XII Physics Rotational Motion

Rotational MotionProf. Sameer Sawarkar

Contents

• Rigid Body• Rotational Motion• Cause & Consequence• Moment of Inertia• Kinetic Energy• Angular Momentum• Conservation Principle• Parallel & Perpendicular Axes Theorems• Radius of Gyration• Rolling Motion

Prof. Sameer Sawarkar 2

Rigid Body: A body which does not undergo any appreciable deformation under the

action of external forces, i.e. the intermolecular distances remain constant when

subjected to external forces.

3Prof. Sameer Sawarkar





A

B

C





No body is truly rigid nor elastic or plastic. The state is always referred to as rigid/elastic/plastic in context with the magnitude and range of external forces.

A

B

C


Rotational Motion: A body is said to be purely rotating

when all the constituents of the body are moving in circular motions, with centers of their paths lying on a

fixed straight line called axis of rotation.

,


Rotational Motion: A body is said to be purely rotating

when all the constituents of the body are moving in circular motions, with centers of their paths lying on a

fixed straight line called axis of rotation.

,

A

B


The axis of rotation may lie within the body or without the body

,


Examples: Motion of table/ceiling fan bladesMotion of Turbine rotorMotion of gear wheelsSpinning Motion of planetsOpening of doors/window panelsMotion of hands of clock etc.


CAUSE & CONSEQUENCEin Rotational Motion


Force produces translationi.e. linear acceleration, ‘a’

Couple Moment produces rotation i.e. angular acceleration, ‘’

F

a

F

F

d


• Rigid body subjected to

torque

• Rotating about a fixed axis

with angular acceleration


1

n

2

R1

R2

Rn

• Consider ‘n’ particles of the

body in circular motion with

masses m1, m2, … , mn.

• R1, R2, … , Rn are the radii.


R1

R2

Rn

a1

an

a2

• Linear tangential accelerations of constituents; a1, a2, … , an

• Using aT = R

a1 = R1 a2 = R2 … … … …… … … …an = Rn

(1)


R1

R2

Rn

a1

an

a2

F1

F2

Fn

a1 = R1 , a2 = R2 , … , an = Rn _(1)

• Using Newton’s II Law; F = ma

F1 = m1a1 = m1R1 F2 = m2a2 = m2R2 … … … … … … … …… … … … … … … …Fn = mnan = mnRn

(2)


F1

F2

Fn

R1

R2

Rn

a1

an

a2

1

n

F1

F2

Fn

F1 = m1R1 , F2 = m2R2 , … … … Fn = mnRn _(2)

• Using definition of torque; = d*F1 = R1F1 = R1(m1R1)

1 = m1R12

2 = m2R22

… … … … … … … … … …n = mnRn

2

(3)


F1

F2

Fn

R1

R2

Rn

a1

an

a2

1

n

F1

F2

Fn

1 = m1R12, 2 = m2R2

2, … … … n = mnRn

2 _(3)

• Sum of all individual constituent torques must be equal to the externally applied original torque.

= 1 + 2 + … + n

= m1R12 + m2R2

2 + … + mnRn2

= (miRi2)

i = 1, 2, … , n.


Translational Motion Rotational Motion

= (miRi2)*F = m*a

F a

m miRi2

Quantity miRi2 is called as Moment of Inertia of rotating body

about the defined axis of rotation.


Moment of Inertia (miRi2 ) about a given axis of rotation is

defined as the sum of product of

mass of each constituent and

square of its distance from the axis of rotation.

Moment of Inertia (abbreviated as MI, denoted by I) represents

inertia in rotational motion i.e. reluctance of a rigid body to

undergo angular acceleration. Larger the MI, more difficult it is

to change the state of the body (to accelerate/decelerate).


With regular geometric boundaries,

where division in discrete shapes is

possible, MI is expressed as;

I = miRi2

With irregular geometric boundaries,

where division in elemental shapes is

necessary, MI is expressed as;

I = R2 dm


• I = (mi, Ri2)

• MI represents mass distribution of the rotating rigid body.

• Rotational motion depends not just upon total mass but upon

mass distribution!!


Moment of Inertia

I = miRi2 or I = R2 dm

Unit: kg-m2

Dimensions: [L2 M1 T0]


KINETIC ENERGYin Rotational Motion


• Rigid body rotating about a

fixed axis with angular

velocity


1

n

2

R1

R2

Rn






R1

R2

Rn

V1

Vn

V2

• Linear tangential velocities of constituents; V1, V2, … , Vn

• Using V = R

V1 = R1 V2 = R2 … … … …… … … …Vn = Rn

(1)


R1

R2

Rn

V1

Vn

V2

V1 = R1 , V2 = R2 , … , Vn = Rn _(1)

• KE = ½ mV2 = ½ m(R22) of each constituent.

