CLASSICAL ENGINEERING MECHANICS

R. Syam Sudhakar Rao | Associate Professor | Guru Nanak Institutions Technical Campus

CLASSICAL ENGINEERING MECHANICS

Short Answer Questions

Prepared by

R. Syam Sudhakar Rao

Associate Professor

DEPARTMENT OF MECHANICAL ENGINEERING

GURU NANAK INSTITUTIONS TECHNICAL CAMPUS IBRAHIMPATNAM, RANGA REDDY DISTRICT -501506, TELANGANA, INDIA.


UNIT-1

(1) Define Rigid body.

Ans: A Rigid body may be defined as a body

whose size and shape are unaffected by the

forces acting on it and the distance between

any two points of it is invariable. But, in

practice no solid body fulfils this

condition.

(2) State the theorem of transmissibility of

a force.

Ans: If a force acts at any point on a rigid

body, it may also be considered to act at

any other point on its line of action,

provided this is rigidly connected with the

body.

(or)

The point of application of a force may be

transmitted along its line of action without

changing the effect of the force on any rigid

body to which it may be applied.

(3) State the parallelogram law of forces.

Ans: If two coplanar concurrent forces

acting at a point be represented in magnitude

and direction by the adjacent sides of a

parallelogram, then their resultant is

represented in magnitude and direction by

the diagonal of the parallelogram passing

through that point.

R = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝐶𝑜𝑠 𝛼

Θ = 𝑡𝑎𝑛−1 (𝑄 𝑆𝑖𝑛 𝛼

𝑃+𝑄 𝐶𝑜𝑠 𝛼)


(4) The resultant of two forces is equal to

each of the force. Compute the angle between

them.

Ans: R = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝐶𝑜𝑠 𝛼 Given that R = P = Q

P = √𝑃2 + 𝑃2 + 2𝑃. 𝑃 𝐶𝑜𝑠 𝛼 1 + Cos α = 0.5

α = 120°

(5) If two forces P and Q act at right

angles, then R = _________ and θ = _______.

Ans: α = 90°, Cos α = 0

R = √𝑃2 + 𝑄2

Θ = 𝑡𝑎𝑛−1 (𝑄 𝑆𝑖𝑛 90

𝑃+𝑄 𝐶𝑜𝑠 90) = 𝑡𝑎𝑛−1 (

𝑄

𝑃)

(6) What is the law of Superposition?

Ans: The law of Superposition states that

“the action of a given system of forces on

a rigid body will in no way be changed if we

add or subtract from them another system of

forces in equilibrium.

(7) Define equilibrium.

Ans: When two or more forces act upon a body,

sometimes they may neutralize the effect of

one another and the body continues in its

state of rest. The body under such a system

of forces is said to be in equilibrium.

Similarly, a body will remain in equilibrium

if the resultant of forces acting upon it is

zero.


(8) Differentiate Resultant and Equilibrant.

Ans: Resultant: It is the single force which

produces same effect on the body as is

produced by a number of forces acting

together on the same body.

Equilibrant: The single Force that keeps the

body in equilibrium is called equilibrant.

(9) Explain non-coplanar concurrent forces

non-coplanar non-concurrent forces with

examples.

Ans: Non-coplanar concurrent forces: All

forces do not lie in the same plane, but

their lines of action pass through a single

point.

Ex: A tripod carrying a camera.

Non-coplanar non-concurrent forces: All

forces do not lie in the same p lane and

their lines of action do not pass through a

single point.

Ex: Forces acting on a moving bus.

(10) State Triangle law of forces.

Ans: Statement: “If two forces acting

simultaneously on a particle represented in

magnitude and direction by the two sides of

a triangle taken in order, their resultant

may be represented in magnitude and

direction by the third side of the triangle

taken in opposite order.”


(11) State Polygon law of forces.

Ans: Statement: “If a number of forces acting

simultaneously on a particle be represented

in magnitude and direction, by the side of

a polygon taken in order, then the resultant

of all these forces may be represented in

magnitude and direction by the closing side

of the polygon taken in opposite order.”

(12) Explain law of action and reaction.

Ans: Action and Reaction: When two bodies

are in contact, they exert force on each

other. One of these forces is called Action

and the other as Reaction. Always action and

reaction will be equal and when the surface

is smooth, they act normal to the surface in

contact.

Law: “Any pressure on a support causes an

equal and opposite pressure from the support

so that action and reaction are two equal

and opposite forces.”

(13) Write the principle of resolution of

forces.

Ans: “The algebraic sum of the resolved parts

of a number of forces in a given direction

is equal to the resolved part of their

resultant in the same direction.”


(14) Explain the term “Resultant force”

Ans: If a number of forces P, Q, R …. etc.

are acting simultaneously on a particle, it

is possible to find out a single force which

could replace them i.e. which would produce

the same effect as produced by all the given

forces. This single force is called

‘Resultant force’ and the given forces P, Q,

R …. etc. are called component forces.

(15) The resultant of two collinear forces

is equal to their _______

Ans: Algebraic sum.

(16) Explain the term Constraint with an

example.

Ans: Restriction to the free motion of a body

in any direction is called Constraint.

Ex: Ball will move down

Freely if there is

no string. So, here

String is a Constraint.

(17) Differentiate kinematics and kinetics.

Ans: If a body in motion due to a system of

forces, then the study of motion of the body

without considering the forces which cause

this motion is called kinematics.

If the forces and also their effects are

considered along with the motion of the body,

that branch of science is called kinetics.

(18) State equilibrium law.

Ans: Two forces can be in equilibrium only

if they are equal in magnitude, opposite in

direction and collinear in action.


(19) What is the limitation of the theorem

of transmissibility of a force?

Ans: The use of the theorem of

transmissibility of a force is limited to

those problems of statics in which we are

interested only in the conditions of

equilibrium of a rigid body and not with the

internal forces to which it is subjected.

(20) Differentiate Composition of forces and

resolution of force.

Ans: The reduction of a given system of

forces to the simplest system that will be

its equivalent is called the problem of

composition of forces.

The replacement of a single force by several

components which will be equivalent in

action to the given force is called the

problem of resolution of force.

(21) State and explain Lami’s theorem.

Ans: It states that, if three coplanar forces

acting at a point (i.e. concurrent) are in

equilibrium, each force will be proportional

to the sine of the angle between the other

two forces.

Explanation: Suppose three force F1, F2 and

F3 are concurrent at point O.

