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2013-2015

Notes for School Exams

Physics XI

Rotational Motion Pranjal K. Bharti, B. Tech., IIT Kharagpur

© 2007 P. K. Bharti All rights reserved.

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Rotational Motion Pranjal K. Bharti (B. Tech., IIT Kharagpur)

2 Concept, JB 20, Near Jitendra Cinema, City Centre, Bokaro Mb: 7488044834

Rigid body

Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distances between different pairs of such a body do not change on the application of force. Pure Translation or Translation

• A body is said to have pure translational motion, if each particle of it has same velocity/acceleration at a particular instant of time.

Rotational Kinematics

• Kinematics is the study of motion without reference to the forces which cause motion.

• Kinetics is the study of motion which relates the action of forces on bodies to their resulting motions.

• Kinematics and Kinetics together known as Dynamics. • As earlier we have studied about displacement (r),

velocity (v) and acceleration (a) in the chapter kinematics; in the same way we shall study here about angular displacement (θ), angular velocity (ω) and angular acceleration (α) .

Angular Displacement

• Let us consider an arbitrary shaped rigid body rotating in counter-clockwise direction about an axis passing through point O. Here axis of rotation AB is passing through point O and is perpendicular to the plane of rotation.

• Let us consider a particle on the body at a distance r from the axis of rotation. Clearly, this particle is moving along a circle of radius r. Suppose at a time t, this particle was at point P and after a time t + ∆t it is at point P’.

• Clearly, during this time interval, particle P rotates through an angle θ from initial line OP. This angle θ is known as angular displacement of the particle.

• The line OP is known as reference position. • It is clear that all the particle of the body has rotated

through the same angle θ. Therefore, we can say that the whole rigid body has rotated through an angle θ. Hence, we can say that the rigid body has an angular displacement θ.

Important points about angular displacement

• Angular displacement is a scalar quantity. Infinitesimal small angular displacements are vectors. (Refer Resnick, Halliday).

• Denoted by ∆θ or θ • S.I. unit: radian (rad)

1800 = π rad

Angular Velocity (Ω)

• Average angular velocity (∆ω): It is defined as the time rate of change of angular displacement over a time interval.

2 1

2 1t t tθ θθω

−∆∆ = =

∆ −

• Instantaneous angular velocity (ω): It is defined as the time rate of change of angular position at an instant of time. Clearly,

0lim

t

dt dtθ θω

∆ →

∆= =

∆ • Angular velocity is a vector quantity. • Direction of ω is obtained using right hand thumb rule:

Curl your fingers of right hand in such a manner that direction of curl is in the direction of sense of rotation (i.e., clockwise or anticlockwise). Your extended thumb will give the direction of angular velocity ω.

Important points about angular velocity

• Angular velocity is a vector quantity. • Its direction is obtained by right hand thumb rule • Denoted by ∆ ω (average) or ω (instantaneous) • S.I. unit: rad/s • Another unit: rpm = revolution per minute.

Clearly, a particle has an angular displacement of 2π radian in one revolution.

Therefore, 21rpm /60

rad sπ=

v v v

At time t = t At time t = t + ∆t

O P

A

B

P

P’ A

B

O θ

x

y

z

x

y

z

Anticlockwise rotation ω coming out of paper i.e. along +ve z-axis

Clockwise rotation ω going into paper i.e. along –ve z-axis

ω

ω



Angular Acceleration (α)

• Average angular acceleration (∆α): It is defined as the time rate of change of angular velocity over a time interval.

2 1

2 1

=t t t

ω ωωα−∆

∆ =∆ −

• Instantaneous angular acceleration (α): It is defined as the time rate of change of angular velocity at an instant of time. Clearly,

0lim

t

dt dtω ωα

∆ →

∆= =

∆

Important points about angular acceleration

• Angular acceleration is a vector quantity. • Direction of α:

Case I: When angular speed is increasing. In this case angular acceleration is positive. Therefore, direction of angular acceleration is along angular velocity. Case II: When angular speed is decreasing. In this case angular acceleration is negative. Therefore, direction of angular acceleration is opposite to that of angular velocity.

• Denoted by ∆ α (average) or α (instantaneous) • S.I. unit: rad/s2

Uniform angular acceleration

• In chapter Kinematics we have studied about uniformly accelerated motion. Here we shall study about uniform angular acceleration motion. For uniform angular acceleration α = constant.

• Let us consider a particle on the rotating rigid body. Let at time t = 0, its initial angular position θ = 0 & initial angular velocity ω = ωo.

• Again, let at a time t = t, its final angular position θ = θ & final angular velocity ω = ω.

