Preface - UGC POINT · Preface Over the period of ... If the vector sum of the forces acting an a...

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Preface

Over the period of time the CSIR-NET/JRF, GATE, JEST examination has become more

challenging due to increasing number of students, and new questions pattern.

Though every candidate has ability to succeed but competitive environment, in-depth

knowledge, quality guidance and good source of study is required to achieve high level

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Classical Mechanics -

NET/JRF, GATE and JEST examination. The whole book has been divided into topic-wise

sections.

I have true desire to serve student community by way of providing good source of study and

quality guidance. I hope this book will be proved in important tool to succeed in above

examinations.

Any suggestion from the readers for the improvement of this book are most welcome


Contents

Topic Page No.

1 Newtonian Mechanics

(ii) Gravitation (iii) Centre of mass (iv) Collision (v) Moment of inertia (vi) Rotational dynamics

(vii) Simple Harmonic motion

2 Lagrangian

3 Hamiltonion

4 Special Theory of relativity

5 Small Oscillation

6 Central Force

7 Rigid Body Dynamics & Inertia Tensor

8 Poission Bracket & Canonical Transformation

9

10 Phase space


NEWTONIAN MECHANICS CHAPTER -1

(1) Draw the problem of write what u find, hat is given.

(2) Solve problem symbolically.

(3) Check dimension.

(4) Fermi problem do the

(5) Limiting case or sp. Case (related to Phy) involved in the prob.

Newtonian Mechanics:

st Law of Motion: If the vector sum of the forces acting an a particle is zero then and only then the particle remains unaccelerated.[ i.e. remains at rest or moves with constant velocity]

i.e. iff Note: st Law is not applicable for all frames of references. Frame of references: st Law at valid is called an inertial frame of references Eq. All frame of reference moving with constant speed or at rest.

-inertial frame of references. Eq: all accelerated frame of reference.

The rate of change of linear moment is proportioned to applied force. i.e or In most cases K=1

i.e dp dv

F m.dt dt

If m is independent of force

If a body A exerts a force on another Body B, then B exerts a

forces - on A then it is

Note: In all Bodies single force cannot be exists i.e. all forces in pairs. [ eq. is exists will also

exists which simple ampli cation of IIIrd Law of Motion]

All force is nature except gravitational force are electrical forces.


Que. A black of mass the inclined on a plane and force Mg act on it given µ = µs.

Find normal and frictional force?

Give for what range of will be black in fact remains at rest?

Case I. If block has tendency to move upward.

By force equilibrium

Normal force N = Mg sin + Mg cos

N = Mg (sin + cos )

and =

fr = Mg (cos sin ) ( Frictional force)

fr µN

So µ Mg (sin + cos )

µ

µ


Case II: When block has tendency to move downward.

N = Mg (cos + sin )

fr = Mg (sin cos )

fr µN

µ Mg (cos + sin )

µ

µ

µ

Overall answer :

N = Mg (cos + sin )

fr = Mg | sin cos |

and

SPL Case 1:

(1) If µ = 1

0


Case II. If µ = 0

So force and normal reaction both are balanced so particle at rest. Gravitation Force: Gravitation is the force of attraction between any two point particles in the universe. It is given by:

Where G is universal gravitational constant. The value of 11 2 2G 6.67 10 Nm / kg .

(i) Due to altitude: Acceleration due to gravity at a height h above the surface of earth is given by.

h 2

GM 2hg g 1

(R h) R (for h << R)

(ii) Due to depth: Acceleration due to gravity at a depth h below the surface of earth is given by

'h

hg g 1

R (for all depths)

At h = R (i.e., at the centre of earth): 'hg 0

(iii) Due to rotation of earth: Acceleration due to gravity at latitude is given by: 2 2g g R where angular velocity of the earth

(a) At poles: 2 2

p maxg g R

(b) At equator: 2 2 2eq min

N = Mg N = N = Mg

fr = Mg fr = 0 fr = Mg

2Mg

2


(iv) Due to non-spherical shape of earth: Due to the shape of the earth, g is maximum at poles and minimum at equator.