Let α = Angle between

F2 and F3 Β = Angle between

F1 and F3 γ = Angle between

F2 and F1 According to Lami’s definition,

𝑭𝟏

𝑆𝑖𝑛 𝛼=

𝑭𝟐

𝑆𝑖𝑛 𝛽=

𝑭𝟑

𝑆𝑖𝑛 𝛾 = constant


(22) What is meant by a free-body diagram?

Ans: A diagram drawn by separating a body

from a structure or machine and on which

actions and reactions acting without

physical constraints or supports is called

as free-body diagram.

(23) A man standing on a ladder leaning

against a wall and a floor is an example of

___________ system of forces.

Ans: Coplanar non-concurrent forces.

(24) State the theorem of three forces.

Ans: Three non-parallel forces can be in

equilibrium only when they lie in one plane,

intersect in one point and their free vectors

build a closed triangle.

(25) Write the equations of equilibrium for

a system of concurrent forces in a plane.

Ans:

Σ Xi = 0

Σ Yi = 0

Σ MA = 0

Where A is any point

on the plane.

(26) Explain the moment of a force.

Ans: Moment of a force about a point is the

measure of its rotational effect. Moment is

defined as the product of the force and the

perpendicular distance of the point from the

line of action of the force. The point about

which the moment is considered is called

moment centre and the perpendicular distance

of the point from the line of action of the

force is called moment arm.


(27) State the theorem of Varignon.

Ans: Statement: “The moment of the resultant

of two concurrent forces with respect to a

centre in their plane is equal to the

algebraic sum of the moments of the

components with respect to the same centre.”

Varignon is a French Mathematician (1654-

1722).

(28) Define couple.

Ans: Two parallel forces equal in magnitude

and opposite in direction and separated by

a definite distance are said to form a

couple.

(29) What are the characteristics of a

couple?

Ans: a) A couple consists of a pair of equal

and opposite parallel forces which are

separated by a definite distance.

b) The translational effect of a couple on

the body is zero.

c) The rotational effect (moment) of a couple

about any point is constant and it is equal

to the product of the magnitude of the forces

and the perpendicular distance between the

two forces.

d) The effect of a couple is unchanged if

the couple is rotated through any angle.

e) The effect of couple is unchanged if the

couple is shifted to any other position.

f) The effect of a couple is unchanged if

the couple is replaced by another pair of

forces whose rotational effect is the same.

(30) When can two equal and opposite forces

have zero moment?

Ans: When the two forces are collinear.


(31) Can a couple be balanced by a single

force applied to the body?

Ans: No, a couple can be balanced only by

another couple which is equal in moment

opposite in sign and coplanar in action with

the given couple.

(32) A system of two equal parallel forces

acting in opposite directions and separated

by a definite distance cannot be reduced to

one resultant force. (True/False)

Ans: True

(33) In a certain case, a body subjected to

a force system Σ Fi = 0, Σ M # 0. What is

your conclusion about the condition of the

body?

Ans: It is a body subjected to a couple.

(34) How to resolve a force acting at a point

on a body into a force acting at some other

suitable point on the body and a couple?

Ans:

F is a force acting on a body at A.

Apply equal and opposite forces of magnitude

F at B the system of forces in not disturbed.

Now the original force F at A and opposite

force F at B form a couple of magnitude F.X.

Thus, the given force F at A is replaced by

a force F at B and a couple of magnitude F.X.



a system of parallel forces in one plane.

Ans: Σ Yi = 0 ----- (I); Σ(YiXi) = 0 ---- (II)

When the equation (I) is satisfied, the

possibility of a resultant force vanishes.

When the equation (II) is satisfied, the

possibility of a resultant couple vanishes.

Hence, the conclusion is the system will be

in equilibrium when they are simultaneously

satisfied.

The same conditions of equilibrium can also

be expressed by two moment equations,

Σ(MA)i = 0; Σ(MB)i = 0

Where the line joining the points A and B

should not be parallel to the given force

system.

(36) Classify different types of system of

forces.

Ans: There are six types of force systems.

1. Concurrent force system in a plane.

2. Parallel force system in a plane.

3. General force system in a plane.

4. Concurrent force system in space.

5. Parallel force system in space.

6. General force system in space.

(37) What is meant by a statically

determinate and statically indeterminate

structure?

Ans: If equilibrium equations are sufficient

to analyse for unknown forces, structure is

said to be statically determinate.

If the equilibrium equations are not

sufficient to analyse the structure for the

unknown forces, the structure is said to be

statically indeterminate.


(38) Explain different types of supports and

support reaction.

Ans: (i) Frictionless support: The reaction

acts normal to the surface at the point of

contact as shown in figure.

Fig. Sphere resting on horizontal and inclined planes

(ii) Roller and Knife-edge supports: The

roller and knife-edge restrict the motion

normal to the surface of the beam AB. So,

the reactions RA and RB shall act normal to

the surface at the points of contact A and

B as shown.

Roller Support Knife-edge support

(iii) Hinged or pin supports: The hinge or

pin restricts the motion at the support end

of the beam both in horizontal as well as

vertical directions. Thus, there are two

independent reactions Fh and Fv acting on the

beam at the support end(A). These two


rectangular components can be combined in to

a single force or reaction RA. Therefore, the

reaction at the hinge or pin can be

represented by a single force RA.

Hinge

RA

Hinge or Pin supports

(iv) Built-in or fixed support: If the end

C of the beam CD is embedded in the concrete,

it restricts the motion of the end C in the

horizontal and vertical directions. It also

restricts the rotation of the beam AB about

point C. The reactions 𝑅𝐶𝑥 and 𝑅𝐶𝑦 therefore

shall be exerted both in the horizontal and

vertical directions accompanied by a

reaction couple 𝑀𝐶 as shown below.

C


concurrent forces in space.

Ans: If a system of concurrent forces in

space is in equilibrium, their resultant

must vanish. The equations of equilibrium

for concurrent forces are:

ΣXi = 0; ΣYi = 0; ΣZi = 0


(40) Explain stable, unstable and neutral

equilibrium with examples.

Ans: Stable equilibrium: A system is said to

be in stable equilibrium if, when displaced

from equilibrium, it experiences a net force

or torque in a direction opposite the

direction of the displacement.

Unstable equilibrium: A system is in

unstable equilibrium if, when displaced from

equilibrium, it experiences a net force or

torque in the same direction as the

displacement from equilibrium.