• Then, we have three formula:

1. o tω ω α= + (Analogous to v = u + at)

Proof:

[ ] [ ]00 0

( 0) ...(i)

oo

t tt

o

o

ddt

d dt dt t

tt

ωω

ωω

ωα

ω α α ω α

ω ω αω ω α

=

⇒ = = ⇒ =

⇒ − = −⇒ = +

∫ ∫ ∫

2. 212ot tθ ω α= +

(Analogous to 21 2

s ut at= + )

Proof:

From (i), we have

( ) [ ] [ ]2

0 00 0 0

2

21 ...(ii)2

o

o

ttt

o o

o

td dtdt dt

td t dt t

t t

θθ

ω ω αθ θω α ω

θ ω α θ ω α

θ ω α

= +

⇒ = + =

⇒ = + ⇒ = +

⇒ = +

∫ ∫

3. 2 20 2ω ω αθ= + (Analogous to 2 2 2v u as= + )

Proof: We know that

[ ]0 0

22 20

00

2 20

2 2 2

2 ...(iii)

d d d ddt dtd d

dd

d dωωθ

θ

ω ω

ω ω θ ωα ωθ θ

ωα ωθ

ωω ωα ω ω α αθ θ θ

ω ω αθ

= = ⋅ = ⋅

∴ =

⇒ = ⇒ = ⇒ = −

⇒ = +

∫ ∫

RReellaattiioonn bbeettwweeeenn aanngguullaarr ddiissppllaacceemmeenntt && lliinneeaarr ddiissttaannccee

• Let us consider a particle P at a perpendicular distance r from the axis of rotation on a rotating body. Clearly, this particle moves in a circle of radius r.

• Suppose the angular displacement at an instant is θ. As a result of this rotation, particle P has also travelled a circular arc of distance s. This distance s is the linear distance moved by the particle.

• From basic Trigonometry we have, sr

θ =

RReellaattiioonn bbeettwweeeenn aanngguullaarr vveelloocciittyy && lliinneeaarr vveelloocciittyy

• We know that, s = θ r

• If we differentiate both sides with respect to time t, we get relationship between angular velocity (ω) and linear velocity (v) straight away:

( )ds d dr rdt dt dt

θθ= =

(because r is constant for a particle)

v rω⇒ =

&ds dvdt dt

θ ω = =



I dm

R

RReellaattiioonn bbeettwweeeenn aanngguullaarr aacccceelleerraattiioonn && ttaannggeennttiiaall

aacccceelleerraattiioonn

• We know that, v = ω r

• If we differentiate both sides with respect to time t, we get relationship between angular acceleration (α) and linear tangential acceleration (at) straight away:

( )dv d dr rdt dt dt

ωω= =

(because r is constant for a particle)

ta rα⇒ =

&ds dvdt dt

θ ω = =

• Here, at is tangential acceleration in the tangential direction.

SSuummmmaarryy

• Relation between angular position & linear distance: sr

θ = ( in radian)

• Relation between angular velocity & linear velocity: v rω= (v along tangent)

• Relation between angular accln & linear tangential accln:

tdva rdt

α= = . (at along tangent)

• Radial acceleration = 2

2 Rva r vr

ω ω= = =

(radially inward i.e., towards the axis of rotation)

• NOTE: 1. These eqns. viz. v = ω r, at = αr & ar = ω2

r are not valid for whole body. It is valid for a particle on the rotating body at a perpendicular distance r from the axis of rotation.

2. Be careful of r. Remember r is the perpendicular distance of the point of interest from the axis of rotation.

3. At a particular instant of time angular displacement, angular velocity & angular acceleration of each particle (excluding those on axis of rotation) on a rigid body is same but different particles has different linear displacement, linear velocity & linear acceleration.

4. As s = θ r, v = ω r, at = α r & ar = ω2

r , therefore a point on the rigid body, which is farther away from the axis of rotation, has larger linear displacement, larger linear speed, larger tangential acceleration and larger radial acceleration .

5. We can write these eqns. in vector form as: , , .t Rv r a a r a vω ω= × = × = ×

Moment of Inertia

((RRoottaattiioonnaall IInneerrttiiaa)):: II

• Before going into the details of moment of inertia, let us have a look at concept of mass (m) of a body. Mass gives an idea of difficulty we face to move (translate) a body. Higher the mass, it will be more difficult to translate a body. Therefore mass is also known as Translational Inertia.

• Moment of Inertia (represented by I) or Rotational Inertia represents an analogous concept in rotation as presented by mass in translation.

• Physically, Moment of Inertia gives an idea of difficulty we face to rotate a body. Higher the moment of inertia, it will be more difficult to rotate the body. Therefore, Moment of Inertia is also known as Rotational Inertia.