U1 = ½ m1R122

U2 = ½ m2R222

… … … … … …… … … … … …Un = ½ mnRn

22

(2)


R1

R2

Rn

V1

Vn

V2

U1 = ½ m1R122, U2 = ½ m2R2

22, … … Un = ½ mnRn

22 _(2)

• Total KE of the rotating rigid body;U = U1 + U2 + … + Un

U = ½ m1R122 + ½ m2R2

22 + … + ½ mnRn

22

U = ½ (miRi2)2

U = ½ I2


ANGULAR MOMENTUM in Rotational Motion


V, mV

L

R

m

Angular Momentum: Property possessed

by a rotating body by virtue of its

angular velocity.

Defined as; moment

of linear momentum.

i.e. L = R*P = R*(mV)

Like linear momentum, angular momentum

is a vector.

Unit: kg-m2/s, Dimensions: [L2 M1 T-1]


P

L

R

m

Vector relation between linear momentum and angular momentum:From scalar relation; L = R*Pand using Right-hand rule;

PRL

R


• Rigid body rotating about a

fixed axis with angular

velocity


1

n

2

R1

R2

Rn






R1

R2

Rn

V1

Vn

V2

• Linear tangential velocities of constituents; V1, V2, … , Vn

• Using V = R

V1 = R1 V2 = R2 … … … …… … … …Vn = Rn

(1)


, L

R1

R2

Rn

V1

Vn

V2

V1 = R1 , V2 = R2 , … , Vn = Rn _(1)

• Linear momentum P = mV for each constituent.

• Angular momentum for each constituent; L = R*P = RmV = Rm(R) = mR2

L1 = m1R12

L2 = m2R22

… … … … … …… … … … … …Ln = mnRn

2

(2)


, L

R1

R2

Rn

V1

Vn

V2

L1 = m1R12, L2 = m2R2

2, … … Ln = mnRn

2 _(2)

• Total angular momentum of the rotating rigid body;L = Li, i = 1, 2, … , n.

L = m1R12 + m2R2

2 + … + mnRn2

L = (miRi2)

L = I


PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM


Ldt

dI

dt

ddt

dII

,0 LIf then is constant.

In absence of an external torque, the angular momentum of the system remains constant


Applications of Principle of Conservation of Angular Momentum


PARALLEL AXES THEOREMPERPENDICULAR AXES THEOREM


• Rigid body with mass M

• Purely rotating about an

axis through C.M.

• MI = IG (known)

IG

G


IP

• It is desired that MI

about a parallel axis at a

distance ‘h’ through P

i.e. IP be found.

IG

GP

h


IP

GP

• Assume elemental mass

dm at an arbitrary point

Q.

IG

Q

GP

h


IP

• ConstructionIG

GP

Q (dm)

Sh


IP IG

GP

Q (dm)

Sh

IG = QG2dm

IP = QP2dm

QP2 = PS2 + SQ2

= (PG + GS)2 + SQ2

= PG2 + 2PG*GS + (GS2 +

SQ2)

QP2 = PG2 + 2PG*GS + QG2


IP IG

GP

Q (dm)

Sh

QP2 = PG2 + 2PG*GS + QG2

Multiplying throughout by dm and

integrating;

òQP2 dm = PG2dm + 2PG GSdm

+ QG2dm

òQP2 dm = IP

QG2dm = IG

PG2dm = PG2dm = Mh2

GSdm = 0, G being the center of

mass of the body.


IP

Substituting;

IP = IG + Mh2IG

GP

Q (dm)

Sh

MI of a rigid body about any

axis is equal to sum of its MI

about a parallel axis through

center of mass and product of

mass of body and square of the

distance between two parallel

axes.


• Rigid with mass M

• Laminar body (thickness

very small compared to

surface area)


• System of 3 mutually

perpendicular axes

through any point O.

• X and Y in the plane of

the lamina, Z being

perpendicular to the

plane.

O

XY

Z


• Imagine elemental mass

dm at a distance ‘r’ from

Z axis.O

XY

Z

r

dm


• Moment of inertia of

the lamina @ Z axis;

IZ = r2dmO

XY

Z

r

dm

IZ


• Construction –

perpendiculars on X and

Y axes from elemental

mass.O

XY

Z

r

dm

IZ

xy


• MI of lamina about X

axis;

IX = y2dm

• MI of lamina about Y

axis;

IX = x2dm

O

XY

Z

r

dm

IZ

xy IY

IX


r2 = x2 + y2

Multiplying throughout by

dm and integrating;

r2dm = x2dm + y2dm

Substituting;

O

XY

Z

r

dm

IZ

xy IY

IX

IZ = IX + IY

Moment of inertia of a lamina about an axis perpendicular to its plane is equal to sum of its moments of inertia about two mutually perpendicular axes in the plane of lamina and concurrent with that axis.


RADIUS OF GYRATION


Radius of Gyration (K) w.r.t. the given axis of rotation is the theoretical distance at which, when entire mass of the body is assumed to be concentrated, gives same MI (of idealized point mass system) as that of the original rigid body. If MK2 = R2dm, then K is the radius of gyration.

I = R2dm I = MK2

MK


IG = ½MR2 IG = MK2

KM

REAL SYSTEMS IDEALIZED SYSTEMS

IG = 2MR2/5 IG = MK2

KM

MK2 = ½MR2

K = R/2

MK2 = 2MR2/5

K = R*(2/5)


Thank You!

XII Physics Rotational Motion

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