XII Physics Rotational Motion

October 22, 2017 | Author: Sameer Sawarkar | Category: Rotation Around A Fixed Axis, Force, Rotation, Space, Classical Mechanics
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An important concept from Mechanics section of Physics, dealing with fundamentals of rotational motion of rigid bodies....

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Rotational Motion Prof. 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

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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.

Prof. Sameer Sawarkar

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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.

Prof. Sameer Sawarkar

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B A C

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.

Prof. Sameer Sawarkar

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B A C

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.

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.

Prof. Sameer Sawarkar

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, 

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.

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, 

A

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 B paths lying on a fixed straight line called axis of rotation.

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, 

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

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Examples: Motion of table/ceiling fan blades Motion of Turbine rotor Motion of gear wheels Spinning Motion of planets Opening of doors/window panels Motion of hands of clock etc.

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CAUSE & CONSEQUENCE in Rotational Motion

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Force produces translation i.e. linear acceleration, ‘a’

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

F



d F a

F

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 

• Rigid body subjected to

torque  • Rotating about a fixed axis with angular acceleration 

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• Consider ‘n’ particles of the



body in circular motion with

 R1 2

1

masses m1, m2, … , mn.

R2

• R1, R2, … , Rn are the radii.

Rn n

Prof. Sameer Sawarkar

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a1



• Using aT = R

R1 R2 an

a2

Rn

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

a1 = R 1  a2 = R 2  ………… ………… an = R n 

Prof. Sameer Sawarkar

(1)

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a1 = R1 , a2 = R2 , … , an = Rn  _(1) • Using Newton’s II Law; F = ma



a1

 R1

F2 R2

an

a2

F1

Rn

F1 = m1a1 = m1R1  F2 = m2a2 = m2R2  …………………… …………………… Fn = m nan = m nR n 

(2)

Fn

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F1 = m1R1 , F2 = m2R2 , … … … Fn = m nR n  _(2)  

1

F2

• Using definition of torque;  = d*F 1 = R1F1 = R1(m1R1)

a1 R1

R2 an

a2

F1

Rn n Fn

1 = m1R12 2 = m2R22 …………… …………… n = mnRn2

Prof. Sameer Sawarkar

(3)

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1 = m1R12, 2 = m2R22, … … … n = mnRn2 _(3)  

1

F2

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

a1 R1

R2 an

a2

F1

Rn n Fn

 = 1 + 2 + … + n  = m1R12 + m2R22 + … + mnRn2

 = (miRi2)  i = 1, 2, … , n.

Prof. Sameer Sawarkar

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Translational Motion

Rotational Motion

F

F = m*a

a

 = (miRi2)*

m  miRi2 Quantity miRi2 is called as Moment of Inertia of rotating body about the defined axis of rotation. Prof. Sameer Sawarkar

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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). Prof. Sameer Sawarkar

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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

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• I = (mi, Ri2) • MI represents mass distribution of the rotating rigid body. • Rotational motion depends not just upon total mass but upon mass distribution!!

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Prof. Sameer Sawarkar

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Moment of Inertia I = miRi2 or I = R2 dm Unit: kg-m2 Dimensions: [L2 M1 T0]

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KINETIC ENERGY in Rotational Motion

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• Rigid body rotating about a

fixed axis with angular velocity 

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• Consider ‘n’ particles of the



body in circular motion with R1 2

1

masses m1, m2, … , mn.

R2

• R1, R2, … , Rn are the radii.

Rn n

Prof. Sameer Sawarkar

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V1

• Using V = R

R1 R2 Vn

V2

Rn

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

V1 = R1  V2 = R2  ………… ………… Vn = Rn 

Prof. Sameer Sawarkar

(1)

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V1 = R1 , V2 = R2 , … , Vn = Rn  _(1) 

V1 R1

R2 Vn

V2

Rn

• KE = ½ mV2 = ½ m(R22) of each constituent. U1 = ½ m1R122 U2 = ½ m2R222 ……………… ……………… Un = ½ mnRn22

Prof. Sameer Sawarkar

(2)

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V1

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

R1 R2 Vn

V2

Rn

U1 = ½ m1R122, U2 = ½ m2R222, … … Un = ½ mnRn22 _(2)

U = ½ m1R122 + ½ m2R222 + … + ½ mnRn22 U = ½ (miRi2)2

U = ½ I2

Prof. Sameer Sawarkar

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ANGULAR MOMENTUM in Rotational Motion

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Angular Momentum: Property possessed by a rotating body by virtue of its

angular velocity.

L V, mV

Defined as; moment of linear momentum.