• Moment of inertia depends on mass and distance from axis of rotation. Higher the mass, higher will be moment of inertia. Larger the distance from axis of rotation, larger will be moment of inertia.

Moment of Inertia of a particle

• Suppose a particle of mass m is rotating at a perpendicular distance r from the axis of rotation. Then, its moment of Inertia about axis of rotation is defined as :

I = mr2 (Moment of Inertia of a particle)

• MI for system of particles about an axis is: 2

i iI m r= ∑ (MI for system of particles)

• MI for a rigid body about an axis is: 2I r dm= ∫ (MI of a rigid body)

Moment of inertia of a RING about an axis through its

centre and perpendicular to its plane

• Consider a ring of mass M, radius R and centre O. Consider an elementary portion of the ring of dm and length dx.

• Clearly, moment of inertia of this elementary potion is given by 2dI R dm=

• The moment of inertia (I) of the ring about this axis can be found by integrating

2 2

2

I R dm R dm

I MR

∴ = =

⇒ =

∫ ∫



x

C

I

dx L/2

L/2

Moment of inertia of a DISC about an axis through its

centre and perpendicular to its plane

• Consider a circular disc of mass M, radius R and centre O.

Then, mass per unit area = 2

MRπ

• The disc can be assumed to be made up of a large number of concentric rings. Consider one such ring of radius x and an infinitesimally small thickness dx. Then, Area of ring = circumference × width = 2 π x dx

• Therefore, mass of ring

( )2 22 M Mdm xdx x dxR R

ππ π = =

• Using the moment of inertia of this ring about the axis passing through its centre and perpendicular to its plane,

( ) 2 2 32 2

2 2 M MdI dm x x dx x x dxR R

= = =

• The moment of inertia I, of the whole disc is the sum of the moment of inertia of all such rings, which have radii between x = 0 and x = R and is according obtained by integrating the above between the limits x = 0 and x = R. Therefore,

4 43

2 2 20 0 0

2

2 2 2 0 4 4 4

2

RR R M M x M RI dI x dxR R R

MRI

= = = = −

⇒ =

∫ ∫

Moment of inertia of a ROD about an axis through its centre and perpendicular to it

• Consider a uniform straight rod AB of length L, mass M

and centre C. Then mass per unit length of the rod = ML

• Consider a small element of mass dm and length dx of the rod at a distance x from the point C. The mass of

elementary portion of the rod, dm = M dxL

The moment of inertia (I) of the rod about this axis can

be found by integrating between the limits 2Lx = − to

2Lx =

/ 2/2 /2 /2 32 2 2

/2 /2 /2 /2

3 3 3

2

3

2 3 2 2 3 8

12

LL L L

L L L L

M M M xI x dm x dx x dxL L L

M L L M LL L

MLI

++ + +

− − − −

∴ = = = =

− = − = ×

⇒ =

∫ ∫ ∫

• Quick Exercise: Find the MI of the rod about an axis

passing through its edge and perpendicular to rod.

Ans: 2

12

MLI =

Hint: Integrate between the limits 0x = to x L= Moment of inertia of the CYLINDER about its axis of

symmetry

• Consider a cylinder of mass M, radius R and length L.

Then, mass per unit volume of the cylinder = 2

MR Lπ

• To find the moment of inertia of the cylinder, imagine that the cylinder is made of a large number of coaxial cylindrical shells. Consider one such cylindrical shell having internal radius x and external radius x + dx.

• The cross-section of the cylindrical shell is a circular ring of radius x and thickness dx. Therefore, cross-sectional area of cylindrical shell = circumference × thickness = 2πxdx

• The volume of the cylindrical shell, V = (cross-sectional area of the shell) × (length of the cylinder)

V = (2πx dx) × L V = 2 π L x dx • Now, mass of the cylindrical shell,

dm = V × (mass per unit volume of the cylinder)

( ) 2

2

2

2

Mdm LxdxR L

Mdm x dxR

ππ = ×

⇒ =

• Since the whole mass of the cylindrical shell is distributed at the same distance x from the symmetry axis, the moment of inertia of the cylindrical shell about the axis of symmetry,

I x

dx

R

I

R x

dx



2 22

32

2

2

MdI dm x xdx xR

MdI x dxR

= = ×

⇒ =

• The moment of inertia of the cylinder (I) about the symmetry axis can be obtained by integrating dI for all such cylindrical shells into which the cylinder can be divided i.e. by integrating dI between the limits x = 0 to x = R. Thus,

4 43 3

2 2 20 0 0 0

4

2

2

2 2 24

2 04

1 2

RR R M M M xI dI x dx x dxR R R

M RIR

I MR

= = =

⇒ = −

⇒ =

∫ ∫ ∫

Moment of inertia of a SOLID SPHERE about an axis through its centre

• Consider a sphere of mass M, radius R and centre O.

