ME2151 ENGINEERING MECHANICS lesson notes

NOTES ON LESSON
ME2151 ENGINEERING MECHANICS
First law: A body does not change its state of motion unless acted upon by a force. This
law is based on observations but in addition it also defines an inertial frame . By
definition an inertial frame is that in which a body does not change its state of motion
unless acted upon by a force. For example to a very good approximation a frame fixed in
a room is an inertial frame for motion of balls/ objects in that room. On the other hand if
you are sitting in a train that is accelerating, you will see that objects outside are
changing their speed without any apparent force. Then the motion of objects outside is
changing without any force. The train is a non-inertial frame.
Second law: The second law is also part definition and part observation. It gives the
force in terms of a quantity called the mass and the acceleration of a particle. It says that
a force of magnitude F applied on a particle gives it an acceleration a proportional to the
force. In other words
F = ma , (1)
where m is identified as the inertial mass of the body. So if the same force - applied
either by a spring stretched or compressed to the same length - acting on two different
particles produces accelerations a1 and a2, we can say that
m1 a1 = m2 a2
or (2)
In the first part of the course i.e. Statics we consider only equilibrium situations. We will
therefore not be looking at F = ma but rather at the balance of different forces applied on
a system. In the second part - Dynamics - we will be applying F = ma extensively.
Third Law: Newton's third law states that if a body A applies a force F on body B , then
B also applies an equal and opposite force on A . (Forces do not cancel such other as
they are acting on two different objects)
Figure 1
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Thus if they start from the position of rest A and B will tend to move in opposite
directions. You may ask: if A and B are experiencing equal and opposite force, why do
they not cancel each other? This is because - as stated above - the forces are acting on
two different objects. We shall be using this law a lot both in static as well as in
dynamics.
Equality of Vectors: Since a vector is defined by the direction and magnitude, two
vectors are equal if they have the same magnitude and direction. Thus in figure 2 vector
is equal to vector and but not equal to vector although all of them have the same
magnitude.
Thus we conclude that any two vectors which have the same magnitude and are parallel
to each other are equal. If they are not parallel then they cannot be equal no matter what
their magnitude.
Adding and subtracting two vectors (Graphical Method): When we add two vectors
and by graphical method to get , we take vector , put the tail of on the
head of .Then we draw a vector from the tail of to the head of . That vector
represents the resultant (Figure 4). I leave it as an exercise for you to show that
. In other words, show that vector addition is commutative.
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Let us try to understand that it is indeed meaningful to add two vectors like this. Imagine
You may now ask: can I multiply by a negative number? The answer is yes. Let us see
what happens, for example, when I multiply a vector by -1. Recall from your school
mathematics that multiplying by -1 changes the number to the other side of the number
line. Thus the number -2 is two steps to the left of 0 whereas the number 2 is two steps to
the right. It is exactly the same with vectors. If represents a vector to the right,
would represent a vector
To represent vectors in terms of their x,y and z components, let us first introduce
the concept of unit vector. A unit vector in a particular direction is a vector of
magnitude '1' in that direction. So a vector in that particular direction can be written as a
number times the unit vector . Let us denote the unit vector in x-direction as , in ydirection
as and in z-direction as . Now any vector can be described as a sum of
three vectors , and in the directions x, y and z, respectively, in any order (recall
that order does not matter because vector sum is commutative). Then a vector
Further, using the concept of unit vectors, we can write , where Ax is a number.
Similarly and . So the vector above can be written as
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where Ax, Ay and Az are known as the x, y, & z components of the vector. For example a
vector would look as shown in figure 9.
It is clear from figure 9 that the magnitude of the vector is
. Now when we add two vector, say and
, all we have to do is to add their x-components, y-components and
the z-components and then combine them to get
Similarly multiplying a vector by a number is same as increasing all its components by
the same amount. Thus
=
How about the multiplying by -1? It just changes the sign of all the components. Putting it
all together we see that
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Scalar product (Dot Product) of two vectors can also be written in another form involving
the magnitudes of these vectors and the angle between them as
where are the magnitudes of the two vectors, and è is the angle between them.