Neutral or Indifferent equilibrium: A system

is in neutral equilibrium if its equilibrium

is independent of displacements from its

original position.


(41) Define free vector, sliding vector,

bound vector, equal vectors, equivalent

(equipollent)vectors.

Ans: Free vector: Vector that can be placed

anywhere in space and moved parallel to

itself at will is called a free vector. We

do this when we add two vectors head to tail

using triangle law.

Sliding vector: Vector that may be moved

along its line of action so that its

magnitude, direction and line of action

remains the same is called a sliding vector.

A force acting on a rigid body can be treated

as a sliding vector.

Bound vector: A bound vector is a vector with

a well-defined point of application. A bound

vector is therefore specified by magnitude,

direction and its point of application. A

vector representing a force acting on a

particle can be treated as a bound vector.

Equal vectors: Two vectors are equal if they

have the same magnitude and direction.

Equivalent or equipollent vectors: Two

vectors are said to be equivalent or

equipollent if they produce the same effect

on a rigid body in a certain sense.

(42) The point of application of a force, F

= 5i + 10j – 15k is displaced from the point

(i+3k) to the point (3i-j-6k). Find the work

done by the force.

Ans: Displacement, d

= (3-1)i+(-1-0)j+(-6-3)k = 2i – j – 9k

Work done = F.d = (5i+10j–15k).(2i–j–9k)

= 10 – 10 + 135 = 135


(43) A force given by F = 3i+2j-4k is applied

at the point (1,-1,2). Find the moment of

the force F about the point (2,-1,3).

Ans: Moment, M = r x F

r = (1-2)i+(-1+1)j+(2-3)k = -i–k

M = (-i-k)x(3i+2j-4k) = (2i-7j-2k)


the general case of a system of forces in

space.

Ans: In the general case of a system of

forces in space, equilibrium can exist only

if both the resultant force R and the

resultant couple M vanish.

ΣXi = 0; ΣYi = 0; ΣZi = 0

Σ(Mx)i = 0; Σ(My)i = 0; Σ(Mz)i = 0

(45) Write the conditions of equilibrium for

any system of parallel forces in space.

Ans: ΣZi = 0; Σ(ZiYi) = 0; Σ(ZiXi) = 0

(46) A force of 22 N acts through the point

A (4,-1,7) in the direction of vector 9i+6j-

2k. Find the moment of the force about the

point O(1,-3,2).

Ans: Unit vector in the direction of vector

9i+6j-2k is, 9𝑖+6𝑗−2𝑘

√92+62+(−2)2=

9𝑖+6𝑗−2𝑘

11

Force vector, F = 22 x 9𝑖+6𝑗−2𝑘

11 = 18i+12j-4k

Position vector, r = (4-1)i+(-1+3)j+(7-2)k

= 3i+2j+5k

Therefore, M = rxF

= (3i+2j+5k)x(18i+12j-4k)

= (-68i+102j)


(47) A force F = 2i+4j-3k is applied at a

point P (1,1,-2). Find the moment of the

force F about the point (2,-1,2).

Ans: Position vector, r = -i+2j-4k

Moment, M = rxF = 10I-11J-8K

(48) What is the angle between the space

diagonals of a cube?

Ans: cos-1(1

3)

(49) Find the shortest distance from the

origin to the line passing through A(-2,1,3)

and B(4,5,0).

Ans: 3.04 units

(50) Find the moment about the point

O(-2,3,5) of the force represented in

magnitude and direction by the line joining

P(1,-2,4) and Q(5,2,3).

Ans: F = 4i+4j-k; r = 3i-5j-k; M = 9i-j+32k

(51) A force 300N acts through a point

P(1,6,-5) and directed towards Q(0,4,3).

Find the moment of the force about a point

A(1,0,-1).

Ans: 400i+400j+600k


UNIT-2

(1) What is meant by friction?

Ans: When a body moves or tends to move over

another body, a force opposing the motion

develops at the contact surfaces. This force

which opposes the movement or the tendency

of movement is called frictional force of

friction.

(2) Differentiate dry friction and fluid

friction.

Ans: The friction between dry surfaces in

contact is called dry friction. It is also

called coulomb friction. The major cause of

such friction is believed to be the

interlocking of microscopic protuberances

(i.e. minute projections on the surfaces)

which oppose the relative motion.

The friction between two surfaces in the

presence of fluid is called fluid friction.

(3) Differentiate limiting friction, static

friction and dynamic friction.

Ans: The maximum value of frictional force

when the motion is impending is known as

limiting friction. It may be noted that when

the applied force is less than the limiting

friction, the body remains at rest and such

a frictional force is called static friction

which may have any value between zero and

the limiting friction. If the value of the

applied forces exceeds the limiting

friction, the body starts moving over the

other body and the frictional resistance

experienced by the body while moving is known

as dynamic friction.


(4) Write the laws of dry friction.

Ans: 1. The total friction that can be

developed is independent of the magnitude of

the area of contact.

2. The total friction that can be developed

is proportional to the normal force.

3. For low velocities of sliding, the total

friction that can be developed is

practically independent of the velocity.

(5) Explain Angle of friction and Angle of

limiting friction.

Ans:

Consider the block shown in the above figure

subjected to a pull P. Let F be the

frictional force developed and R the normal

reaction. F and R can be combined graphically

to get the reaction which acts at an angle

θ to normal reaction.

This angle θ, called the angle of friction

and is given by tan θ = 𝐹

𝑅

As the frictional force increases, the angle

θ increases and it can reach the maximum

value (α) when limiting value of friction is

reached. At this stage, tan α = 𝐹

𝑅 and this

value of α is called angle of limiting

friction.


(6) What is meant by Angle of repose?

Ans: Angle of repose is the maximum angle of

inclination that an inclined plane may have

with the horizontal before a body lying on

the plane begins to slide down under the

action of its own weight.

(7) Differentiate sliding friction and

rolling friction.

Ans: Sliding friction: It is the frictional

force experienced by a body when the body

slides over another body.

Ex: When a piston slides within the cylinder,

the friction experienced by the piston is

sliding friction.

Rolling friction: It is the frictional force

which opposes or tends to oppose the rolling

of one body over another body.

Ex: The frictional force experienced by the

balls of a ball bearing is the rolling

friction.

(8) What is meant by coefficient of friction?