Then, mass per unit volume of the sphere 343

MRπ

=

• Let the moment of inertia of the sphere about the axis ZZ’ passing through its centre O be I. Consider the portion of the sphere between two planes perpendicular to the axis at distance x and x + dx from its centre. This elementary portion of the sphere will be a disc, whose centre is C; radius is AC=x and thickness is dx. Therefore, volume of the elementary portion of the cylinder, V = area × thickness

( ) ( )( )

22 2 2

2 2

V AC dx R x dx

V R x dx

π π

π

= × × = × − ×

= −

• Mass of the elementary portion of the sphere, dm = V × mass per unit volume of the sphere

•

( )

( )

2 2

3

2 23

43

34

Mdm R x dxR

Mdm R x dxR

ππ

= − ×

= −

• The moment of inertia of the elementary portion (disc) of the sphere about the axis,

( )

( ) ( )

22

22 2 2 2 2 2 23 3

1 1mass radius2 21 3 3)2 4 8

dI dm AC

M MdI R x dx R x R x dxR R

= × = ×

= × − × − = −

• The moment of inertia of the sphere (I) about this axis

can be obtained by integrating dI between the limits x = – R to x = + R. Thus,

( ) ( )22 2 4 2 2 43 3

2

3 3 28 8

25

R R R

R R R

M MI dI R x dx R R x x dxR R

I MR

+ +

− − −

= = − = − +

⇒ =

∫ ∫ ∫

Moment of inertia of a few rigid bodies

Body Axis Moment of Inertia Thin rod of length L

Passing through centre and ⊥ to the rod

2

12MLI =

Circular ring or hoop of radius R

Passing through its centre and ⊥ to its plane

I = M R2

Circular ring or hoop of radius R

Diameter

2

2MRI =

Circular disc of radius R

Passing through its centre and ⊥ to its plane

2

2MRI =

Circular disc of radius R

Diameter 2

4MRI =

I

L

I R

I

R

I R

I

R

C O

R

A

I

dx x



Body Axis Moment of Inertia Right circular solid cylinder of radius R and length L

Symmetry axis

2I MR=

Right circular hollow cylinder of radius R and length L

Symmetry axis

2

2MRI =

Solid sphere of radius R

Diameter 225

I MR=

Hollow sphere of radius R

Diameter 223

I MR=

Rectangular plane sheet

Symmetry axis ( )2 2

12M L B+


Passing through centre and parallel to its breadth

2

12MLI =


Passing through centre and parallel to its length

2

12MLI =

Square plane sheet

Symmetry axis

2

6Ma

Theorem of parallel axes

• It states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes. According to the theorem of parallel axes,

2cmI I Mh= +

• Proof. Suppose the rigid body is made up of n particles of masses m1, m2,m3…mn at perpendicular distance r1, r2,r3,… rn respectively from the axis KL passing through centre of mass C of the body.

• If ri is the perpendicular distance of a particle of mass mi from KL, then

2

1 ...(i)

i n

cm i ii

I m r=

=

= ∑

The perpendicular distance of ith particle from the axis AB = (ri + h)

( ) ( )2 2 2

1 1

2 2

1 1 1

2

2 ...(ii)

i n i n

i i i i ii ii n i n i n

i i i i ii i i

I m r h m r h r h

I m r h m h m r

= =

= =

= = =

= = =

= + = + +

⇒ = + +

∑ ∑

∑ ∑ ∑

As the body is balanced about the centre of mass, the algebraic sum of the moments of the weights (mg) of all particles about an axis passing through C must be zero

1 1

1

1

2

( ) 0 0

As 0, 0

Also, total mass of the body.

From eqn.(i) &(ii), I

i n i n

i i i ii i

i n

i ii

i n

ii

cm

m g r g m r

g m r

m M

I Mh

= =

= =

=

=

=

=

∴ = ⇒ =

≠ ∴ =

= =

∴

= +

∑ ∑

∑

∑

This proves the theorem of parallel axes.

A K

B L

mi ri h O C

I

I

I

I

I

I

L

I B

I



Theorem of perpendicular axes

It states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body. According to the theorem of perpendicular axes,

Z x YI I I= +

Proof: Suppose the lamina consists of masses m1, m2, m3, …, mn at perpendicular distance r1, r2 ,r3, …, rn respectively from the axis OZ. Suppose the corresponding perpendicular distances of these particles from the axis OY are x1, x2, x3, …., xn and from the axis OX are y1, y2, y3, …, yn respectively.