Gravitational Intensity: In case of a solid or hollow sphere of mass M and radius R:

(a) For an external point (r > R): I0 = (GM/r2)

(b) For an internal point (r > R)

(i) of a spherical shell: 0iI

(ii) of a solid sphere: 3(GM/R )riI

Gravitational potential: In case of a solid or hollow sphere:

(a) For an external point (r > R): 0

(b) For an internal point :

(i) of a spherical shell: i constant

(ii) of a solid sphere:

Escape Velocity: (i) It is minimum speed with which a body must be projected away from the surface of the earth so

that it may never return to the earth

(ii) Escape velocity of a body from the surface of earth is given by: es

Geostationary satellite: (a) A satellite which appears to be stationary for a person on the surface of the earth is called

geostationary satellite.

(b) It revolves in the equatorial plane from west to east with a time period of 24 hours

(c) It height from the surface of the earth is nearly 35600 km and radius of the circular orbit is nearly

42000 km

(d) The orbital velocity of this satellite is nearly 3.08 km/sec

(e) The relative velocity of geostationary satellite with respect to earth is zero

(f) The orbit of geostationary satellite is called as parking orbit

(i) All planets move around the sun in elliptical orbits, with the sun being at rest at one focus of the

orbit (ii) The position vector from the sun to the planet sweeps out equal area in equal time, i.e., area

velocity of a planet around the sun always remains constant. this gives that the angular momentum or moment of momentum remain constant.

(iii) The square of the time period of a planet around the sun is proportional to the cube of the semi-

major axis of the ellipse or mean distance of the planet from the sun, i.e. 2 3 where a is the semi major axis of the ellipse.


CENTRE OF MASS

1. Discrete System: The position vector of the centre of mass is

1 1 2 2

1 2

......

.....n n

c

n

m r m r m rr

m m m

Where 1 2, ....... nr r r are the position vectors of masses 1 2 nm ,m .......m respectively.

The components of the position vector of centre of mass are defined as

2. Continuous system: The centre of mass of a continuous body is defined as

In the component form

3. Centre of Mass of Some Common System:

(i) A system of two point masses. The centre of mass lie closer to the heavier mass

(ii) A circular cone

(iii) A semi-circular ring

(iv) A semi-circular disc


(v) A hemispherical shell

(vi) A solid hemisphere

Motion of the centre of mass:

(i) Velocity: The instantaneous velocity of the centre of mass is defined as

(ii) Acceleration: The acceleration of the centre of mass is defined as

(iii) Momentum: The total momentum of a system of particle is cp Mv

(iv) Kinetic Energy: The kinetic energy of a system of particles consists of two parts. c ,

where , Kinetic energy due to motion of c.m. relative to the fixed origin O and

, kinetic energy of the particles relative to the c.m.

Note that term may involve translational, rotational or vibrational energies relative to the centre of mass.

The first and second laws of motion for a system of particles are modified as First law: The centre of mass of an isolated system is at rest or moves with constant velocity.

Second law: Thhe net external force acting on a system of total of mass M is related to the acceleration of centre of mass of the system

ext cmF Ma

Projectile motion :

Validity of projectile motion: Motion is always in const acceleration.

(2) Direction as well as magnitude of velocities changes at each & every

point.

(3) High range projectile are elliptical in nature.


Equation of motion :

v = u + at

S =

v2 = u2 + 2as

Motion on a flat ground

Equation of paths

v =

In horizontal direction u cos =

t =

In vertical direction y =

y = 2

2 2 by putting value of t.

Max height : v2 = u2 + 2gh

v = 0 = u2 sin2 2gh

2 2

Range

Time of flight :

v = u + at

u sin = u sin gt

2 2u sin

Rg


2 sin = gt

Que. For what value of range will be max.