Ans: From the laws of static friction we know

that the limiting friction is directly

proportional to the normal reaction between

the two bodies.

i.e. F α R, where F = limiting friction,

R = Normal reaction

Therefore, F = μR, where μ is a constant.

and μ = 𝐹

𝑅

This constant is known as coefficient of

friction. Hence, the coefficient of friction

is defined as the ratio of the limiting

friction to the normal reaction.


(9) Explain cone of friction and its

significance.

Ans:

For a pair of body in contact with each

other,

Let ϕ = Angle of friction

N = Normal reaction

F = Limiting force

R = Resultant reaction i.e. resultant of

N and F

If we describe a right circular cone of apex

angle 2ϕ about the line of action N, the cone

thus formed is known as cone of friction.

The significance of the cone of friction is

that in which ever direction the body may

tend to move under the action of external

force, the line of action of resultant

reaction (R) will always lie on the slant

surface of the cone of friction.


(10) Give some examples indicating useful

and harmful effects of friction.

Ans: Useful effects: (a) When we write

anything on a sheet of paper or anywhere

else, it is due to friction that the pen does

not slip from the grip of the hand.

(b) It is due to friction that we can walk

on road.

(c) A ladder placed with its one end against

the ground and the other end against a

vertical wall, does not slip due to friction

at the ground and at the vertical wall.

(d) It is due to friction that motion and

power can be transmitted from one pulley to

another by means of belt drive.

(e) It is due to friction that the car wheels

advance forward while revolving.

Harmful effects: (a) It is due to friction

that bearings heated and their longevity

shortened.

(b) Considerable amount of power is lost in

friction in different parts of a machine. As

a result, output power of a machine is always

much less than its input power.

(c) It is due to friction in different parts

of an internal combustion engine that the

parts get worn out and fuel consumption goes

on increasing with the years of service.

(11) Define a simple machine.

Ans: A simple machine will have single point

of application, each effort and load and has

a simple mechanism to perform the work.

Ex: Levers, Pulleys, Wheels and axles.


(12) Define Mechanical Advantage.

Ans: The ratio of load lifted and the effort

applied is known as mechanical advantage.

Mechanical Advantage = 𝐿𝑜𝑎𝑑 𝑙𝑖𝑓𝑡𝑒𝑑

𝐸𝑓𝑓𝑜𝑟𝑡 𝑎𝑝𝑝𝑙𝑖𝑒𝑑

(13) Define velocity ratio.

Ans: It is the ratio between the distance

moved by the effort (D) and the distance

moved by the load (d).

Velocity ratio = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑒𝑓𝑓𝑜𝑟𝑡

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑

(14) What is an ideal machine?

Ans: If the friction of a machine is

negligible, the machine is known as ideal

machine. In ideal machine, the efficiency is

100% and output is equal to input. In actual

practice, no machine is ideal. A part of the

work done is always lost in overcoming the

friction and hence the work done on the

machine is always greater that the work done

by the machine.

(15) Define efficiency of a machine.

Ans: It is the ratio of output and input of

the machine. It is denoted by ‘η’ and

expressed as percentage.

Efficiency, η = 𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡 x 100

(16) Rolling friction is ______ proportional

to the radius of the rolling body.

Ans: Inversely.


(17) What is a wedge?

Ans: The wedge is a simple contrivance

commonly used for raising or lowering heavy

blocks, machinery, precast beams etc. The

weight of the wedge is very small as compared

to the weight lifted. Hence, the weight of

wedges will be neglected in most of the

problems.

(18) Define centre of gravity and centroid.

Ans: Centre of gravity can be defined as the

point through which the resultant of force

of gravity(weight) of the body acts.

(19) Differentiate centre of gravity and

centroid.

Ans: (a) The term centre of gravity applies

to bodies with mass and weight and centroid

applied to plane areas such as triangles,

circles, quadrilateral etc., which have only

areas and no masses.

(b) The centre of gravity of a body is a

point through which the resultant

gravitational force(weight) acts for any

orientation of the body whereas the centroid

is a point in a plane area such that the

moment of area about any axis through that

point is zero.

(20) Define first moment of an area about an

axis.

Ans: First moment of an area (or) centroid

of a plane figure is defined as the point

through which the entire area of the plane

figure is assumed to be concentrated.


(21) Locate the centroid of a quarter circle

of radius 'r'.

Ans.

Centroid = (4𝑟

3𝜋,4𝑟

3𝜋)

(22) Locate the centroid of a half and

quarter circular arcs.

Ans:

Centroid,

�̅� = 2𝑟 𝑆𝑖𝑛 𝛼

2𝛼

For a semicircular arc,

2α = π �̅� = 2𝑟

𝜋

For a quarter circular arc,

Along the line of symmetry:

�̅� = 2𝑟 𝑆𝑖𝑛 𝛼

2𝛼 =

2𝑟 𝑆𝑖𝑛 45

2 𝑥 (𝜋 4⁄ )=

4𝑟 √2⁄

𝜋=

2𝑟 √2

𝜋

Along the vertical axis:

�̅� Sin45 = 2𝑟 √2

𝜋 x

1

√2 =

2𝑟

𝜋


(23) State Pappus Theorems.

Ans: (a) The area of the surface generated

by rotating any plane curve about a non-

intersecting axis in its plane is equal to

the product of the length L of the curve and

the distance traveled by its centroid.

(b) The volume of the solid generated by

rotating any plane figure about a non-

intersecting axis in its plane is equal to

the product of the area A of the figure and

the distance traveled by its centroid.

(24) Write the co-ordinates of centre of

gravity for the general spandrel shown in

figure.

Ans: Area = 𝑎 ℎ

(𝑛+1)

�̅� = (𝑛+1)

(𝑛+2) a

�̅� = (𝑛+1)

4𝑛+2) h


(25) Differentiate centre of parallel forces

and centre of gravity.

Ans: If number of parallel forces F1, F2,…..Fn

applied at given points A1, A2,…… An

respectively, we conclude that there is one

and only point through which the resultant

always passes regardless of the direction in

which the parallel forces act through their

given points of application. This point is

called the centre of parallel forces for the

given system of forces applied at the given

system of points.

The centre of gravity of a body is that point

through which the resultant of the

distributed gravity forces passes regardless

of the orientation of the body in space. From

this definition, it follows that the centre

of gravity of a rigid body is the centre of

parallel gravity forces acting on the

various particles of the body.


UNIT-3

(1) Explain the term Moment of Inertia.

Ans:

Consider the area shown in figure. dA is an

elemental area with coordinates as x and y.