2 2 2 2 21 1 2 2 3 3 3

1

2 2 2 2 21 1 2 2 3 3

1

2 2 2 2 21 1 2 2 3 3 3

1

... ...(i)

... . ..(ii)

... ...(iii)

i n

x n i ii

i n

y n n i ii

i n

z n i ii

I m y m y m y m y m y

I m x m x m x m x m x

I m r m r m r m r m r

=

=

=

=

=

=

∴ = + + + =

= + + + =

= + + + =

∑

∑

∑

( )2 2 2 2

1 1 12 2 2

2

1

Adding eqns. (i) and (ii) ,we get

is clear from figure,

Hence ,

i n i n i n

x y i i i i i i ii i i

i i ii n

x y i i zi

z x y

I I m y m x m y x

As r x y

I I m r I

I I I

= = =

= = =

=

=

+ == + == +

= +

∴ + = =

= +

∑ ∑ ∑

∑

which proves the theorem of perpendicular axes.

Radius of Gyration (k)

• Radius of gyration (k) is the distance from the required axis such that if the total mass of the body were concentrated at that distance, the moment of inertia would be the same as for the actual body (which could have an arbitrary distribution of mass).

• The radius of gyration (k) of a body of mass m about an axis, is defined by the equation

I = mk2 Ikm

⇒ =

where, I = MI of the object about the given specified axis and k = radius of gyration.

Torque acting on a particle

• Let us consider an inertial reference frame with origin at O. Suppose at any instant, particle has a position vector r and force F is acting on it. Then, torque acting on the particle with respect to the origin is defined as: r Fτ = ×

• Torque is a vector quantity. Its direction is obtained by

right hand thumb rule. Its S. I. unit is Nm.

Torque acting on a rigid body

• Let us consider a rigid body rotating about a fixed axis of rotation with origin O. Suppose we have to find out the torque of a force F having position vector r about a point O. Then, torque acting on the particle with respect to the origin is defined as:

r Fτ = ×

(Torque)

• Let the angle between r & F be θ. Therefore, magnitude of torque is, r Fτ = ×

sinrFτ θ⇒ =

• We can write this as τ = F (r sinθ). • If we draw a line perpendicular to force F from point O,

we find its length is also r sinθ using little Trigonometry. • This perpendicular line of length r⊥ = r sinθ is also known

as lever arm or moment arm of torque. Therefore,

F rτ ⊥= × (Torque)

where r⊥ = lever arm

• Therefore, to find torque, we draw a perpendicular from

the point (about which torque is acting) to the line (or extended line) of action of force. Then we find its length (lever arm). Finally we take product of lever arm and magnitude of force to get the magnitude of torque.

Newton’s 2nd Law of Rotation

• Newton’s 2nd law of rotation is quite analogues to the Newton’s 2 nd law of motion (translation). Newton’s 2 nd law of rotation states that the total external torque acting on a rigid body about an axis of rotation in an inertial frame of reference is directly proportional to the

i. Moment of Inertia I of the rigid body about that axis, and

ii. Angular acceleration α of the particle .

Iτ α∴ =

(Newton’s 2nd Law of Rotation)

ri O

mi

yi

xi

z

x

y

r

F

O θ

r⊥



• Important points to note about Iτ α=

:

1. Frame of reference should be inertial. 2. It is valid only for rigid body. Not valid for particle or

system of particles. 3. τ = total external torque acting on a rigid body about an

axis of rotation. 4. This eqn. can be applied only about two axis in a rigid

body: a) About axis of rotation (if known) b) About an axis passing through centre of mass (if axis

of rotation is unknown). 5. Torque τ and Moment of Inertia I should be calculated

about same axis of rotation.

KKiinneettiicc eenneerrggyy iinn ffiixxeedd aaxxiiss ooff rroottaattiioonn

• Kinetic Energy of the rigid body rotating about fixed axis is given by:

KE = ½ Iω2 (Kinetic Energy)

• KE = ½ Iω2 is not a new kind of kinetic energy. Actually, it is the sum of kinetic energies ½ mv2 of all the particles on the rigid body.

Work & Power

• Work done by applied torque:

Suppose a rigid body is undergoing fixed axis rotation. If it undergoes an angular displacement from initial angular position θ1 to the final angular position θ2 under the action of torque τ, then work done by this torque is given by:

2

1

W dθ

θ

τ θ= ∫

Special case: Work done by a constant torque:W τθ=

• Average Power: Work done

timeavgP = P = work done / time taken

• Instantaneous power (power) is given by dWPdt

τω= =

• Work Kinetic Energy Theorem:

W = K2 – K

1 = ½ Iω

2

2 – ½ Iω

1

2

Angular momentum of a particle

• Let us consider an inertial reference frame with origin at O. Suppose a particle of mass m at position vector r

is

moving with velocity v

with at an instant. Then, angular momentum L

of the particle with respect to the origin is

defined as:

L r p= ×

(Angular momentum)

• Angular momentum is a vector quantity. Its direction is obtained by right hand thumb rule.