Rmax = 2 2u sin (45 ) u

g g

= 45°

or = 0

22u cos 2

g = 0

2u2 cos 2 = 0

cos 2 = cos 90°

2 = 90°

= 45°

ARE UNDER THE PROJECTILE

A =

b

a

y dx

=

=

2 2

=

= 4 2

2 2


=4 2 2

2 2

= 4 2

2 2

= 4 2

2 2

3sin 2 sin 2

u sin 2 22g 3 cos

=

LENGTH OF THE PROJECTILE

dl = 2 2dx dy

l = 2 2dx dy

or l =

2u sin 2 2Rg

20

dy1 dx

dx

=

2R2

2 2 2 20

Let = u

2 2

= dy

dx = 2 2u cos

dyg

When x = 0, y = tan

x = 2u sin 2

g

y =

y = tan


So

tan2 22 2

tan

u cos y 1y 1 log y y 1

g 2 2

2 2u cos

g

2 22 2

= 2 2

=

IInd method :

2 2

dx dydt

dt dt = 2 2

T T2 2

x y0 0

v v dt (u cos ) (u sin gt) dt

=

T2 2 2 2 2 2

0

(u cos u sain g t 2gt v sin t) dt

=

T 22

20

u 2vtg t sin dt

gg

=

T 22 22

2 2

0

=

T 2 22

20

u ug t sin cos dt

g g

Let = x

dt = dx

When t = 0, x = u

sing

When t = 2u sin u sin

T xg g


=

u sin

g 22 2

2u sin

g

ug t cos dx

g

=

u sin

g2 2 2 22 2 2 2

2 2u sin

g

x u u u cosg x cos cos log x x

2 gg 2g

= 2 2 2 2 2 2 2

2 2 22 2 2

u sin u u u u sin u sin u cosg sin cos cos log

2g g g gg g 2g

2 2 2 2 2 22

2 2 2 2

u sin u sin u cos u u sin ucos log

2g gg g 2g g

=

=

2u sinL

g

When particle is thrown from same height to ground

In horizontal

...(1)

In vertical

21h gt

2

In horizontal

...(1)

In vertical

In horizontal

..(1)

In vertical

Collision :


1D Collision is an event in which a large amount of force is exerted on a colliding particle for a very

short period of time. It is not necessary that two particle touch each other total linear momentum of the

particle does not change.

Momentum Conservation:

1 1 1 2 10m u mv m v

1 1 1 1 2 2m u m v m v

Elastic Collision:

Conservation of Energy

From (1)

1 1 1 1 1 2 2

& from (2) 2

1 1 1 1 1 2 2m u v u v m v

1 1 2

In general (if ) we get

i.e.

Inelastic Collision

In inelastic collision

i.e. 2 1 1 2

e coef of restitution

Let us consider


Solving we get

Case-1: A heavy body hits a light body from behind

1 2

1 2

1 2

1 2

22

m

m m

1

1 2

20

m

m m

From equation

1 2 21 1 2

1 2 1 2

1 2 12 1 2

1 2 1 2

2

2

m m mv u u

m m m m

m m mv u u

m m m m

We have

1 1

2 1 22

v u

v u u

If second (light) body at rest i.e.

1 1

2 1

i.e. is heavy body hits a light body (which is at rest ) then after collision heavy body moves as same

speed as before collision, but rest body (light body) moves twice the first moving body

Case-2: 1 22 1

1 2

: 1m m

m mm m


2

1 2

22

m

m m

1

1 2

20

m

m m

1 1 22v u u

1 2

Let us (Second body (Heavy body)) is at rest

This implies heavy body (rest body) is remains at rest after collision but moving body (light body m1) is

moving with opposite direction to direction of before collision

Case-3: Elastic collision of two bodies of equal mass (m1 = m2)

i.e.

when two bodies of equal mass collides elastically their velocities are mutually interchanged

2-D Collision:

In Laboratory frame of reference

Conservation of momentum and energy

The x-component of linear momentum before collision is equal to x-component of angular momentum

after collision and same for y-component.

i.e.

x-component 1 1 1 1 1 2 2 2cos cosm u m v m v

y-components 1 1 1 2 2 20 sin sinm v m v

Energy Conservation:


(3)

Collision is C.M. frame:

1 1 2

1 2

0cm

m u mv

m m

and '1 1 cmu u v

i.e. ' 2 11

1 2

m uu

m m

and '2 0 cmu v

' 1 22

1 2

m uu

m m

From (5) and (6)

' ' 1 2 1 1 2 11 1 2 2

1 2 1 2

0m m u m m u

m u m um m m m

i.e ' '

1 1 2 2m u m u

After Collision: ' '

1 1 2 2 0m v m v

' '1 1 2 2m v m v

Energy Conservation in C.M. frame


2 2'

'2 '2 '1 1 11 1 2 1 1 2 1

2 2

1 1 1 1

2 2 2 2

m u mm u m m v m v

m musing equation (7) & (8)

we get

'2 '2 ' '1 1 1 1u v u v

& '2 '2 ' '2 2 2 2u v u v

In C.M. frame of reference speed of the particle does not change after collision.

