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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