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ENES100-0702
Introduction to Engineering Design
R. J. Phaneuf
Fall 2002
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Forces, Work and Energy
A force is a push or pull which alters the state of motion of an object. From Newton’s
observations we know that the net force applied to an object is proportional to its
acceleration, that is, the change in its velocity per unit time:


 F  ma
(1).
This is referred to as Newton’s second law.
You are most likely familiar also with the concept of work. The work done by a
force is defined as :
 
W  F  r
(2),
that is, by the dot product of the force and the change in position. Notice that only that
component along the displacement does work. The component of a force perpendicular
to the displacement does no work. If the force is opposite the motion (for example
frictional forces always oppose motion) then it does negative work.
Work and Kinetic Energy
Now consider the work done by a constant horizontally-applied force in accelerating an
object, initially at rest to a final velocity vx = vf, over a time period t:
 
W  F  r  Fx  max
(3)
To calculate x we note that the average velocity <v> = ½vf, since for a constant applied
force the acceleration is also constant,
x  v  t 
1
v f t
2
(4),
so that the work can be written:
1

W  ma v f t 
2

(5),
where
v f  at
(6),
or
a
vf
(7)
t
and substituting this into equation (5) finally yields:
 v  1
 1
2
W   m f  v f t   mv f

t
2
2



(8)
The quantity ½ mv2 has a special significance. We associate it with the energy of a
moving object, the kinetic energy,
KE = ½ mv2
(9).
Thus the work done in accelerating the object from its original condition at rest is equal
to the kinetic energy. More generally, the work done in accelerating an object with an
initial horizontal speed v0 to a final horizontal speed vf is the difference in kinetic energy:
W = KE = ½ mvf2 – ½ mv02 .
(10)
Rotational Motion
Even if the position of the center of mass of an object is fixed, it can often be set
into rotational motion if the application of force results in a torque about the axis of
rotation:



  r F .
(11)
It is possible to describe the rotational motion of the object using an equation analogous
to Newton’s second law:


  I
(12)

where  is the angular acceleration, and I is the moment of inertia of the object about its
rotational axis. The angular acceleration is equal to the time rate of change of the angular
velocity, which in turn is equal to the time rate of change of the angular position:



t
(13)


,

t
(14)



where  is a vector whose magnitude is the change in the angular position of the object
and whose direction is given by the right hand rule. Note that the units for angle are
radians, with 2π radians in one rotation. Thus the angular velocity ω is measured in
radians/s. There is a related quantity, the rotational frequency which is proportional to
the angular velocity:
   (1rotation / 2radian ) ,
(15)
which is measured in rotations/s or more commonly rotations/minute (RPM):
 [ RPM ]  (1rotation / 2radians )60s / 1min .
(16)
The moment of inertia for an object made up of a collection of masses mi displaced from

the rotational axis by ri can be written as:
I   mi ri ,
2
(17)
i
which would be replaced by an integral for a continuous distribution of mass. For an
object with a uniform mass density ρ=M/V, the moment of inertia can be easily
determined. For example for a disk of radius R, Idisk=MR2/2. For a ring of inner radius


Ri and outer radius Ro, I ring  M Ri  Ro / 2 , and for a sphere of radius R,
I sphere 
2
2
2
MR 2 .
5
For an object rotating with constant angular velocity ω, the associated kinetic
energy can be written by analogy to equation (9):
KErot 
1 2
I
2
(18)
For the case of an object both rotating and translating
KEtotal  KEtrans  KErot 
1 2 1 2
mv  I .
2
2
(19)
Work and Potential Energy
Now consider a second simple case: that of a vertical force applied on an object,
initially moving upward with a velocity vy. If the applied force is equal and opposite to
that exerted by gravity, the velocity will be constant. The work done by the applied force
in moving the object from its initial position y0 to its final position yf is given by
 
W  F  r  mg( y f  y0 ) .
(20)
In this case there is no change in velocity, and thus no change in kinetic energy. Where
did the work go? We recognize that an object at a height H = yf –y0 initially at rest would
gain kinetic energy if released, due to the action of gravity. There must be energy stored
in raising the height of an object above the ground. We refer to this stored energy as
potential energy. From the above expression for the work we define the difference in
gravitational potential energy as
PE = mg(y – y0).
(21)
There are many other examples of forces which allow potential energy to be stored-the
Hooke’s law force associated with deflection of a spring is another familiar example. For
forces of this type, in the absence of applied external forces the sum of the kinetic and
potential energies is conserved
KE + PE = Constant,
(22)
so that these forces are referred to as conservative forces. The second type of force is
referred to as dissipative, since these forces remove energy from a system. Friction, and
viscous fluid forces are examples. This does not mean that energy is not conserved,
however. Dissipative forces cause heating of the objects within the system and their
surroundings. If we keep track of the heat flow into the surroundings, and consider the
entire system including the surroundings, energy is always conserved.
We can write down the most general form of the work-energy relation

WappliedWdisKE + PE,
(14)
where Wdis is always negative, and equal in magnitude to the heat flow into the objects of
the system (increasing their so called internal energy) and surroundings.
Wapp = KE + PE – Wdis
(15)
Power and Efficiency
The rate at which a force supplies work is referred to as the power:
 
 
W F  r
P

 F v .
t
t
(16)
If the work is instead supplied by a torque, it can be written in an analogous form:
 
P    
(17)
From equation (15) we see that the power which must be generated in moving an object,
goes into increasing its kinetic energy, increasing its potential energy, and in doing work
against dissipative forces. The overall input power Pin, which must be supplied by
applied forces and/or torques is given by:
Pin 
KE PE Wdis


 Pout  Pdis
t
t
t
where the output power is defined as Pout 
dissipative forces is defined as Pdis  
(18)
KE PE

, and the power supplied against
t
t
Wdis
.
t
It is possible to define the efficiency of a process in which power is used to increase the
kinetic and potential energy of an object, as well as to counter dissipative forces:

Pout
Pout

Pin
Pout  Pdis
(19)
Note that since Wdis is always negative, the power supplied against dissipation is always
positive, and thus the efficiency of a process is less than or equal to 1.
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