Notice that although can be negative or positive depending on the
angle between them. Further, if two non-zero vectors are perpendicular, . From
the formula above, it is also apparent that if we take vector to be a unit vector, the dot
product represents the component of in the direction of . Thus the scalar
product between two vectors is the product of the magnitude of one vector with the
magnitude of the component of the other vector in its direction. Try to see it pictorially
yourself. We also write the dot products of the unit vectors along the x, y, and the z axes.
These are and
Equilibrium of bodies
In the previous lecture, we discussed three laws of motion and reviewed some basic
aspects of vector algebra. We are now going to apply these to understand equilibrium of
bodies. In the static part when we say that a body is in equilibrium, what we mean is that
the body is not moving at all even though there may be forces acting on it. (In general
equilibrium means that there is no acceleration i.e., the body is moving with constant
velocity but in this special case we take this constant to be zero).
Let us start by observing what all can a force do to a body? One obvious thing it does is
to accelerate a body. So if we take a point particle P and apply a force on it, it will
accelerate. Thus if we want its acceleration to be zero, the sum of all forces applied on it
must vanish. This is the condition for equilibrium of a point particle. So for a point particle
the equilibrium condition is
where are the forces applied on the point particle (see figure 13)
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That is all there is to the equilibrium of a point particle. But in engineering problems we
deal not with point particles but with extended objects. An example is a beam holding a
load as shown in figure 2. The beam is equilibrium under its own weight W , the load L
and the forces that the supports S1 and S2 apply on it.
To consider equilibrium of such extended bodies, we need to see the other effects that a
force produces on them. In these bodies, in addition to providing acceleration to the
body, an applied force has two more effects. One it tends to rotate the body and two it
deforms the body. Thus a beam put on two supports S1 and S2 tends to rotate clockwise
about S2 when a force F is applied downwards (figure 3).
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The strength or ability of a force to rotate the body about a point O is given by the
torque generated by it. The torque is defined as the vector product of the displacement
vector from O to the point where the force is applied. Thus
This is also known as the moment of the force. Thus in figure 3 above, the torque about
S2 will be given by the distance from the support times the force and its direction will be
into the plane of the paper. From the way that the torque is defined, the torque in a given
direction tends to rotate the body on which it is applied in the plane perpendicular to the
direction of the torque. Further, the direction of rotation is obtained by aligning the thumb
of one's right hand with the direction of the torque; the fingers then show the way that the
body tends to rotate (see figure 4). Notice that the torque due to a force will vanish if the
force is parallel to .
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We now make a subtle point about the tendency of force to rotate a body. It is that even
if the net force applied on a body is zero, the torque generated by them may vanish i.e.
the forces will not give any acceleration to the body but would tend to rotate it. For
example if we apply equal and opposite forces at two ends of a rod, as shown in figure
5, the net force is zero but the rod still has a tendency to rotate. So in considering
equilibrium of bodies, we not only have to make sure that the net force is zero but can
also that the net torque is also zero.
Varignon's theorem.
If there are many forces applied on a body then the total moment about O is the vector
sum of all other moments i.e.
As a special case if the forces are all applied at the same point j then
This is known as Varignon's theorem. Its usefulness arises from the fact that the torque
due to a given force can be calculated as the sum of torques due to its components.
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Moment of force about an axis: So far we have talked about moment of a force about
a point only. However, many a times a body rotates about an axis. This is the situations
you have bean studying in you 12th grade. For example a disc rotating about an axis
fixed in two fixed ball bearings. In this case what affects the rotation is the component of
the torque along the axis, where the torque is taken about a point O (the point can be
chosen arbitrarily) on the axis as given in figure 11. Thus
where is the unit vector along the axis direction and is the vector from point O on the
axis to the force .
