altitude the escape velocity is
10.9km/sec.
h
R
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
- The rate of work done (or power expended) by a linear actuator that applies forces A
F ,
B
F to the
objects attached to its two ends can be calculated as A A B B
P F v F v , where A
v and B
v are
the velocities of the two ends.
- The rate of work (or power expended) by a rotational motor that applies moments A
M ,
B
M to
the objects attached to its two ends is A A B B
P M ω M ω.
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 EN
An electric motor applies a moment , or torque to an object
that is coupled to its output shaft. The moment is developed
by electromagnetic forces acting between permanent
magnets in the housing and the electric current flowing
through the winding of the motor. In the following
discussion, we shall assume that the body of the motor is
stationary A
ω 0 , and its output shaft rotates with angular
velocity B
ω n where n is a unit vector parallel to the
shaft and is its angular speed. In addition, we assume
that the output shaft exerts a moment B
M T n.
motor
M
B
M
B
n
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
0 0 0
0
I T^ I^ T
T
I T
^
where 0
T 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
0 0 0
0
V
T V R T
T R R
V R T
This relationship is sketched in the figure – this is called the
torque curve for the motor. The two most important points on
the curve are
- The `stall torque’ 0
s
T V R T
- The `No load’ speed
2
0 0
nl
V T R R
The torque curve can be expressed in terms of these quantities
as
s
nl
T T
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, , 0
T 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.
Stall torque T
s
Rotation rate
Moment
No load speed
nl