Moment of Inertia of the area about x-axis,

𝐼𝑥𝑥 = Σ𝑦2dA

Moment of inertia of the area about y-axis,

𝐼𝑦𝑦 = Σ𝑥2dA

The term 𝑥2dA (or) 𝑦2dA is called the second moment of area.

(2) Define polar moment of inertia.

Ans: The polar moment of inertia of any plane

figure with respect to a point O in its plane

is equal to the sum of moments of inertia of

the figure with respect to two orthogonal

axes through that point and also in the plane

of figure.

Ex: Polar moment of inertia of a circular

area with respect to its centre is,

J = 𝜋 𝑑4

32 , where d is diameter.


(3) What is meant by radius of gyration?

Ans: Dividing moment of inertia with respect

to a certain axis by the cross-sectional area

of the figure, a quantity having the

dimension of length to the second power is

obtained. This length is called the radius

of gyration of the figure with respect to

that axis. Using the notation K for radius

of gyration, we have

𝐾𝑥 = √𝐼𝑥

𝐴 𝐾𝑦 = √

𝐼𝑦

𝐴

(4) Write parallel-axis theorem for moments

of inertia of plane figures.

Ans: The moment of Inertia of a plane area

with respect to any

axis on its plane is

equal to the moment of

inertia with respect to

a parallel centroidal

axis plus the product

of the total area and

the square of the

distance between the

two axes.

𝐼𝐴𝐵 = 𝐼𝐺 + Aℎ2

(5) Define mass moment of inertia of a body

about an axis.

Ans: The mass moment of inertia of a body

about a particular axis is defined as the

product of the mass and the square of the

distance between the mass centre of the body

and the axis.


(6) Explain Product of Inertia.

Ans:

Consider the area shown in figure. dA is an

elemental area with coordinates as x and y.

The Product of inertia with respect to x and

y axes is given by

𝐼𝑥𝑦 = ∫ 𝑥𝑦 𝑑𝐴

(7) If a figure has an axis of symmetry, then

the product of inertia is __________.

Ans: Zero

(8) Explain Principal axes and centroidal

principal axes.

Ans: The product of inertia changes

continuously with the rotation of the axes,

there must be certain directions of the axes

for which this quantity vanishes. The axes

taken in these directions are called the

principal axes of the figure.

Usually the centroid of the figure is taken

as the origin of co-ordinates and the

corresponding principal axes are then

centroidal principal axes.


(9) Write parallel axis theorem for product

of inertia.

Ans: If the product of inertia 𝐼�̅�𝑦 for the centroidal axes x and y is known, the product

of inertia 𝐼𝑥𝑦 for parallel axes X and Y is

given by the expression,

𝐼𝑥𝑦 = 𝐼�̅�𝑦 + Aab where a and b are the co-ordinates of the

centroid C with respect to new axes and ‘A’

is the area of the figure.

(10) What is the moment of inertia of a

circle of diameter d about the centroidal

axis?

Ans: 𝐼𝑥𝑥 = 𝜋 𝑑4

64; 𝐼𝑦𝑦 =

𝜋 𝑑4

64; 𝐼𝑧𝑧 =

𝜋 𝑑4

32

(11) Differentiate between "mass moment of

inertia" and "area moment of inertia".

Ans: Mass moment of inertia gives a measure

of the resistance that body offers to change

in angular velocity and accordingly is used

in conjunction with rotation of rigid

bodies.

Area moment of inertia is essentially a

measure of resistance to bending and is

applied while dealing with the deflection or

deformation of members in bending.

(12) Determine the mass moment of inertia of

a rectangular plate of size a x b and

thickness t about its centroidal axis.

Ans: 𝐼𝑥𝑥 = 𝑀𝑏2

12; 𝐼𝑦𝑦 =

𝑀𝑎2

12; 𝐼𝑧𝑧 =

𝑀(𝑎2+𝑏2)

12

where a = width, b = depth, t = thickness

and M = mass of the plate = density x volume


(13) What is the moment of inertia of a

sphere?

Ans: 𝐼𝑥𝑥 = 𝐼𝑦𝑦 = 2 𝑀 𝑅2

5; 𝐼𝑧𝑧 =

4 𝑀 𝑅2

5

where M = mass of the sphere, R = radius.

(14) What is the mass moment of inertia of

a hollow cylinder with outer radius, inner

radius and length as R, r and L respectively?

Ans: 𝐼𝑥𝑥 = 𝐼𝑦𝑦 = 𝑀(𝑅2+𝑟2)

4; 𝐼𝑧𝑧 =

𝑀(𝑅2+𝑟2)

2

where M = mass of cylinder.

(15) State perpendicular axis theorem.

Ans: Perpendicular axis theorem: The moment

of inertia of a plane area about an axis

perpendicular to the plane of the area is

equal to the sum of the moments of inertia

of the lamina about the two axes at right

angles to each other and intersecting each

other at the point where the perpendicular

axis passes through it.

(16) Why is moment of inertia called as

second moment of area?

Ans: Second moment of area = The moment of

area is multiplied by the perpendicular

distance between the centre of gravity of

the area and axis. = A r x r = Ar2

Moment of inertia is also equal to Ar2.

Therefore, moment of inertia called as

second moment of area.


UNIT-4 & UNIT-5

(1) The slope of displacement-time diagram

at any instant gives _________ of the

particle at that instant.

Ans: Velocity.

(2) The slope of velocity-time curve at any

instant gives ________ of the particle.

Ans: Acceleration.

(3) The area under velocity-time curve

between the given interval of time gives the

___ by the particle during the same interval.

Ans: The change in the displacement or the

distance travelled.

(4) The area under acceleration-time curve

between the given interval of time gives the

___ of the particle during the same interval.

Ans: Change in velocity.

(5) If the displacement-time curve is a

polynomial of degree ‘n’, then the velocity-

time curve will be a polynomial of degree

_______ and the acceleration-time curve will

be a polynomial of degree ______.

Ans: (n-1); (n-2)

(6) What is the principle of conservation of

energy?

Ans: Energy cannot be created or destroyed,

it can be transformed or transferred from

one form to another. The total amount of

energy in a closed system doesn't change.


(7) Differentiate the kinematics and

kinetics.

Ans: Kinematics describes the motion of the

bodies and deals with finding out velocities

or accelerations for various objects.

Kinetics deals with the forces or torque

applied on a body.

(8) The equation of motion of a body moving

in a straight line is given in terms of

displacement and time. How the (i) velocity

at time t (ii) acceleration at time t and

(iii) maximum velocity are determined?