Angular momentum of a rigid body

• Let us consider a rigid body rotating about a fixed axis of rotation in an inertial frame of reference. Let us consider the angular velocity of rigid body at a particular instant be ω. Also, assume MI of the rigid body about axis of rotation be I.

• Then, angular momentum L of the rigid body about this fixed axis is defined as:

L = I ω (Angular momentum)

• Clearly, angular momentum L direction is along the direction of ω.

• Important things to note about L = I ω : 1. Frame of reference should be inertial. 2. This eqn. will be applicable about a fixed axis of rotation. 3. Angular momentum L and Moment of Inertia I should be

calculated about same fixed axis of rotation.

Torque & angular momentum

• We know that total external torque acting on a rigid body is given by:

Iτ α=

• But, ddtωα =

• Therefore, we have

( ) d dI I Idt dtωτ α ω= = =

d Ldt

τ⇒ =

(because, angular momentum L Iω=

) • Hence, total external torque acting on a rigid body is

equal to time rate of change in angular momentum of the rigid body.

Conservation of angular momentum

• It states that if no external torque acts on a system, the angular momentum of the system remains constant.

• Consider a system of two bodies on which no external torque acts. As such, the system is said to be isolated from the surroundings. Thus, if L

be the angular momentum of

the two bodies at any instant, then in absence of external torque

1 2

constantL

L L

=

⇒ =

Proof:

We know that d Ldt

τ =

.

Therefore, in the absence of external torque, 0τ =

, we

have, 0d Ldt

τ = =

constantL⇒ =



r r+ ∆

r∆

r

X

O

Z

r∆

C

Y

B

A

Physical meaning of Angular Momentum

Let us consider a particle rotating in XY plane about Z-axis. Suppose that at any instant, particle is at point A, whose

position vector is OA r=

and after a small time interval Δt;

it reaches point B, whose position vector is r r+ ∆

. The displacement of the particle is small time Δ t is given by

( )AB r r r r= + ∆ − = ∆

If v

is velocity of particle at point A, then the small displacement covered in time Δ t may be expressed as

r v t∆ = ∆

Draw a line OC parallel and equal to AB. Then,

OC r= ∆

Area of parallelogram OABC r r= ×∆

or area of triangle 12

OAB r r= ×∆

The area of the triangle OAB represents the area swept by the position vector of the particle in time interval Δ t. If the area swept is represented by ,A∆

then

( ) ( )1 12 2

A r r r v t∆ = ×∆ = × ∆

If p

is linear momentum of the particle, then pvm

=

Therefore, the above equation becomes

( )1 1 2 2

p AA r t r pm t m

∆∆ = × ∆ ⇒ = ×

∆

Since ,r p L× =

angular momentum of the particle about Z-axis, we have

2

2

A Lt m

AL mt

∆=

∆∆

⇒ =∆

Now, At

∆∆

= time rate, at which the area is swept by the

position vector of the particle and is called areal velocity of the particle. Thus, geometrically, the angular momentum of a particle is equal to twice the product of its mass and areal velocity i.e. angular momentum = 2 (mass × areal velocity)

Deduction of Kepler’s second law of planetary motion

• Consider a planet moving around the sun in elliptical

orbit. Let r

be position vector of the planet w.r.t. the sun

and F

be gravitational force on the planet due to the sun. • The force on the planet always acts along the line joining

the centres of the planet and the sun and is directed

towards the sun. As a result, the vectors r

and F

are

parallel vectors and c hence 0.r Fτ = × =

• Hence angular momentum is conserved. We know that,

2 AL mt

∆=

∆

• Since angular momentum of the planet is a constant vector,

2A Lt m

∆=

∆

= a constant vector

• Therefore, areal velocity of the planet must always remain constant. It is exactly what Kepler predicted about planetary motion in 1602, i.e. the line joining the planet to the sun sweeps out equal areas in equal intervals of time. It is known as the Kepler’s second law of planetary motion.

General Motion (Combined Motion)

• General Motion (Combined Motion) is a motion composed of both translation as well as rotation.

• Suppose we are studying a General Motion of a rigid body in an inertial frame whose origin is at a point O. Suppose at a particular instant of time, it has angular velocity ω

and linear velocity of its

centre of mass with respect to O is cmv

. Let us assume that the position vector of its centre of mass wrt to O be r

and MI about point O be I0.

