Download - XII Physics Rotational Motion
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Rotational MotionProf. Sameer Sawarkar
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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
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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.
3Prof. 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.
4Prof. 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.
5Prof. Sameer Sawarkar
A
B
C
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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.
6Prof. Sameer Sawarkar
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
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7Prof. Sameer Sawarkar
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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8Prof. Sameer Sawarkar
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
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9Prof. Sameer Sawarkar
The axis of rotation may lie within the body or without the body
,
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10Prof. Sameer Sawarkar
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.
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Prof. Sameer Sawarkar 11
CAUSE & CONSEQUENCEin Rotational Motion
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12Prof. Sameer Sawarkar
Force produces translationi.e. linear acceleration, ‘a’
Couple Moment produces rotation i.e. angular acceleration, ‘’
F
a
F
F
d
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13Prof. Sameer Sawarkar
• Rigid body subjected to
torque
• Rotating about a fixed axis
with angular acceleration
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14Prof. Sameer Sawarkar
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.
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15Prof. Sameer Sawarkar
R1
R2
Rn
a1
an
a2
• Linear tangential accelerations of constituents; a1, a2, … , an
• Using aT = R
a1 = R1 a2 = R2 … … … …… … … …an = Rn
(1)
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Prof. Sameer Sawarkar 16
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)
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Prof. Sameer Sawarkar 17
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)
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Prof. Sameer Sawarkar 18
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.
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Prof. Sameer Sawarkar 19
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.
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Prof. Sameer Sawarkar 20
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).
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Prof. Sameer Sawarkar 21
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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Moment of Inertia
I = miRi2 or I = R2 dm
Unit: kg-m2
Dimensions: [L2 M1 T0]
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Prof. Sameer Sawarkar 25
KINETIC ENERGYin Rotational Motion
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26Prof. Sameer Sawarkar
• Rigid body rotating about a
fixed axis with angular
velocity
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27Prof. Sameer Sawarkar
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.
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28Prof. Sameer Sawarkar
R1
R2
Rn
V1
Vn
V2
• Linear tangential velocities of constituents; V1, V2, … , Vn
• Using V = R
V1 = R1 V2 = R2 … … … …… … … …Vn = Rn
(1)
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29Prof. Sameer Sawarkar
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)
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30Prof. Sameer Sawarkar
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
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Prof. Sameer Sawarkar 31
ANGULAR MOMENTUM in Rotational Motion
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Prof. Sameer Sawarkar 32
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]
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Prof. Sameer Sawarkar 33
P
L
R
m
Vector relation between linear momentum and angular momentum:From scalar relation; L = R*Pand using Right-hand rule;
PRL
R
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34Prof. Sameer Sawarkar
• Rigid body rotating about a
fixed axis with angular
velocity
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35Prof. Sameer Sawarkar
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.
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36Prof. Sameer Sawarkar
R1
R2
Rn
V1
Vn
V2
• Linear tangential velocities of constituents; V1, V2, … , Vn
• Using V = R
V1 = R1 V2 = R2 … … … …… … … …Vn = Rn
(1)
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37Prof. Sameer Sawarkar
, 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)
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38Prof. Sameer Sawarkar
, 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
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Prof. Sameer Sawarkar 39
PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM
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Prof. Sameer Sawarkar 40
Ldt
dI
dt
ddt
dII
,0 LIf then is constant.
In absence of an external torque, the angular momentum of the system remains constant
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Prof. Sameer Sawarkar 41
Applications of Principle of Conservation of Angular Momentum
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Prof. Sameer Sawarkar 42
PARALLEL AXES THEOREMPERPENDICULAR AXES THEOREM
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Prof. Sameer Sawarkar 43
• Rigid body with mass M
• Purely rotating about an
axis through C.M.
• MI = IG (known)
IG
G
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Prof. Sameer Sawarkar 44
IP
• It is desired that MI
about a parallel axis at a
distance ‘h’ through P
i.e. IP be found.
IG
GP
h
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Prof. Sameer Sawarkar 45
IP
GP
• Assume elemental mass
dm at an arbitrary point
Q.
IG
Q
GP
h
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Prof. Sameer Sawarkar 46
IP
• ConstructionIG
GP
Q (dm)
Sh
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Prof. Sameer Sawarkar 47
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
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Prof. Sameer Sawarkar 48
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.
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Prof. Sameer Sawarkar 49
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.
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• Rigid with mass M
• Laminar body (thickness
very small compared to
surface area)
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Prof. Sameer Sawarkar 51
• 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
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Prof. Sameer Sawarkar 52
• Imagine elemental mass
dm at a distance ‘r’ from
Z axis.O
XY
Z
r
dm
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Prof. Sameer Sawarkar 53
• Moment of inertia of
the lamina @ Z axis;
IZ = r2dmO
XY
Z
r
dm
IZ
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Prof. Sameer Sawarkar 54
• Construction –
perpendiculars on X and
Y axes from elemental
mass.O
XY
Z
r
dm
IZ
xy
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Prof. Sameer Sawarkar 55
• 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
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Prof. Sameer Sawarkar 56
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.
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Prof. Sameer Sawarkar 57
RADIUS OF GYRATION
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58Prof. Sameer Sawarkar
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
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Prof. Sameer Sawarkar 59
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)
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Prof. Sameer Sawarkar 60
Thank You!