Using vector identities (exercise at the end of Lecture 1), it can also be written as
Thus the moment of a force about an axis is the magnitude of the component of the
force in the plane perpendicular to the axis times its perpendicular distance from the
axis. Thus if a force is pointing towards the axis, the torque generated by this force about
the axis would be zero. This can be understood as follows. When a force is applied,
forces are generated at the ends of the axis being held on a one place. These forces
together with generate the torque when components along the axis by responsible for
rotation of the body about the axis, in the same manner, the couple about the axis is
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given by the component of the couple moment in the direction for the axis. You can work
it out; it is actually equal to the component of the force in the plane perpendicular to the
axis times the distance of the force line of action from the axis. One point about the
moment about an axis, it is independent of the origin since it depends only on the
distance of the force the axis.
Free Body Diagram. I have actually been using it without calling it so. Now, let us
formalize it.
In talking about the equilibrium of a body we consider all the external forces applied on it
and the interaction of the body with other objects around it. This interaction produces
more forces and torques on the body. Thus when we single out a body in equilibrium,
objects like hinges, ball-socket joint, fixed supports around it are replaced elements by
the corresponding forces & torques that they generate. This is what is called a free-body
diagram. Making a free-body diagram allows us to focus our attention only on the
information relevant to the equilibrium of the body, leaving out unnecessary details. Thus
making a free-body diagram is pretty much like Arjuna - when asked to take an aim on
the eye of a bird - seeing only the eye and nothing else. The diagrams made on the right
side of figures 1, 2 and 3 are all free-body diagrams.
In the coming lecture we will be applying the techniques learnt so far to a very special
structure called the truss. To prepare you for that, in the following I consider the special
case of a system in equilibrium under only two forces. For completeness I will also take
up equilibrium under three forces.
When only two forces are applied, no matter what the shape or the size of the object in
equilibrium is, the forces must act along the same line, in directions opposite to each
other, and their magnitudes must be the same. That the forces act in directions opposite
to each other and have equal magnitude follows from the equilibrium conditions
, which implies that . Further, if the forces are not along the same
line then they will form a couple that will tend to rotate the body. Thus implies that
the forces act along the same line, i.e. they be collinear (see figure 7).
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Similarly if there are three forces acting on a body that is in equilibrium then the three
forces must be in the same plane and concurrent. If there are not concurrent then they
must be parallel (of course remaining in the same plane). This can be understood as
follows. Any two members of the three applied forces form a plane. If the third force is
not in the same plane, it will have a non-vanishing component perpendicular to the
plane; and that component does not get cancelled. Thus unless all three forces are in
the same plane, they cannot add up to zero. So to satisfy the equation , the
forces must be in the same plane, i.e. they must be coplanar. For equilibrium the torque
about any point must also be zero. Since the forces are in the same plane, any two of
them will intersect at a certain point O. These two forces will also have zero moment
about O. If the third force does not pass through O, it will give a non-vanishing torque
(see figure 8). So to satisfy the torque equation, the forces have to be concurrent. Zero
torque condition can also be satisfied if the three forces are parallel forces (see figure 8);
that is the other possibility for equilibrium under three forces.
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Friction
Whatever we have studied so far, we have always taken the force applied by one
surface on an object to be normal to the surface. In doing so, we have been making an
approximation i.e., we have been neglecting a very important force viz., the frictional
force. In this lecture we look at the frictional force in various situations.
In this lecture when we talk about friction, we would mean frictional force between two
dry surfaces. This is known as Coulomb friction. Frictional forces also exist when there is
a thin film of liquid between two surfaces or within a liquid itself. This is known as the
viscous force. We will not be talking about such forces and will focus our attention on
Coulomb friction i.e., frictional forces between two dry surfaces only. Frictional force
always opposes the motion or tendency of an object to move against another object or
against a surface. We distinguish between two kinds of frictional forces - static and
kinetic - because it is observed that kinetic frictional force is slightly less than maximum
static frictional force.
