Transcript File

ROTATIONAL MOTION
Mrs. CHITRA JOSHI
PGT (Physics)
CENTRE OF MASS AND
ROTATIONAL MECHANICS
System: A collection of any number of particles
interacting with one another are said to
form a system.
Concept of centre of mass: Centre of mass of
a body or a system of bodies as the point that moves
as though all of the were concentrated there and
all external forces were applied there.
Centre of mass of a two particle
Let m1, m2 are masses
R1,R2 position vector of mass m.
R of centre of mass is given by
Centre of mass of two particle system always
lies between the particle (m1&m2)
coordinate of m1 are x1y
coordinate of m2 are x2y2
coordinate of C.M.
x = (m1x1 + m2x2)/m1+m2
y = (m1y1 + m2y2)/m1 + m2
m1= m2= m
If m1 = m2
Then x = m(x1 + x2)/2m
= (x1 + x2)/2
C.M lies at the mid point of the line joining m1,m2
(when m1= m2= m)
Momentum conservation of centre of mass equation
of motion of centre of mass given by Fext = dmv/dt
fext = 0
d(mv) = 0
dt
mv = constant
If no external force is act on system momentum is
constant is called law of conservation of momentum.
Eg. A special type of cracker which when ignited goes
up in the sky and then expands in the mid air into many
firely pieces. As the cracker expands under the action of
internal forces the centre of mass of 4 fragment should
continue moving along the same parabolic path. The
centre of mass of 4 fragments will move on initial
parabolic path PBC, which is the continuation of the
initial parabolic path AP of the cracker.
Rigid body : A body is taken as rigid, where change in
inter particle distances under the effect of external forces
can be ignored.
Center of mass of a rigid body : The center of mass
of a rigid body is defined as a point where the entire mass
of the rigid body is supposed to be concentrated. The
nature of motion of the rigid body shall remain
unaffected, if all the forces acting on the body were
applied directly on the center of mass of the body.
s.
no.
Body
Position of center of
mass
1.
2.
3.
4.
5.
6.
Uniform hollow sphere
Uniform solid sphere
Uniform circular ring
Uniform circular disc
Uniform rod
A plane lamina in the form
of a square or a rectangle or
a parallelogram.
Triangular plane lamina
Center of the sphere
Center of the sphere
Center of the ring
Center of the disc
Center of the rod
Points of intersection of diagonals.
Points of intersection of the
medians of the triangle.
Points of intersection of the
diagonals.
Middle point of the axis of the
cylinder.
Middle point of the axis of the
cylinder.
On the axis of the cone at a point
distant 3h/4 from the vertex O
where h=OA is height of the cone.
7.
8.
9.
Rectangular or cubical
block.
Hollow cylinder
10.
Solid cylinder
11.
Cone or pyramid
Rotational motion of a particle in a plane
and concept of Torque
Torque:- It is a product of force and perpendicular
distance.
τ = F*d
Unit of torque is Nm
[τ ] = [ML2T-2]
Expression of torque in Cartesian coordinate system
W = F.dr or W = τ . dθ
= (Fxî +Fyĵ) * (dxî + dyĵ)
W = Fxdx + Fydy --------------(1)
[(d/dθ)*cosθ = - sinθ]
[(d sinθ/dθ) = cosθ]
x = r cosθ
dx/dθ = -r sinθ
dx = -ydθ ------------(2) (because y = r sinθ)
dy/dθ = r cosθ
dy = xdθ -------------(3)
From (1),(2) &(3)
W = F*dy + Fydy
= F*(-ydθ) + Fy(xdθ)
= (xFy – yFx)dθ -------(4)
dW = τdθ -----------------(5)
From (4)&(5)
τdθ = (xfy – yfx)dθ
dω = τdθ (rotational motion)
dω/dt = τdθ/dt (linear motion)
P = τω
If we apply force in x,y plane torque will be along z axis.