Ans: (i) Velocity at time t = 𝑑𝑥

𝑑𝑡;

(ii) Acceleration at time t = 𝑑2𝑥

𝑑𝑡2

(iii) Maximum velocity = 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡

𝑡𝑖𝑚𝑒

(9) Write the principles of dynamics.

Ans: First law: Every object continues in

its state of rest or of uniform motion in a

straight line except in so far as it may be

compelled by force to change that state. This

is also called law of inertia.

Second law: The acceleration of a given

particle is proportional to the force

applied to it and takes place in the

direction of the straight line in which the

force acts.

Third law: To every action there is always

an equal and contrary reaction, or the mutual

actions of any two bodies are always equal

and oppositely directed.


(10) Define plane motion of a rigid body.

Ans: If the motion of a rigid body that a

certain plane in the body always remains in

a fixed plane, then the path of each particle

of the body is a plane curve parallel to the

fixed plane and the body is said to have

plane motion. (or) A body is said to have

plane motion if it possesses translation and

rotation simultaneously.

Ex: (i) A wheel rolling on a straight line

and (ii) A rod sliding against a wall at one

end the floor at the other end.

(11) Classify different types of plane

motions.

Ans. A body undergoes three types of plane

motions:

1) Translation: The particle constituting

the rigid body move in parallel planes and

travel the distance.

2) Rotation: The body rotates about a fixed

point and all the particles constituting the

body move in a circular path.

3) General plane motion: It is combined

motion of translation and rotation.

(12) In plane motion of a rigid body, the

velocity of any point P in the body is

obtained as the geometric sum of the velocity

of the chosen pole A and the relative _______

of P with respect to A.

Ans: Rotational velocity (�̅�𝑝 = �̅�𝑎 + �̅�𝑝/𝑎)

(13) A bus is travelling at a uniform speed

of 30 kmph. What is the velocity of the point

of contact of the tyre with the road?

Ans: Zero


(14) State the principle of conservation of

momentum.

Ans: The principle of conservation of

momentum may be stated as, the momentum is

conserved in a system in which the resultant

force is zero. (or) In a system, if the

resultant force is zero then the initial

momentum will remain equal to final

momentum. It must be noted that conservation

of momentum applies to entire system and not

to individual elements of the system.

(15) Explain D’Alembert’s principle in

rectilinear translation.

Ans: The equation of motion of the particle

ΣF = ma can be written in the form ΣF–ma =0

which means that the resultant of the

external forces (ΣF) and the force (-ma) is

zero. The force (-ma) is called the inertia

force. The equation of dynamic equilibrium

of the particle is ΣF + (-ma) = 0

D’Alembert’s Principle: To write the

equation of dynamic equilibrium of a

particle add a fictious force equal to the

inertia force to the external forces acting

on the particle and equate the sum to zero.

This concept is known as D’Alembert’s

principle.

(16) Write the equation of rectilinear

motion.

Ans: The differential equation of

rectilinear motion of the particle is given

by 𝑊

𝑔 �̈� = 𝑋; where X = The resultant force

acting, �̈� = Acceleration.


(17) Write the work-energy equation in

rectilinear translation and its statement.

Ans: 𝑊

𝑔 �̇�2

2 -

𝑊

𝑔 �̇�02

2 = ∫ 𝑥 𝑑𝑥

𝑥

𝑥0

Statement: “The work-energy principle may be

stated as the work done by a system of forces

acting on a body during a displacement is

equal to the change in kinetic energy of the

body during the same displacement”

(18) A man weighing W Newton entered a lift

and started moving with an acceleration of

a m/s2. Find the force exerted by the man on

the floor of the lift when (i) the lift is

moving down and (ii) the lift is moving up.

Ans:

The lift is moving up

ΣFy = 0

FN -W + 𝑊

𝑔 a = 0

FN = W(1 - 𝑎

𝑔)

The lift is moving down

ΣFy = 0

FN -W - 𝑊

𝑔 a = 0

Lift-up Lift-down FN = W(1 + 𝑎

𝑔)

(19) Differentiate between rectilinear and

curvilinear motion?

Ans: A particle is said to be a rectilinear

motion, if the path traced by it is a

straight line.

When a moving particle describes a curved

path, it is said to have curvilinear motion.


(20) What is D’Alembert’s principle in

curvilinear motion.

Ans: In case of any rigid body that has

curvilinear translation, all the particles

of the body have the same motion and hence

the same acceleration. If the inertia force

is added to each particle, the resultant of

these inertia forces balances the resultant

of the active forces external to the body

and we have a system of forces in

equilibrium.

Σ𝐹𝑥 + (-m𝑎𝑥) = 0; Σ𝐹𝑦 + (-m𝑎𝑦) = 0

This is D’Alembert’s principle in

curvilinear motion.

(21) Write the differential equations of

curvilinear motion.

Ans: 𝑊

𝑔 �̈� = X;

𝑊

𝑔 �̈� = Y

where X, Y are components parallel to the

co-ordinate axes x, y of the resultant acting

force and �̈�, �̈� are the corresponding

components of acceleration.

(22) Normal acceleration (𝒂𝒏) of a particle at a point is equal to the square of its ____

divided by the radius of curvature of the

path at that point.

Ans: Speed

Explanation: 𝑎𝑛 = 𝑣2

𝜌

The direction of normal acceleration is

always directed towards the centre of

curvature of the path. This normal

acceleration is also called centripetal

acceleration or centre-seeking acceleration.


(23) The change in speed is accounted for by

the ________ acceleration alone, while

change of direction of motion is accounted

for by the ________ acceleration alone.

Ans: Tangential, Normal.

(24) What is trajectory?

Ans: The path traced by the projectile is

called as its trajectory.

(25) Write the equation of trajectory for a

projectile.

Ans. y = a tan α - 𝑔 𝑠𝑒𝑐2 𝛼

2 𝑣02 𝑥2

(26) For a given initial velocity 𝒗𝟎, the maximum range is obtained when the angle of

projection α = _______ and the maximum range

is _________.

Ans: α = 45°; Maximum range = 𝑣02

𝑔

(27) A body is projected at such an angle

that the horizontal range is three times the

maximum height. Find the angle of

projection.