1 1 cmv v v

1 1 cm

From figure (1) & (2)

From fig(3)

'

1 1 1cos cos cmv v v

and '1 1 1sin sinv v

or

'1

1 '1 cm


' 2 11

1 2

m uv

m m &

1'1 2

cmv m

v m

From (9)

or 1

1

2

sintan

cos

cm

lab

cm

m

m

Equation 9] (9) & (9) represents the relationship between angle of scattering in lab from and C.M.

frame.

Relation between angle fo recoil with C.M. frame

From fig (1) and (2)

From fig (4)


'2 2 2

'2 2 2

'2

2 '2

2

'2

cos cos

sin sin

sintan

cos

sintan

cos

cm

cm

cm

v v v

v v

v

v v

v

v

but

2

sintan

1 cos

PROBLMES

Prob.(1): Show that in lab system two particles of same mass will move at right angle to each other

after collision one of these is at rest before collision.

Soln: 11

2

sintan

cosm

m

Given that 1 2m m

i.e.

1

sintan tan tan 2

cos 1


and 2

sintan cot

1 cos 2

From fig (1) and (2)

i.e. after collision two particles makes angle

Prob.(2):

with a nucleus of mass No-A. then prove that

Where is scattering angle in C.M. from

Soln:

2

1

1

After in Lab frame


From fig

'1 1 cmv v v

'1 1 1cos coscmv v v

& '1 1 1sin ( sin )v v

From (2) and (3) squaring and adding

2 2 '2 '1 1 12 coscm cmv v v v v

From equation (A)

But 2 1A.m m

2

2 2

1 1

2

1

1 2 cos'

1

m m

m mE

E m

m

proved

Moment of Inertia

* Inertia depend on mass

* Property of matter

Moment of inertia

* Depend on mass as well as distribution on mass


* Property of material about an axis.

Physical significance of M.I.

In thermodynamics entropy

I = mr2 (for point mass)

I = 2mr (for discrete point mass)

mr

I = 2r dm (for continuous body)

RING :

dI = r2dm

I = 2r dm

I = mr2

Piece of ring of mass m is

I = mr2


If ring of mass m is cut then this mass is

I = mr2

Circular cylinder shell: Cylinder is made of ring. If first we cut ring then take small part of this ring

Circular disc :

I = 2

Solid cylinder : Made of disc

I = 21

2Mr

Spherical shell : (Very small thickness as Balloon) mass is distributed on surface so take surface

element

I = 22

3mR

Solid sphere :

I = 22

5mR

Hollow sphere :

I = 5 52 13 32 1

2 ( )

5 ( )

R Rm

R R

Uniform Rod : 2

Theorem of L axis: Applicable of for plane body only.

Ix = 2 2


Iz =

Iz = y xI I

Iz = x yI I

Exp.

=

I = 27

5mR

Exp.

mr2 =

So about x & y axis

,x yI I = 2

Radius of Gyration :

K = I

M

I = MK2


Ex. M.I. of solid sphere about an axis passing through centre.

I =

Ex. Moment of inertia of triangular disc about O. I the plane of disc.

length of element = 2x tan

mass of element dm = . 2x tan dx

where = 1

(21sin ) (cos )2

M

MI. of element =

M.I. of triangular disc

cos cos23 3

0 0

(2 tan ). 2 tan tan 2

12

L L

x x dx x dx

= 3 4 4(2 tan ) ( cos ) 2( cos )

tan12 4 4

L L

= 4 2

=


=2 2 2

2

= 2 2 2 2

2

= 2

2 22 sin 6 cos12

ML

= 2 2

2 2 2[1 5 cos ] [sin 3 sin ]12 6

ML ML

I = 2

2[3 2 sin ]6

ML

Now calculate M.I. for

I =

2

266 (3 2 sin 30 )

6

Mr

= 2

Polygon of n sides :

angle =

I = 2

23 2 sin6

Mr

N

2 2 2 22 2

5 5Mr mr mr MR =


ROTATION DYNAMICS Type of motion: 1. Translational: If all the particles have same velocity at any given time than it is translational

motion. Also rigid body moves parallel to itself in translational motion.