R

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

m

Like linear momentum, angular momentum is a vector. Unit: kg-m2/s, Dimensions: [L2 M1 T-1]

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

L

P

L  RP

R m R

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• Rigid body rotating about a

fixed axis with angular velocity 

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• Consider ‘n’ particles of the



body in circular motion with R1 2

1

masses m1, m2, … , mn.

R2

• R1, R2, … , Rn are the radii.

Rn n

Prof. Sameer Sawarkar

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V1

• Using V = R

R1 R2 Vn

V2

Rn

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

V1 = R1  V2 = R2  ………… ………… Vn = Rn 

Prof. Sameer Sawarkar

(1)

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V1 = R1 , V2 = R2 , … , Vn = Rn  _(1)

, L

V1 R1

R2 Vn

V2

Rn

• Linear momentum P = mV for each constituent. • Angular momentum for each constituent; L = R*P = RmV = Rm(R) = mR2 L1 = m1R12 L2 = m2R22 ……………… ……………… Ln = mnRn2 Prof. Sameer Sawarkar

(2)

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, L

V1 R1

R2 Vn

V2

Rn

L1 = m1R12, L2 = m2R22, … … Ln = mnRn2

_(2)

• Total angular momentum of the rotating rigid body; L = Li, i = 1, 2, … , n. L = m1R12 + m2R22 + … + mnRn2 L = (miRi2)

L = I

Prof. Sameer Sawarkar

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PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM

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d   I  I dt d d   I   L  dt dt If

  0,

then

L

is constant.

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

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Applications of Principle of Conservation of Angular Momentum

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PARALLEL AXES THEOREM PERPENDICULAR AXES THEOREM

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IG

• Rigid body with mass M

• Purely rotating about an axis through C.M. • MI = IG (known)

G

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IP

IG

• It is desired that MI

about a parallel axis at a distance ‘h’ through P i.e. IP be found.

G

P h

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IP

IG

• Assume elemental mass Q

dm at an arbitrary point Q.

PP

G h

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IP

IG

• Construction Q (dm)

P

h

G

S

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IP

IG

IG = QG2dm 2dm I = QP P Q (dm)

QP2 = PS2 + SQ2 = (PG + GS)2 + SQ2 P

h

G

S

= PG2 + 2PG*GS + (GS2 + SQ2) QP2

Prof. Sameer Sawarkar

= PG2 + 2PG*GS + QG2

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QP2 = PG2 + 2PG*GS + QG2 IP

Multiplying throughout by dm and

IG

integrating; Q (dm)

QP2 dm =  PG2dm + 2PG  GSdm +  QG2dm

P

h

G

S

QP2 dm = IP  QG2dm = IG  PG2dm = PG2dm = Mh2

 GSdm = 0, G being the center of mass of the body. Prof. Sameer Sawarkar

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Substituting; IP

IG

IP = IG + Mh2 Q (dm)

MI of a rigid body about any

axis is equal to sum of its MI about a parallel axis through P

h

G

S

center of mass and product of mass of body and square of the distance between two parallel axes.

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• Rigid with mass M

• Laminar body (thickness very small compared to surface area)

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Z

• System of 3 mutually perpendicular axes through any point O. O

• X and Y in the plane of

Y

the lamina, Z being

perpendicular to the

X

plane.

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Z

• Imagine elemental mass dm at a distance ‘r’ from Z axis. O r X

Y dm

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Z

• Moment of inertia of IZ

the lamina @ Z axis; IZ =  r2dm

O r X

Y dm

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Z

• Construction – IZ

perpendiculars on X and Y axes from elemental

O

mass. r

X

y

x

Y

dm

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Z

• MI of lamina about X IZ

axis; IX =  y2dm

O IX X

• MI of lamina about Y r

y

x dm

Y IY

Prof. Sameer Sawarkar

axis;

IX =  x2dm

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Z

r2 = x 2 + y 2 Multiplying throughout by

IZ

dm and integrating;

 r2dm =  x2dm +  y2dm

O IX X

r y

x dm

Y IY

Substituting;

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. Prof. Sameer Sawarkar

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RADIUS OF GYRATION

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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

K

Prof. Sameer Sawarkar

M

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IG = ½MR2

IG = MK2

K

REAL SYSTEMS IG = 2MR2/5

M

MK2 = ½MR2

 K = R/2

IDEALIZED SYSTEMS

IG = MK2

MK2 = 2MR2/5 K

Prof. Sameer Sawarkar

M

 K = R*(2/5)

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Thank You!

Prof. Sameer Sawarkar

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