Transcript File - Physics Made Easy

WELCOME TO
SYSTEM OF
PARTICLES AND
ROTATIONAL
MOTION
CENTRE OF MASS AND
ROTATIONAL MECHANICS
Pure Translation Motion
In such a motion, every particle of the
body has the same velocity at a particular
instant of time.
In such a motion, a rigid body rotates
about a fixed axis. Every particle of the
body moves in a circle, which lies in a
plane perpendicular to the axis, and has its
centre on the axis.
Combination of translational and
rotational motion.
The motion of a rigid body, which is not
pivoted or fixed in some way is either a
pure translation or a combination of
translation and rotation.
(ii) The motion of a rigid body, which is pivoted
or fixed in some way is rotation.
(i)
Concept Of Centre Of Mass

We may define centre of mass of a body
or a system of bodies as a point at which
the entire mass of the body system of
bodies, is supposed to be concentrated.
Centre of Mass of a Two
Particle System
Centre of Mass of a Two Particle
System






OA = r1
OB = r2
(m1 + m2) r = m1r1 + m2r2
For Center of Mass
X = m1x1 + m2x2
m1 + m2
Y = m1v1 + m2v2
m1 + m2
Acceleration and Velocity of center of
mass
Vcm = m1v1 + m2v2
m 1 + m2 Acm = m1a1 + m2a2
m1 + m2 -
- - + mnvn
- + m2
- - mnan
- mn
Derivation





f1 = external force applied on particle of mass
m1
F2 = external force applied on particle of mass
m2
F12 = internal force on m1 due to m2
F21=internal force on m2 due to m1
Linear momentum of particle m1 is
P1 = m1v1
F12 = -f21
F12 + f21=0
Acc. To Newton’s Second law
I = dp
dt
For m1:F 10= dp1
= d(m1 v1)
- - - - (1)
dp
dt
for M2 :F12 + F21 =
dp2
dt
=
d(m2 v2)
dt
(1) +
(2) :F1 + F 2 =
d (m1v1+m2v2)
dt
-- -- (2)
F (total) =
d [ m1 dv1 + m2 dv2
dt
dt1
dt2
F(total) = d (d (m1r1 + m2r2)
dt dt
f (total) = d2 (m1r1 + m2r2)
dt2
F(total) = m1 + m2 d2 (m1r1 + m2r2)
dt2
 F(total) = m1 + m2 d2
dt2
F = ma
= mdr
dt
t = md
dr
dt
dt
F = md2r
dt2
= r = m1r1 + m2r2
m1 + m2
m1r1 + m2r2
m1 + m2
Momentum Conservation and
motion at the centre of mass
 F(total) =
d (mivi)
dt
f(t) = d (m1v1 + m2v2 - - - - mnvn)
dt
Total force = 0
d (m1v1 + m2v2 - - - mnvn) = 0
dt
d (const.) = 0
dt
m1v1 + m2v2 + m3v3 - - - - mnvn = const.
For centre of mass also
 F = ma
 F=m
F=0
mdv = 0
dt
m≠0
so, dv = 0
dt
dv
dt
(only when v = constant)
Examples Of The Motion
Of Centre Of Mass
• A special type of cracker which when
ignited goes up in the sky and then
explodes into many firry pieces. The
centre of mass of the fragment will
move on its initial parabolic path.
Sun Moon System
 Moon revolves around the earth in a circular
orbit and the earth moon system goes round the
sun in elliptical orbit (as shown), Both Earth &
moon move along the circular paths about their
cm such that they are always on opp. Sides of it.
It is the cm of earth
 moon system that exactly
follow the elliptical
path around the sun.
Sun Moon System
Radioactive Nucleus

When a radioactive nucleus initially at
rest, decays into two fragments which
fly apart obeying the laws of
conservation of momentum ,centre of
mass of the two fragments. Remain at
rest.
Angular Velocity
For rotation about a fixed axis, the
angular velocity vector lies along the
axis of rotation and points out in which
the tip of right handed screw would
advanced if the head of the screw is
rotated with the body.
ω = dθ
dt
Position of Centre of Mass
#Analog between Linear &
Rotational Motion:
Analog between Linear & Rotational
Motion:
Rigid body
Is regarded as a assembly of unit masses. The
mutual distances among the distances
among the different point masses do not
change during the motion of rigid body.
Angular Acceleration



Angular acceleration of an object in
circular motion is defined us the time
rate of its angular velocity. It is
represented by Greek letter α (alpha).
i.e., angular acceleration
= change in angular velocity
Time taken
Derivation Of Equations
ω = ω0 + αt
• α = dω
dt
• Putting integration :
• αα (t-0) = ω (ω-ω0)
• α t =-ωo+ αt
_ _
• dt = dω]
• ω = ω0 + αt
(1)
θ = w0 t +(α t)2
2
• ω = dθ
dt
• from (1):ω = + ωo+ αt
• dθ = (ωo+ αt) dt
• dθ = ωodt+ αtdt
• putting
integration
• (θ- θo) = ωo (t - 0) = α(t2)
2
• θ = ωot + (αt)2
2
θ = ωot + (αt)2
2
• α = dω
dt
(multi divide by dθ)
• α = dw X dω
dt
dθ
• α = dω X dθ
dθ
dt
• α = dω X ω
dθ
• αdθ = ωdω
• Putting integration
• α (θ – θ0) = ω2 – ω02
2
• ω2 = ω02 + 2 α θ
Law of Conservation of
Angular Momentum
• T = dL
dt
• T=0
• dL = 0
dt
• since L = const.
• L=rxp
• = r x m1v
• = r x m x rω
• L = mr2ω
• L=I ω
So, I ω = const.
KINETIC ENERGY OF ROLLING
MOTION
K = KT = KR
 If m is the mass of the body & v cm is
velocity
 of centre of mass of the body, then
 KE of translation, KT = 1/2 mv2
 KE of rotation, kr = ½ Iω2
 Where I = moment of inertia