• Then, we can express KE and angular momentum of the rigid body in combined motion in this frame as:

2 2

1 1 2 2cm cmKE I mvω= +

cmcmL I r mvω= + ×



Pure Rolling Motion

• A combined motion where point of contact remains at rest with respect to the contacting surface is known as Pure Rolling Motion or Rolling without slipping.

• Pure rolling is a kind of combined motion where point of contact does not moves wrt to the surface of contact but still, the body as a whole translates with some velocity.

• The motion where point of contact has some velocity with respect to the contacting surface is known as slipping or skidding.

• As the point of contact O has zero velocity wrt to surface, therefore disc has an instantaneous axis of rotation which passes through lowest point O. Thus we can virtually think rolling as a fixed axis rotation about lowest point O. Now, we can apply all equations about this instantaneous axis of rotation as same as that rotation about a fixed axis.

• Fundamental Equation of Pure Rolling:

Let us assume surface is stationary as in the most of the practical cases. Clearly the velocity of centre of mass of the disc wrt point O is v

cm = ω R

Similarly, acceleration of centre of mass acm

=α R

• Fundamental Equation of Pure Rolling

vcm

= ω R

acm

=α R

• The velocity and acceleration of a point increases as it moves upward from the surface during pure rolling pmotion.

• As we can assume virtually Rolling as a fixed axis of rotation about the lowest point, therefore, we can use same eqn. of KE and angular momentum as that of fixed axis of rotation.

• Kinetic energy in pure rolling:

Kinetic energy of rolling disc (or cylinder or sphere) is 2 2 2

1 1 1 2 2 2o cm cmKE I I mvω ω= = +

• Angular momentum pure rolling:

Angular momentum of rolling disc (or cylinder or sphere) about point O (i.e., about axis of rotation) is

o cmo cmL I I r mvω ω= = + ×

where, Io = Moment of Inertia about axis passing through point O.

and, r

= position vector of centre of mass wrt O.

ω

vcm

O

ω, α

vcm , acm

R

O



Physics Classes by Pranjal Sir (Admission Notice for XI & XII - 2014-15)

Batches for Std XIIth

Batch 1 (Board + JEE Main + Advanced): (Rs. 16000) Batch 2 (Board + JEE Main): (Rs. 13000) Batch 3 (Board): (Rs. 10000) Batch 4 (Doubt Clearing batch): Rs. 8000

About P. K. Bharti Sir (Pranjal Sir)

• B. Tech., IIT Kharagpur (2009 Batch) • H.O.D. Physics, Concept Bokaro Centre • Visiting faculty at D. P. S. Bokaro • Produced AIR 113, AIR 475, AIR 1013 in JEE -

Advanced • Produced AIR 07 in AIEEE (JEE Main)

Address: Concept, JB 20, Near Jitendra Cinema, Sec 4, Bokaro Steel City Ph: 9798007577, 7488044834 Email: [email protected] Website: www.vidyadrishti.org

Physics Class Schedule for Std XIIth (Session 2014-15) by Pranjal Sir

Sl. No.

Main Chapter Topics Board level JEE Main Level JEE Adv Level

Basics from XIth Vectors, FBD, Work, Energy, Rotation, SHM

3rd Mar to 4th Apr 14

1. Electric Charges and Fields

Coulomb’s Law 5th & 6th Apr 5th & 6th Apr 5th & 6th Apr Electric Field 10th & 12th Apr 10th & 12th Apr 10th & 12th Apr Gauss’s Law 13th & 15th Apr 13th & 15th Apr 13th & 15th Apr Competition Level NA 17th & 19th Apr 17th & 19th Apr

2. Electrostatic Potential and Capacitance

Electric Potential 20th & 22nd Apr 20th & 22nd Apr 20th & 22nd Apr Capacitors 24th & 26th Apr 24th & 26th Apr 24th & 26th Apr Competition Level NA 27th & 29th Apr 27th & 29th Apr, 1st, 3rd

& 4th May PART TEST 1 Unit 1 & 2 4th May NA NA

NA 11th May 11th May 3.

Current Electricity Basic Concepts, Drift speed, Ohm’s Law, Cells, Kirchhoff’s Laws, Wheatstone bridge, Ammeter, Voltmeter, Meter Bridge, Potentiometer etc.