Let us now perform the following experiment. Put a block on a rough surface and pull it
by a force F (see figure 1). Since the force F has a tendency to move the block, the
frictional force acts in the opposite direction and opposes the applied force F. All the
forces acting on the block are shown in figure 1. Note that I have shown the weight and
the normal reaction acting at two different points on the block. I leave it for you to think
why should the weight and the normal reaction not act along the same vertical line?
It is observed that the block does not move until the applied force F reaches a maximum
value Fmax. Thus from F = 0 up to F = Fmax, the frictional force adjusts itself so that it is
just sufficient to stop the motion. It was observed by Coulombs that F max is proportional
to the normal reaction of the surface on the object. You can observe all this while trying
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to push a table across the room; heavier the table, larger the push required to move it.
Thus we can write
where µs is known as the coefficient of static friction. It should be emphasize again that
is the maximum possible value of frictional force, applicable when the object is about to
stop, otherwise frictional force could be less than, just sufficient to prevent motion. We
also note that frictional force is independent of the area of contact and depends only on
N .
As the applied force F goes beyond F , the body starts moving now experience slightly
less force compound to. This force is seem to be when is known as the coefficient of
kinetic friction. At low velocities it is a constant but decrease slightly at high velocities. A
schematic plot of frictional force F as a function of the applied force is as shown in figure
2.
Values of frictional coefficients for different materials vary from almost zero (ice on ice)
to as large as 0.9 (rubber tire on cemented road) always remaining less than 1.
A quick way of estimating the value of static friction is to look at the motion an object on
an inclined plane. Its free-body diagram is given in figure 3.
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Since the block has a tendency to slide down, the frictional force points up the inclined
plane. As long as the block is in equilibrium
As è is increased, mgsinè increases and when it goes past the maximum possible value
of friction fmax the block starts sliding down. Thus at the angle at which it slides down we
have
Let us now solve a couple of simple standard examples involving static friction/kinetic
friction.
Properties of plane surfaces I: First moment and centroid of area
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Having deal with trusses and frictional forces, we now change gears and go on to
discuss some properties of surfaces mathematically. Of course we keep connecting
these concepts to physical situations.
The first thing that we discuss is the properties of surfaces. This is motivated by the fact
is general the forces do not act at a single point but are distributes over a body. For
example the gravitational force pulling an object down acts over the entire object.
Similarly a plate immersed in water, for example has the pressure acting on it over the
entire surface. Thus we would like to know at which point does the force effectively act?
For example in the case of an object in a gravitational field, it is the centre of gravity
where the force acts effectively. In this lecture we develop important mathematical
concepts to deal with such forces. Let us start with the first moment of an area and the
centroid .
First moment of an area and the centroid: We first consider an area in a plane; let us
call it the X-Y plane (see figure 1).
The first moment MX of the area about the x-axis is defined as follows. Take small area
element of area ÄA and multiply it by its y-coordinate, i.e. its perpendicular distance from
the X-axis, and then sum over the entire area; the sum obviously goes over to an
integral in the continuous limit. Thus
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Similarly the first moment MY of the area about the y-axis is defined by multiplying the
elemental area ÄA by its x-coordinate, i.e. its perpendicular distance from the Y-axis,
and summing or integrating it over the entire area. Thus
This is shown in figure 2.
Centroid: Centroid of a bounded area is a point whose x-coordinates XC and ycoordinate
YC are defined as
where A is its total area. We now solve some examples of calculating these quantities
for some simple areas.
Application to mechanics: As the first simple application of the methods developed let
us consider beams which are externally loaded. We consider only those situations where
beams are supported externally so that the external reactions can be calculated on the
basis of statics alone. As in the case of trusses, such beams are called statically
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determinate beams. Now one such beam is loaded externally between X1 and X2 as
shown in figure 9.
In the figure the function f(x) is the load intensity which is equal to load per unit length.
Thus force over a length dx is given by dF = f(x) dx . The total load R therefore is
Next question we ask is where is the total load located? This is determined by finding the
Moment (torque) created by the load, which is given by
Thus the location of the load is given by the centroid of the area formed by the load
curve and the beam, taking beam as the x-axis. Let us now take some examples.