2. Pure Rotational: A rigid body is said to be in pure rotation if every particle of the body moves in a circle and the centers of all the circles lie on a straight line called the axis of rotation.

3. Combined translation & rotational: A general plane motion is a combination of translation and motion.

Relation between linear and angular variables

(A) Distance

(B) Acceleration (rate of change of speed) (tangentially-directed)

2

2r

Va

r (directed towards centre)

Angular velocity of a point on rigid body wrt other point on rigid body All points are rotating with angular velocity with respect to each other.

ENERGY OF A RIGID BODY ABOUT A FIXED AXIS

Gravitational P.E. U = mg hcm Kinetic energy

MOMENT OF INERTIA

Definition: Property of a rigid body by virtue of which it opposes change in its rotational velocity

(angular velocity) is known as MI.

This is always written wrt a axis of rotation.This plays same role in rotational motion as mass plays in

translation motion.Difference between mass & MI is that mass is property of body and is independent of

any reference from choosen but MI dependen on the

(i) axis of rotation

(ii) shape of the body Mass depends only on these two things

(iii) size of the body

(v) density of the material of the body

1. MI of a point of mass:

2. MI for point mass distribution: 2 2 21 1 2 2 3 3Iyy ' m r m r m r .......

For continuous mass distribution

where r is perpendicular distance of any element from axis of rotation.

Important point MI of two or more than two bodies can be added or subtracted only when all MI are written w.r.t. same axis.

Uniform ring I = mr2


(about an axis passing through centre and perpendicular to the plane)

Disc 2MR

2

(about axis passing through centre and perpendicaulr to plane of disc)

Hollow cylinder (about yy axis)

Note: Independent of length of cylinder 2MR

2

of same mass

Hollow sphere (about at diameter) 2

Solid sphere (about a diameter ) 3

Uniform rod mass M length 2

3

ml(about axis passing through one end & perpendicular)

Uniform rod 2

12

ml

(about an axis passing at from one end and perpendicular

I = 0 uniform rod (about axis passing through rod)

Parallel-axis theorem:

Used find moment of inertia about an axis which parallel to the axis passing through C.M. 3

P CMI I Md

CMI MI of the rigid body about an axis through CM

PI MI of the rigid body about an axis which is parallel to the about axis through CM and is

distance d from the axis through CM


Theorem of perpendicular axis : Iz = Ix + Iy

This theorem is applicable only for the laminar bodies (i.e. plane bodies) (e.g. ring, disc, not sphen)

Ix & Iy are MI of body about a common pt. O in two mutually perpendicular directions in the palne of

body Iz is MI of body about a axis perpendicular to C and Y axis and passing through pt. O

Radiaus of gyration: I is MI about the axis, for which k is written.

Torque:

Vector representation and understanding of torque:

Torque due to force of gravity: cm

Relation between torque and angular acceleration of a rigid body:

ext where and I are both written about axis of rotation.

Angular momentum of a particle:

Angular momentum of a system of particle:

(a) For collection of point masses i

(b) For rigid body axis axis

Here axisL means component of angular momentum along the axis

Relation between torque and angular momentum: , axisext axis

dLdL

dt dt

Conservation of angular momentum:

If external torque about a axis is zero then angular momentum of the system about that axis will remains

conserved. So if axis axis constant

COMBINED ROTATION AND TRANSLATION

s a

combination of translationl motion of the centre of mass and rotation about an axis through the centre of


Kientic Energy :

Angular momentum

Theorem: Angular momentum of a body about any point or axis of body about C.M. (parallel to

axis A) + of C.M. about axis A body cm cmL L L

Instantaneous Point/Axis of rest

Axis about which combined rotation and translational motion of a body can be represented as pure

rotation

Note:

1. the axis is perpendicular to the plane of motion

2. can not be used for acceleration

3.