KE of rolling body,
K = ½ mv2cm+ 1/2Iω2
Vcm = Rω
ω = Vcm
R
K = 1/2mv2cm + ½ mk2 Vcm
R
K = ½ mv2cm = [1+ k2]
R2
Laws of Rotational Motion
 Corresponding to Newton’s three laws of translational motion,
we can state three laws of rotational motion as follows:
 A body continues to be in a state of rest or in a state of uniform
rotation about a given axis unless an external torque is applied
on the body.
 The rate of change of angular momentum of a body about a
given axis is directly proportional to the external torque applied
on the body.
 When a rigid body A exerts a torque on another rigid body B in
contact with it, then the body B would exert an equal and
opposite torque on the body A.
Moment of Inertia
 This force varies directly as the mass of the
body. Hence mass of a body is a measure of
inertia of the body in linear motion.
 A quantity that measures the inertia of
rotational motion of the body is called
rotational inertial or moment of inertia of the
body.
 Kinetic Energy of Rotation
 Kinetic energy of rotation of a body is the
energy possessed by the body on account of
its rotation about a given axis.
 This force varies directly as the mass of the
body. Hence mass of a body is a measure of
inertia of the body in linear motion.
 This force varies directly as the mass of the body.
Hence mass of a body is a measure of inertia of the
body in linear motion.
 A quantity that measures the inertia of rotational
motion of the body is called rotational inertial or
moment of inertia of the body.
 Kinetic Energy of Rotation
 Kinetic energy of rotation of a body is the energy
possessed by the body on account of its rotation
about a given axis.
A quantity that measures the inertia of
rotational motion of the body is called
rotational inertial or moment of inertia of
the body.
Kinetic Energy of Rotation
Kinetic energy of rotation of a body is the
energy possessed by the body on account
of its rotation about a given axis.
Radius of Gyration
 The radius of gyration of a body
about a given axis is the
perpendicular distance of a point P
from the axis, where if whole mass of
the body were concentrated, the body
shall have the same moment of
inertia as it has with the actual
distribution of mass.
 K =  r12 + r22 + …r n2
n
 Radius of gyration of a body about a
given axis is equal to root mean
square distance of the constituent
particles of the body from the given
axis.
Relation between Torque and Angular
Momentum of a Rigid Body.
 L=In
 Differentiating body sides w.r.t. t, we get
 dL = I dw = Iα
dt
dt
 where α = dw = angular acceleration of the body.
dt
 But torque, T= Iα
 From above equation
 T = dl
dt
Principle of Conservation of
Angular Momentum
 When no external torque acts on a system
of particles, then the total angular
momentum of the system remains always a
constant.
  When no external torque acts on the
system,
 d
dt
 or
THEOREM OF PARALLEL AXES:
 According to this theorem, momentum of
inertia of a rigid body about any axis AB is
equal to moment of inertia of the body
about another axis KL passing through
centre of mass C of the body in a direction
parallel to AB, plus the product of total
mass M of the body and square of the
perpendicular distance between the two
parallel axes.
 IAB = IKL + Mh2
THEOREM OF PARALLEL AXES:
Proof. Suppose the rigid body is made up of n
particles of masses m1, m2, m3 …..mn at
perpendicular distances 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
• The perpendicular distance of ith
particle from the axis KL = (ri + h)
Theorem of Perpendicular
Area
• According to this theorem, the moment of
inertia of a plane lamina (i.e. a two
dimensional body of any shape/size) about
any axis OZ perpendicular to the plane of
the lamina is equal to sum of the moments of
inertia of the lamina about any two mutually
perpendicular axes OX and OY in the plane
of the lamina, meeting at a point where the
given axis OZ passes through the lamina.
Suppose the lamina is in XY plane, Let
•
•
•
•
•
•
IX = moment of inertia of the lamina about OX
Iy = moment of inertia of the lamina about OY
Iz = moment of inertia of the lamina about OZ.
According to the theorem of perpendicular axes,
Iz = 1x + 1y
Proof. Suppose the lamina consists of n particles
of masses m1, m2, m3, ….mn at
KINETIC ENERGY OF ROLLING MOTION












K = KT = KR
If m is the mass of the body & v cm is velocity
of centre of mass of the body, then
KE of translation, KT = 1/2 mv2
KE of rotation, kr = ½ Iω2
Where I = moment of inertia
KE of rolling body,
K = ½ mv2cm+ 1/2Iω2
Vcm = Rω
ω = Vcm
R
K = 1/2mv2cm + ½ mk2 Vcm
R
K = ½ mv2cm = [1+ k2]
R2
Laws of Rotational Motion
 Corresponding to Newton’s three laws of translational motion, we
can state three laws of rotational motion as follows:
 A body continues to be in a state of rest or in a state of uniform
rotation about a given axis unless an external torque is applied on
the body.
 The rate of change of angular momentum of a body about a given
axis is directly proportional to the external torque applied on the
body.
 When a rigid body A exerts a torque on another rigid body B in
contact with it, then the body B would exert an equal and opposite
torque on the body A.
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