Uniform loading: This is shown in figure 10 along with the total load R acting at the
centroid of the loading intensity curve. The uniform load intensity is w .
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The total load in this case is and the load acts at the centroid
Properties of surfaces II: Second moment of area
Just as we have discussing first moment of an area and its relation with problems in
mechanics, we will now describe second moment and product of area of a plane. In this
lecture we look at these quantities as some mathematical entities that have been defined
and solve some problems involving them. The usefulness of related quantities, called the
moments of inertia and products of inertia will become clear when we deal with rotation
of rigid bodies.
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Let us then consider a plane area in xy plane (figure 1). The second moments of the
area A is defined as
That is given a plane surface, we take a small area in it, multiply by its perpendicular
distance from the x-axis and sum it over the entire area. That gives IXX . Similarly IYY is
obtained by multiplying the small area by square of the distance perpendicular to the yaxis
and adding up all contributions (see figure 2).
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The product of area is defined as
where x and y are the coordinates of the small area dA . Obviously IXX is the same as IXY
.
Let us now solve a few examples.
Example1: Let us start with a simple example of a square of side a with its center of the
origin (see figure 2).
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Figure 3
To calculate this, we choose the elemental area as shown in figure 4 and integrate. Then
dA = ady
so that
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Similarly for calculating IYY we choose a vertical elemental area and calculate
Let us also calculate the product of inertia. Choose on elemental area dxdy and
calculate (see figure 5)
As noted earlier, IYX is equal to IXY and therefore it also vanishes.
Transfer theorem: Let the centroid of an area be at point ( x0 y0 ) with respect to the set
of axes (xy). Let ( x' y' ) be a parallel set of axes passing through the centroid. Then
But by definition
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which gives
This is how the moment of area of a plane about an axis is related to the moment of the
same area about another axis parallel to the previous one but passing through the
centroid. Similarly it is easily shown that
.
and
We now solve an example to show the application of this theorem
Method of Virtual Work
So far when dealing with equilibrium of bodies/trusses etcetera, our strategy has been to
isolate parts of the system (subsystem) and consider equilibrium of each subsystem
under various forces: the forces that we apply on the system and those that the
surfaces, and other elements of the system apply on the subsystem. As the system size
grows, the number of subsystems and the forces on them becomes very large. The
question is can we just focus on the force applied to get it directly rather than going
through each and every subsystem. The method of virtual work provides such a scheme.
In this lecture, I will give you a basic introduction to this method and solve some
examples by applying this method.
Let us take an example: You must have seen a children's toy as shown in figure 1. It is
made of many identical bars connected with each other as shown in the figure. One of
the lowest bars is connected to a fixed pin joint A whereas the other bar is on a pin joint
B that can move horizontally. It is seen that if the toy is extended vertically, it collapses
under its own weight. The question is what horizontal force F should we apply at its
upper end so that the structure does not collapse.
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To see how many equations do we have to solve in finding F in the structure above, let
us take a simple version of it, made up of only two bars, and ask how much force F do
we need to keep it in equilibrium (see figure 2).
Let each bar be of length l and mass m and let the angle between them be è. The freebody
diagram of the whole system is shown above. Notice that there are four unknowns
- NAx , NAy , NBy and F - but only three equilibrium equations: the force equations
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and the torque equation
So to solve for the forces we will have to look at individual bars. If we look at individual
bars, we also have to take into account the forces that the pin joining them applies on
the bars. This introduces two more unknowns N1 and N2 into the problem (see figure 3).
However, there are three equations for each bar - or equivalently three equations above
and three equations for one of the bars - so that the total number of equations is also six.
Thus we can get all the forces on the system.