Equilibrium & Toppling

Necessary conditions for equilibrium of an object

1. The resultant external force must equal zero.

2. The resultant external torque about any axis must be zero:

SIMPLE HARMONIC MOTION

Oscillatory Motion:

A body is said to posses oscillatory motion if it moves back and forth repeatedly about the mean

position. For example swinging pendulum, vibratory motion of a mass attached to a spring.

Periodic Motion

A motion which repeats itself after equal intervals of time is called a periodic motion. Each oscillatory

motion is a periodic motion. A body moving in a circle is also an example of periodic motion but it is

not oscillatory motion.

Simple Harmonic Motion

A system is said to execute simple harmonic motion if the magnitude of the forces acting on it is directly

proportional to the magnitude of its displacement from the mean position and the force is always

directed towards the mean position. For example: motion of a simple pendulum, vibrating tuning fork,

loaded spring, etc., are all S.H.M.s

Analysis of S.H.M

Consider a particle of mass m executing S.H.M along X-axis with mean position at origin

Displacment from mean position X coordinate

The force acting on the particle must be towards mean position (i.e., opposite in sign to x) and its

magnitude must be proportianl to magnitude of x.

where k is a constant

If a is acceleration then:


2

2

k

m

d xx

dt

This equation is called as the differential equation of S.H.M. The general expression for satisfying

the equation is:

0

Substituting (ii) in (i), we get

k

M

Amplitude:

The amplitude of a particle executing S.H.M. is magnitude of its maximum displacement on either side

of the mean position.

Consider the equation of S.H.M. i.e.,

0

Maximum value of so

A is amplitude of the particle.

Time Period:

Time period of a particle executing S.H.M. is the time taken to complete one cycle and is denoted by T.

Consider the equation of S.H.M i.e,

0

If t is increased by then

0

0

0

0

2( ) sin

sin 2

( ) sin 2

( ) sin

x t A t

x t A t

x t A t

x t A t

Thus, the function repeats itself after an interval of therefore

Time Period (T) =

2

T 2m k

ask m

Frequency:

The frequency of a particle executing S.H.M. is equal to the number of oscillations completed in one

second. It is measured in cycles per second (Hertz) and is denoted by


Phase: The phase of a particle executing S.H.M. at any instant is its state as regard to its position and direction

of motion at that instant. It is measured as argument (angle) of sine in the equation of S.H.M

0

Phase 0

At phase 0 ; the constant is called initial phase of the particle or phase constant.

Initial Conditions:

The amplitude A and initial phase are determined by the initial conditions (position and velocity at

0t ) for the oscillating particle.

When the motion starts at initial position is:

0(0) sinx A and the initial velocity is:

0(0) cosv A

For example

If 0 / 2 then . Hence the particle starts moving at 0t from the right extreme

position (with zero speed)

Angular Frequency:

The quantity is called the angular frequency and is equal to times the frequency of the particle

executing S.H.M.

i.e.,

Energy Consideration in S.H.M.

Consider a mass m executing simple harmonic motion given by the equation

0

where, A is the displacement amplitude.

is the angular frequency and is the initial phase.

1. Kinetic Energy

At any instant of time, if v is the velocity of particle, its kinetic energy (K.E.) is given by

Hence,

(i) The kinetic energy varies periodically.

(ii) K.E. is maximum at mean position, where velocity is maximum.

(iii) K.E. is minimum, i.e., zero at the extreme position, where velocity is zero.

2. Potential Energy

The potential energy (P.E.) of the particle at time t depends on the displacement . The P.E. is

given by the amount of the applied force must be equal and opposite to the restoring force.


i.e. appliedF kx

The work done required to produce an infinitesimal displacement is

( ) ( )appdw F dx

dw kx dx

Total work done 2

0

1

2

x

kx dx kx

i.e. gain in P.E. = 21 2Kx at a distance from mean position.

If we take P.E. = 0 at mean position then:

2

In general,

3. Total Energy: (TE)

T.E. = K.E. + P.E.

T.E. = 2 2

T.E. = 2 2 2 2 20 01 2 cos 1 2 sin (0)mA t kA t U

Using we get:

T.E.

T.E.