6th, 8th, 10th, 13th May



Competition Level NA 15th & 16th May 15th, 16th, 17th, 18th & 19th May

PART TEST 2 Unit 3 18th May NA NA NA 20th May 20th May

SUMMER BREAK 21st May 2013 to 30th May 2013 4. Moving charges and

Magnetism Force on a charged particle (Lorentz force), Force on a current carrying wire, Cyclotron, Torque on a current carrying loop in magnetic field, magnetic moment

31st May, 1st & 3rd Jun



Biot Savart Law, Magnetic field due to a circular wire, Ampere circuital law, Solenoid, Toroid

5th, 7th & 8th Jun 5th, 7th & 8th Jun 5th, 7th & 8th Jun

Competition Level NA 10th & 12th Jun 10th, 12th, 14th & 15th Jun PART TEST 3 Unit 4 15th Jun NA NA

NA 22nd Jun 22nd Jun 5. Magnetism and Matter 17th, 19th & 21st

Jun 17th, 19th & 21st Jun

Not in JEE Advanced Syllabus

6. Electromagnetic Faraday’s Laws, Lenz’s Laws, 24th, 26th & 28th 24th, 26th & 28th 24th, 26th & 28th Jun

mailto:[email protected]�

http://www.vidyadrishti.org/�



Induction A.C. Generator, Motional Emf,

Induced Emf, Eddy Currents, Self Induction, Mutual Induction

Jun Jun

Competition Level NA 29th Jun & 1st Jul 29th Jun, 1st, 3rd & 5th Jul PART TEST 4 Unit 5 & 6 6th Jul NA NA

NA 13th Jul 13th Jul 7. Alternating current AC, AC circuit, Phasor,

transformer, resonance, 8th, 10th & 12th Jul 8th, 10th & 12th Jul 8th, 10th & 12th Jul

Competition Level NA 15th July 15th & 17th July 8. Electromagnetic Waves 19th & 20th July 19th & 20th July Not in JEE Advanced

Syllabus PART TEST 5 Unit 7 & 8 27th Jul 27th Jul 27th Jul

Revision Week Upto unit 8 31st Jul & 2nd Aug

31st Jul & 2nd Aug

31st Jul & 2nd Aug

Grand Test 1 Upto Unit 8 3rd Aug 3rd Aug 3rd Aug

9.

Ray Optics

Reflection 5th & 7th Aug 5th & 7th Aug 5th & 7th Aug Refraction 9th & 12th Aug 9th & 12th Aug 9th & 12th Aug Prism 14th Aug 14th Aug 14th Aug Optical Instruments 16th Aug 16th Aug Not in JEE Adv

Syllabus Competition Level NA 19th & 21st Aug 19th, 21st, 23rd, 24th Aug

10.

Wave Optics

Huygens Principle 26th Aug 26th Aug 26th Aug Interference 28th & 30th Aug 28th & 30th Aug 28th & 30th Aug Diffraction 31st Aug 31st Aug 31st Aug Polarization 2nd Sep 2nd Sep 2nd Sep Competition Level NA 4th & 6th Sep 4th, 6th, 7th, 9th, 11th Sep

PART TEST 6 Unit 9 & 10 14th Sep 14th Sep 14th Sep REVISION ROUND 1 (For JEE Main & JEE Advanced Level): 13th Sep to 27th Sep

Grand Test 2 Upto Unit 10 28th Sep 28th Sep 28th Sep

DUSSEHRA & d-ul-Zuha Holidays: 29th Sep to 8th Oct

11.

Dual Nature of Radiation and Matter

Photoelectric effect etc 9th & 11th Oct 9th & 11th Oct 9th & 11th Oct

Grand Test 3 Upto Unit 10 12th Oct 12th Oct 12th Oct

12.

Atoms 14th & 16th Oct 14th & 16th Oct 14th & 16th Oct

13. Nuclei 18th & 19th Oct 18th & 19th Oct 18th & 19th Oct X-Rays NA 21st Oct 21st & 25th Oct

PART TEST 7 Unit 11, 12 & 13 26th Oct NA NA 14. Semiconductors Basic Concepts and Diodes,

transistors, logic gates 26th, 28th, 30th Oct & 1st Nov

26th, 28th, 30th Oct & 1st Nov

Not in JEE Adv Syllabus

15.

Communication System 2nd & 4th Nov 2nd & 4th Nov Not in JEE Adv Syllabus

PART TEST 8 Unit 14 & 15 9th Nov 9th Nov NA Unit 11, 12 & 13 Competition Level NA 8th, 9th & 11th Nov 8th, 9th, 11th, 13th & 15th

Nov PART TEST 9 Unit 11, 12, 13, X-Rays NA 16th Nov 16th Nov

Revision Round 2

(Board Level)

Mind Maps & Back up classes for late registered students

18th Nov to Board Exams



Revision Round 3

(XIth portion for JEE)

18th Nov to JEE 18th Nov to JEE 18th Nov to JEE

30 Full Test Series Complete Syllabus Date will be published after Oct 2014

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