The free-body diagrams of the two bars are shown in figure 3. To get three more
equations, in addition to the three above, we can consider equilibrium of any of the two
bars. In the present case, doing this for the bar pinned at B appears to be easy so we
will consider that bar. The force equations for this bar give
And taking torque about B , taking N1 = 0 , gives
This then leads to (from the force equation above)
Substituting these in the three equilibrium equations obtained for the entire system gives
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Looking at the answers carefully reveals that all we are doing by applying the force F is
to make sure that the bar at pin-joint A is in equilibrium. This bar then keeps the bar at
joint B in equilibrium by applying on it a force equal to its weight at its centre of gravity.
The question that arises is if we have many of these bars in a folding toy shown in figure
1, how would we calculate F ? This is where the method of virtual work, to be developed
in this lecture, would come in handy. We will solve this problem later using the method of
virtual work. So let us now describe the method. First we introduce the terminology to be
employed in this method.
1. Degrees of freedom: This is the number of parameters required to describe the
system. For example a free particle has three degrees of freedom because we require x,
y , and z to describe its position. On the other hand if it is restricted to move in a plane,
its degrees of freedom an only two. In the mechanism that we considered above, there is
only one degree of freedom because angle è between the bars is sufficient to describe
the system. Degrees of freedom are reduced by the constraints that are put on the
possible motion of a system. These are discussed below.
2. Constraints and constraint forces: Constraints and those conditions that we put on
the movement of a system so that its motion gets restricted. In other words, a constraint
reduces the degrees of freedom of a system. Constraint forces are the forces that are
applied on a system to enforce a constraint. Let us understand these concepts through
some examples.
A particle in free space has three degrees of freedom. However, if we put it on a plane
horizontal surface without applying any force in the vertical direction, its motion is
restricted to that plane. Thus now it has only two degrees of freedom. So the constraint
in this case is that the particle moves on the horizontal surface only. The corresponding
force of constraint is the normal reaction provided by the surface.
Work and Energy
You have been studying in your school that we do work when we apply force on a body
and move it. Thus performing work involves both the application of a force as well as
displacement of the body. We will now see how this definition comes about naturally
when we eliminate time from the equation of motion.
The question that immediately comes to mind is why should we eliminate time from the
equation of motion. This is because when we follow the motion of a particle, we are
usually interested in velocity as a function of position. Secondly, if we write the equation
of motion in terms of time derivatives, it may make the equation difficult to solve. In such
cases eliminating time from the equation of motion helps in solving the equation. Let us
see this through an example.
Example: Consider the motion of a particle in a gravitational field of mass M .
Gravitational force on a mass m is in the radial direction and is given as
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Since the force in the radial direction, it is better to write the equation of motion in
spherical polar coordinates. For simplicity we consider the motion only along the radial
direction so that the equation of motion is written as
As you can see, integrating this equation to get r(t) as a function of time is very difficult.
On the other hand, let us eliminate time from the equation by using chain rule of
differentiation to get
,
where is the velocity in the radial direction. This changes the equation of motion
to
This equation is very easy to integrate and gives as a function of r, which can
hopefully be further integrated to get r as a function of time. Now we go back to what I
had said earlier that the definition of work and energy arises naturally when we eliminate
time from the equation of motion. Let us do that first for one dimensional case and
analyze the problem in detail.
Work and energy in one dimension
The equation of motion in one-dimension (taking the variable to be x, and the force to be
F ) is
Let us again eliminate time from the left-hand using the technique used above
to get
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On integration this equation gives
where xi and xf refer to the initial and final positions, and vi and vf to the initial and final
velocities, respectively. We now interpret this result. We define the kinetic energy of a
particle of mass m and velocity v to be
and the work done in moving from one position to the other as the integral given above
With these definitions the equation derived above tells us that work done on a particle
changes its kinetic energy by an equal amount; this known as the work-energy theorem .