T.E.

i.e., total energy remains constants (or conserved), if no energy is dissipated.

Important Relations

1. Positions

If mean position is at origin the position (X coordinate) depends on time is general as:

0

at mean position

at extremes

2. Velocity

at any time instant 0, ( ) cos

at any position x, 2 2

This can be obtained by eliminating t between the expression for and sign indicates that

particle passes every position with equal and opposite velocities.

velocity is minimum at extremes because the particles is at rest

i.e. at extreme position


velocity has maximum magnitude at mean position.

maxv A at mean position

3. Acceleration

at any instant 20, ( ) sint a t A t

at any position 2, ( )x a x x

acceleration is always directed towards mean position.

the magnitude of acceleration is minimum at mean position and maximum at extremes

min at mean position.

at extremes.

4. Force on the particle

at any instant t, 20( ) sin

at any position x, 2( )F x m x or ,

force is always directed towards mean position.

the magnitude of force is maximum at extrme position. 2

maxat extremes

There is no force on the particle at mean position. (The mean position is always a position

of stable equilibrium for the oscillating particle)

5. Energy

Kienetic energy

K is maximum at mean position and minimum at extremes

at mean position

min 0K at extremes

6. Potential Energy

If potential energy is taken as zero at mean position then at any position x

U is maximum at extremes

U is minimum at mean position

Total energy

T.E. 2

and is constant at all time instants and at all positions.


SOLVED PROBLEM

Prob.1: Moment of inertia of a solid cylinder of mass m, height h and radius r about an axis

(shown in figure by dashed line) passing through its centre of mass and perpendicular to its

symmetry axis is

(a) (b)

(c) (d)

Soln: Volume density of mass

Mass of volume element

2

Using parallel axes theorem

2 2

2 2 2 2

2 2

2 2

1

4

1 1

4 12

h h

h h

I r r dx r x dx

mr mh

Prob.2: A rain drop falling vertically under gravity gathers moisture from the atmosphere at a rate

given by where m is the instantnesou mass, is time and k is a constant. The

equation of motion of the rain drop is

If the drop starts falling at with zero initial velocity and initial mass (given:

20 2 , 12m gm k gm s and 2 ), the velocity (v) of the drop after one second

is

(a) (b) (c) (d)


Soln: 2

Equation of motion of drop is given by

If =

3 3

43

.

4

v t gt dt c

gtvt c

At

43

4

4

gtvt

gtv

Prob.3: In case of an ineleastic collision which one of the following is true

(a) Total energy is not conserved

(b) Momentum is not conserved

(c) Kinetic energy is conserved

(d) Kinetic energy is not conserved

Soln: Ans.(d)

Prob.4: An ideal fluid is flowing through a tube of cylindrical cross section with smoothly varying

is given by . The velocity at a point where cross sectiona area is

is given by

(a) (b)

(c) (d)

Soln: Ans.(b)

According to equation of continuity

1 1 2 2

2 0.005v m s

Prob.5: Moment of inertia of a uniform circular disk of radius R and mass M about the tangential

axis parallel to its diameter is

(a) 2

(b) 2

(c) 25

4

MR (4)

23

2

MR

Soln: Ans. (c)


Momoent of inertia of disc about an axis passing through centre and also along diameter

2

4cm

MRI

So, MI of disc about a parallel axis tangent to disc using parallel axis theorem

2

22

2

cm

Prob.6: Four masses, each of mass m, are at the corners of a square of side a as shown in the

figure. The product of inertia Ixy of the system about O is

(a) (b)

(c) (d)

Soln: Ans.(c)

2

xy i i i

Prob.7: Two blocks of masses m and 2m are hung from the ceiling by three cables as shown

The cable in the middle is exactly horizontal and the cable on the left makes an angle of

45o with respect to the vertical. The angle made by the cable on the right with respect to

the vertical is

(a) (b)

(c) (d)

Soln: Ans. (c)


At point A

1

At point B

2 2

sinsin sin2 2

T mg T

T mg

1tan 1 2

Prob.8: A proton of mass m collides with a particle of unknown mass at rest. After the collision,

the proton and the unknown particle are seen moving at an angle of with respect to

each other. The mass of the unknown particle is

(a) (b)