Ans: Range = 𝑢2 sin2𝛼

𝑔;

Maximum height = 𝑢2 𝑠𝑖𝑛2 𝛼

2𝑔

Given that 𝑢2 sin2𝛼

𝑔 = 3 x

𝑢2 𝑠𝑖𝑛2 𝛼

2𝑔

2 sin α cos α = 3 x 𝑠𝑖𝑛2 𝛼

2; α = 53°10l


(28) In what proportion will the maximum

range of a projectile be increased if the

initial velocity is increased by 10 percent?

Ans: 21 percent

(29) Explain impulse and impulsive force

with examples.

Ans: The impulse of a force acting on a body

for any time is the product of the force and

the time during which the force acts. Let F

be the force acting on a mass m for time t,

then the impulse of the force on the mass is

given by I = F x t

If a large amount of force acts on a body

for a very short period of time, then the

force is called Impulsive force.

Ex: a) When a batsman hits a cricket ball,

the time during which the force exerted by

the bat on the ball is very small. This force

is an Impulsive force.

b) When a nail is struck by a hammer, the

force with which the nail is struck by the

hammer is Impulsive force.

c) When a bullet is fired from a gun, the

force with which the bullet is pushed is

Impulsive force.

(30) Write the impulse-momentum equation and

mention its application.

Ans: Impulse-momentum equation is,

The principle of impulse and momentum is

useful when dealing with a system of

particles. Example is a gun firing a bullet.


(31) Write the relation between impulse of

a force and momentum (Impulse-momentum).

Ans: Momentum = mass x velocity.

Impulse = Final momentum – Initial momentum.

The total change in momentum of a particle

during a finite interval of time is equal to

the impulse of the acting force during the

same interval.

(32) Define the terms work and energy

Ans: Work is the product of force and

distance. Energy is the capacity to do work.

(33) Set up an expression for the work done

by torque.

Ans: The work done by torque is given by the

product of torque and the angular

displacement.

Work done = T x ω = T x 2𝜋𝑁

60

(34) Write the work energy equation for plane

motion of a particle.

Ans: Work energy equation is given by,

U1-2 = T2 – T1

Where LHS = Work of the force exerted on the

particle during its displacement from

position 1 to 2 and RHS = The change in

kinetic energy of the particle.

(35) State work-energy principle for a rigid

body in plane motion.

Ans: Work done by a rigid body undergoing

general plane motion = Change in KE of the

rigid body due to linear motion + Change in

KE of the rigid body due to rotary motion.


(36) Is the work a scalar/vector quantity?

Ans: Work is a scalar quantity.

(37) What is the advantage of work-energy

theorem?

Ans: The equation of work-energy theorem is

useful in cases where the acting force is a

function of displacement and where we are

particularly interested in the velocity of

the particle as a function of displacement.

(38) State work and energy equation for

curvilinear motion.

Ans: Statement: “The change in kinetic

energy of the particle between any two

positions on its path is equal to the work

of all forces acting upon it during the

motion between these two points”

𝑊𝑣22

2𝑔 -

𝑊𝑣12

2𝑔 = ∫ 𝑠 𝑑𝑠

𝑠2𝑠1

(39) The angular velocity of a rotating body

is �̇� and angular acceleration �̈�. Calculate tangential and normal components of

acceleration of any point in the body at a

distance r from the axis of rotation.

Ans:

Tangential acceleration, 𝑎𝑡 = 𝑑𝑣

𝑑𝑡 = r �̇� = r �̈�

Normal acceleration, 𝑎𝑛 = 𝑣2

𝑟 = 𝜔2r = r�̇�2

(40) Define moment of momentum

Ans. The product of mass moment of inertia

and the angular velocity of a rotating body

is called moment of momentum or angular

momentum.


(41) Define compound pendulum.

Ans: Any rigid body so suspended that it is

free to rotate about a fixed horizontal axis

through any point O and normal to the

vertical x-y plane is called a compound

pendulum.

(42) Define Impact and Line of Impact.

Ans: Impact: The phenomenon of collision of

two bodies which occurs in a very small

interval of time and during which the two

bodies exert very large force on each other

is called an Impact.

Line of Impact: The common normal to the

surfaces of the two bodies in contact during

the impact is called line of impact.

(43) Differentiate direct impact and

indirect(oblique) impact.

Ans: Direct Impact: When two bodies, moving

along the same line, collide the impact is

called direct impact.

Oblique impact: When two bodies, moving

along different lines, collide the impact is

called oblique impact.

Direct impact Oblique impact


(44) Explain the term direct central impact.

Ans: In case of impact of two spheres, if

the velocities before and after impact are

directed along the line joining the centres

of the two spheres, then impact is called

direct central impact.

(45) Differentiate central and non-central

impact.

Ans: When the mass centres of the colliding

bodies lie on the line of impact, it is

called central impact, otherwise it is

called non-central impact or eccentric

impact.

(46) Explain the period of deformation.

Ans: Just after the impact the two colliding

bodies deform. The time interval from the

first contact to the maximum deformation is

called the period of deformation.

(47) Define coefficient of restitution(e).

Ans: The coefficient of restitution of two

colliding bodies may be taken as the ratio

of impulse during restitution period to the

impulse during the deformation period. This

is also equal to the ratio of relative

velocity of separation to the relative

velocity of approach of the colliding

bodies, the relative velocity being measured

in the line of impact.

Coefficient of restitution,

e = - [𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑒𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ] = - [

𝑣𝑏 𝑙 − 𝑣𝑎

𝑙

𝑣𝑏− 𝑣𝑎]

𝑣𝑎, 𝑣𝑏 are initial velocities of colliding

bodies and 𝑣𝑎𝑙 , 𝑣𝑏

𝑙 are final velocities.


(48) Explain plastic impact, elastic impact

and semi-elastic impact.

Ans: Plastic impact: Consider for an impact

between two putty balls. For Plastic impact

coefficient of restitution, e = 0

Therefore 𝑣𝑏𝑙 = 𝑣𝑎

𝑙 = 𝑣𝑙 Thus, after plastic impact, the final

velocities of both bodies become equal (𝑣𝑙) and they move together as one body.

Elastic impact: Consider for example an

impact between two hardened and polished

steel balls. For Elastic impact coefficient

of restitution, e = 1

In case of an elastic impact, both momentum

and energy are conserved.

Semi-elastic impact: When a body undergoes

semi-elastic collision, K.E is not conserved

and is less than the K.E before collision.

The relative velocity after impact is

smaller than before.