You may ask: how do we know this equation to be true and consistent with our
observations? This is the question that was asked in the early eighteenth century when it
was not clear how to define energy, whether as mv or as mv2 ? The problem with the
definition as mv is that if two particles moving in the opposite directions have their
energies canceling each other and if they collide, they stop and all the energy is lost . On
the other hand, defining it proportional to v2 makes their energies add up and noting is
lost during collision; the energy just changes form but is conserved. Experimental
evidence for the latter was found by dropping weights into soft clay floors. It was found
that by increasing the speed of the weights by a factor of two made them sink in a
distance roughly four times more; increase in the speed by a factor of three made it nine
times more. That was the evidence in favor of kinetic energy being proportional to v2 .
Potential energy: Let us now define another related energy known as the potential
energy . This defined for a force field that may exist in the space, for example the
gravitational field or the electric field. Before doing that we first note that even in one
dimension, there are many different ways in which one can go from point 1 to point 2 .
Two such paths are shown in the figure below.
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On path A the particle goes directly from point 1 to 2 , whereas on path B it goes beyond
point 2 and then comes back. The question we now ask is if the work done is always the
same in going from point 1 to point 2. This is not always true. For example if there is
friction, the work done against friction while moving on path B will be more that on path
A. If for a force the work done depends on the path, potential energy cannot be defined
for such forces. On the other hand, if the work W12 done by a force in going from 1 to 2 is
independent of the path, it can be expressed as the difference of a quantity that depends
only on the positions x1 and x2 of points 1 and 2
(Question: If the work done is independent of path, what will be the work done by the
force field when a particle comes back to its initial position? ). We write this as
and call the quantity U(x) the potential energy of the particle. We now interpret this
quantity. Assume that a particle is in a force field F(x) . We now apply a force on the
particle to keep it in equilibrium and move it very-very slowly from point 1 to 2. Obviously
the force applied by us is - F(x) and the work done by us in taking the particle from 1 to
2, while maintaining its equilibrium, is
Thus for a given force field, the potential energy difference U(x2 ) - U(x1 ) between two
points is the work done by us in moving a particle, keeping it in equilibrium, from 1 to 2 .
Note that it is the work done by us - and not by the force field - that gives the difference
in the potential energy. By definition, the work done by the force field is negative of the
difference in the potential energy. Further, it is the difference in the potential energy that
is a physically meaningful quantity. Thus is we want to define the potential energy U(x)
as a function of x , we must choose a reference point where we take the potential energy
to be zero. For example in defining the gravitational potential energy near the earth's
surface, we take the ground level to be the reference point and define the potential
energy of a mass m at height h as mgh . We could equally well take a point at height h0
to be the reference point; in that case the potential energy for the same mass at height h
would be mg(h - h0 ) . Let us now solve another example.
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Rigid body dynamics III: Rotation and Translation
We have seen in the past two lectures how do we go about solving the rigid body
dynamics problem by considering the rate of change of angular momentum. In the
previous lecture, we concentrated on rotation about a fixed axis and solved problems
involving conservation of angular momentum about that axis. In this lecture we consider
what happens where an external torque is applied and also when the axis is allowed to
translate parallel to itself.
Let us first take the case when the axis is stationary and a torque is applied. Take for
instance your pen or a scale and hold it lightly at one of its ends so as to pivot it there.
Raise the other end so that the scale is horizontal and then leave it. You will see that the
scale swings down. I would like to calculate the speed of its CM when the scale is
vertical after being released from horizontal position (see figure 1). Assume that there is
no loss due to friction. In this case I will solve this problem in two ways and also
comment on a wrong way.
I take the mass of the scale to be m and its length l. Then its moment of inertia about
one of its ends is .
I first solve the problem using energy conservation. Since there is no loss due to friction
the total mechanical energy is conserved. Therefore the total mechanical energy is
conserved. Let us take the potential energy to be zero when the scale is horizontal.
Since the scale starts with zero initial angular speed, its total mechanical energy is zero.
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When the scale reaches the vertical position, its CM has moved down by a distance
so its potential energy is . If its angular speed at that position is ù, then by
conservation of energy
which gives
I now solve the problem by a direct application of torque equation. When the scale
makes an angle è from the horizontal (see figure 2), the torque on it is given as
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