(c) (d)

Soln: Ans. (b)

Using conservation of linear momentum

1 2cos ' cos2

mv mv m v m


1 2cos ' sinmv m v

1 2sin '2

mv m v

2

From (1) and (2)

1 2cos ' sinmv m v mv

1 2

Squaring and adding the both equation

2 2 2 2 21 2'2m v m v m v

From KE conservation

2 2 21 2'mv m v mv

Multiplying (4) by m

2 2 2 2 21 2'm v mm v m v

Subtracting (5) from (3)

So,

Prob.9: A small asteroid is approaching a massive start with a speed v, from a very large distance, at an impact parameter b as shown below.

If the mass of the star is M and its radius is R, then what is the minimum value of b such that the asterioid will miss the star?

(a) (b)

(c) R (d)

Soln: Ans. (a)

Angular momentum

Central force potential

Effective potential2

22e

LV V

mr


2

2

2 2 2

2

2 2

2

2

2

2

GMm L

r mr

GMm m v b

r mr

GMm mv b

r r

The minimum distance is given by root of equation

e

2 2

2

2

1

2 2

GMm mv bmv

r r

Asterioid will just miss that start if

2 2

2

2

1

2 2

GMm mv bmv

R R

2

2

12

2

21

21

GMR

v R

GMb R

v R

Prob.10: A heavy uniform rope of length L and mass per unit length goes over a frictionless

pulley of diameter R, and has two masses M and m attached to its two ends as shown in

terms of the distance x, the equilibrium position is given by

(a) 1

2 2

L RL

m M (b)

(c) (d) 4

Soln: Ans. (d)

For equilibrium

2

Rmg xg Mg L x g


22

Rxg M m g L g

Prob.11: A particle of mass m is located at a distance z along the axis of a uniform disk of mass M

and radius R. The gravitationa force felt by the mass m is given by

(a) 2 2 2 1/2

(b) 2 2 2 1/2

(c) 2 2 2 1/2

(d) 2 2 2 1/2

Soln: Ans (a)

Let us first out the gravitational field due to ring of mass m at a point at a distance z from

its centre. Let us consider an element of mass dm.

Magnitude of gravitational field

2 2

cos( )

Gdmdg

z R

(The vertical component of g will get cancelled out if we consider whole ring)

2 2 3/2 2 2 3/2

.

( ) ( )

Gz Gm zg dm

z R z R

Now, let us find out gravitational field due to disc of radius R and mass M at a point z from

its. Centre. Consider a mass element in the form of ring of radius x and thickness dx

2

2

.2

2

Mdm x dx

R

Mxdx

R

Gravitaion field due to dm at P.


2 2 3/2

2 2 2 3/2

2 2 2 3/2

0

2

2 2 2 3/2

0

2 2 2 3/2

2 2 2 3/2

( )

.2

( )

2

( )

2 1

( )

2 1 1

( )

21

( )

R

Gz dmdg

z x

Gz Mxg dx

R z x

GMz x dx

R z x

GMz

R z x

GMz

R z z R

GM z

R z R

So, force felt by mass at point P

2 2 2 1/2

Prob.12. A simple pendulum of length l is suspended from a hook mounted on a slanted wall. The

wall makes a small angle with the vertical. The pendulum is displaced from the vertical

by a small angle and released. Assuming that the collision of the bob is elastic,

the time period of oscillation is

(a) 2l

g (b) 12 / 2 sin

l

g

(c) (d) 1

Soln: Ans. (b)

For should be 2l

g

Prob.13: A U shaped tube of uniform cross section A contains a liquid of density . The total length

of the column is L. if the fluid is displaced then the frequency of oscillation is

(a) (b)

(c) (d)


Soln: Ans. (a)

Let fluid is displaced by displacement x in upwar direction.

Restoring force acting liquid column in direction opposite to displacement i.e.

downward.

Mass of liquid column

Writing equation of motion

2

2

d xm Axg

dt

2

2

d xAL Axg

dt

2

2

2

2. 0

d x gx

dt L

The angular frequency of oscillation

1

2 2

g

L

gf

L

Preface - UGC POINT · Preface Over the period of ... If the vector sum of the forces acting an a...

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