𝒗𝟏𝒍 − 𝒗𝟐

𝒍 = −𝒆 (𝒗𝟏 − 𝒗𝟐) where ‘e’ is a numerical factor less than unity.

(49) After an elastic impact between two

equal masses, the two masses ______ their

velocities.

Ans: Exchange

(50) In case of an elastic impact, if a body

strikes another body of equal mass at rest

then what will happen?

Ans: The striking ball simply stops after

having imparted its velocity to the other

ball.


(51) Write the equation of motion for a rigid

body rotating about a fixed axis.

Ans: The equation of motion for a rigid body

rotating about a fixed axis is given by,

M = I �̈� Where I = Mass moment of inertia of the body

with respect to the axis of rotation.

M = The resultant moment of all external

forces with respect to the same axis.

�̈� = Angular acceleration of the body.

(52) Write D’Alembert’s principle in

rotation of a body about a fixed axis.

Ans: When a body rotates about a fixed, any

mass particle travels in a circular path of

radius centered on the axis of rotation and

normal thereto. Attaching to each particle

its normal and tangential inertia forces and

in accordance with D’Alembert’s principle,

we equilibrate the real forces and establish

dynamic equilibrium.

(53) When a rigid body rotates about a fixed

axis, write general expressions for the

angular velocity and the angle of rotation.

Ans: The general expression for the angular

velocity is,

�̇� = 𝑀

𝐼 t + 𝜃0̇

The general expression for the angle of

rotation is,

θ = 𝑀 𝑡2

2 𝐼 + 𝜃0̇t + 𝜃0


(54) A body is having motion of translation

as well as motion of rotation. How will you

determine the total kinetic energy of the

body?

Ans: If a body is having motion of

translation as well as motion of rotation,

then total kinetic energy is given by,

Total KE = ½ mv2 + ½ Iω2

(55) Write the energy equation for rotating

bodies.

Ans: The energy equation for rotating bodies

states that the total energy in the kinetic

energy of the rotating body during any angle

of rotation is equal to the total work done

by the external forces during the same angle

of rotation.

Mathematically it can be written as below:

𝐼�̇�2

2−

�̇�02

2= ∫ 𝑀 𝑑𝜃

𝜃

𝜃0

(56) Show that when a particle moves with

simple harmonic motion, it's time for

complete oscillation is independent of the

amplitude of its motion.

Ans: Time taken by a particle to undergo one

complete oscillation is given by, T = 2π /ω,

which is called time period and is

independent of amplitude of its motion.

(57) Find the angular Acceleration of a

flywheel of an engine, which weighs 1500 N

and has a radius of gyration 0.6 m, if the

wheel is subjected to a torque of 2000 Nm.

Take g = 9.8 m/s2

Ans: Torque, T = I α = mk2 α

2000 = 1500

9.8 x (0.6)2 x α; α = 36.29 rad/s2


(58) Explain Instantaneous centre of

rotation in plane motion.

Ans: In plane motion of a rigid body, if a

point I is found such that at any instant at

that particular point, the velocity is zero,

the point is defined as instantaneous centre

of rotation or instantaneous centre of

velocity. Velocity of the rigid body at that

particular instant would consist only of a

rotation of the body about instantaneous

centre of rotation. This point may or may

not lie within the body.

(59) Explain centrode, Instantaneous axis,

axode, space centrode and body centrode.

Ans: Centrode: The instantaneous centre is

not a fixed point. Its location keeps

changing at every instant and the locus of

all such instantaneous centres is known as

centrode.

Instantaneous axis: A line drawn through an

instantaneous centre and perpendicular to

the plane of motion is called instantaneous

axis.

Axode: The locus of instantaneous axes is

known as axode.

Space centrode: The locus of the

instantaneous centre in space during a

definite motion of the body is called space

centrode.

Body centrode: The locus of the

instantaneous centre relative to the body

itself is called body centrode.

(60) Differentiate pole and instantaneous

centre.

Ans: A pole is a point fixed in the body,

but the instantaneous centre is not fixed.


(61) Write the equations of plane motion.

Ans: 𝑊

𝑔 �̈�𝑐 = X

𝑊

𝑔 �̈�𝑐 = Y

𝐼𝑐 �̈� = 𝑀𝑐

where X, Y, 𝑀𝑐 are the applied external

forces.

(62) Explain D’Alembert’s principle in plane

motion.

Ans: To the applied external forces defined

by X, Y, and 𝑀𝑐 add inertia forces (𝑊

𝑔 �̈�𝑐) and

(𝑊

𝑔 �̈�𝑐) and the inertia couple (𝐼𝑐�̈�) to obtain

dynamic equilibrium of the body. The

components of inertia forces must be applied

to the body at its centre of gravity, the

inertia couple anywhere in the plane of

motion. Therefore,

X - 𝑊

𝑔 �̈�𝑐 = 0

Y - 𝑊

𝑔 �̈�𝑐 = 0

𝑀𝑐 - 𝐼𝑐�̈� = 0

This is D’Alembert’s principle in plane

motion.


(63) Write the principle of work and energy

for a rigid body in plane motion.

Ans: In case of rigid body,

Work done =

{

𝑊𝑜𝑟𝑘 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔𝑑𝑢𝑒 𝑡𝑜 𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛

𝑆1 𝑡𝑜 𝑆2 }

+ {

𝑊𝑜𝑟𝑘 𝑟𝑒𝑠𝑢𝑙𝑡𝑖𝑛𝑔𝑑𝑢𝑒 𝑡𝑜 𝑎𝑛𝑔𝑢𝑙𝑎𝑟𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝜃1 𝑡𝑜 𝜃2

}

𝑈1−2 = ∫ (𝐹𝑐𝑜𝑠 𝛼) 𝑑𝑆 𝑆2𝑆1

+ ∫ 𝑀 𝑑𝜃𝜃2𝜃1

𝑈1−2 = F.S + M.θ

𝑈1−2 = 𝑇2 − 𝑇1

where (𝑇2 − 𝑇1) is the change in kinetic

energy of the body resulting from

translation and rotation between the initial

and final positions.

Change in K.E,

= 𝑇2 − 𝑇1

= [1

2 𝑚 𝑣2

2 − 1

2 𝑚 𝑣1

2 ] + [1

2 𝐼 𝜔2

2 − 1

2 𝐼 𝜔1

2]

= [𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐾. 𝐸 𝑖𝑛𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛

] + [𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐾. 𝐸 𝑖𝑛

𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛]

CLASSICAL ENGINEERING MECHANICS

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