Expression of Torque in polar coordinate
system
Fx = F cos α
Fy = F sin α
α=θ+ ø
Ø=α-θ
x = r cosθ
y = r sinθ
τ = xfy – yfx
= r cosθ. Fsinα – r sinθ.F cosα
= rf (sinα.cosθ – cosα. sinθ)
= rf sin (α – θ)
= rf sin ø
= f(r sin ø)
= F*d
Angular momentum: If we can represent torque as the
rate of change of some quantity, that quantity will be
rotational analogue of linear momentum, and it can
appropriately be called the angular momentum.
L = r*p
= r*mv
Mkg*m/s
Unit of angular momentum, L = Kgm2/s
[L] = [ML2T-1]
Expression for angular momentum
In order to express torque as the rate of change of some
quantity , we rewrite expression for torque rotating a
particle in XY plane as
τ = xFy-yFx
If Px = mvx and Py = mvy are the x and y components of
linear momentum of the body, then
According to Newton’s 2nd law of motion,
= Fy
dPy/dt
dPx/dt = Fx
τz = d/dt*(xPy – yPx) ------------(1)
τ = dl/dt
-------------(2)
[L + xPy – yPx] from (1)&(2)
Px = p cosα
py = p sinα
α=θ+Ø
Ø=α+θ
x = r cos θ
y = r sin θ
From above Equation.
L = xPy – yPx
= r cos θ.P sin α – r sin θ.Pcos α
= rP (cosθ.sinα – cosα. Sinθ)
= rP sin (α - θ)
L = rP sinØ
Pθ = P sinØ-----------(a)
Pr = p cosØ
[L = Pθr] -----------(b)
from (b)
L = Pd
Geometrical meanings of angular momentum
Z
M
o
X
Y
K
L
Area of parallelogram on OABC = r + dr
Area of triangle OAB = ½*r*dr
Velocity of particle = ν
Time taken = dt
V = dr/dt
dr = v*dt
dA = ½*r*v*dt
We know that
P = mv
v = p/m
dA = ½ r*P/m*dt
dA/dt = L/2m
[L =r*p]
[dA/dt = Rate of change of are by time is called
Areal velocity.]
L = 2 mdA/dA
Angular momentum is equal to the product of 2
mass and areal velocity.
Rotational inertia or moment of inertia of
the body: A quantity that measures the inertia of
rotational motion of the body.
Units of moment of Inertia
As
I = mass (distance)2
Therefore units of moment of inertia are Kg m2 or g
cm. The dimensions of moment of inertia are [MLT]s
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. This
distance is represented by k.
I = Mk2 ------------(1)
= m1r2 + m2r22 …mnrn2
(m1 + m2 + m3 = M)
I = m(r12 + r22 + r32 +…rn2)
= {m*n (r12 + r22 + r32 +…rn2)/n} ------------(2)
From (1)&(2)
Mk2 = M (r12 + r22 + r32 +…rn2)
K = [(r12 + r22 + r32 +…rn2)/n]1/2
‘K’ is root mean square of distance of the particle from
the axis of rotation
Torque and moment of inertia
consider a rigid body rotating bout a given axis with a
uniform angular α, under the action of a torque.
Let the body consist of particles of masses
m1,m2,m3…mn at perpendicular distances r1,r2,r3…rn
respectively from the axis of rotation
As the the body is rigid , angular acceleration α of all the
particles of the body is same . However there linear
acceleration are different because of different distance of
the particles form the axis.
If a1, a2, a3…an are the respective linear acceleration of
the particles, then
a1 = r1α, a2 = r2 α, a3 =r3 α,…
Force on the particle of mass m is
F = m1α1 = m1r1α
Moment of this force about the axis of rotation
= f1 * r1 = (m1r1 α)*r1 = m1r12α
Similarly, moments of force on other particles about the
axis of rotation are m2r22 α, m3r32 α,…mnrn2 α.
Therefore Torque acting on the body,
τ = m1r2 α + m22r22 α + m3r32 α +…mnrn2α
= (m1r12 + m2r22 + m3r32 +…mnrn2)α
i=n
τ = ( Σmir12 ) α
i=1
or τ = I α
i=n
where I = ( Σ mir12 )moment of inertia of the body
i=1
about the given axis of rotation. If α = 1, τ = I*1 or I = τ
Hence moment of inertia of a body about a given the
axis is numerically equal to torque acting on the body
rotating with unit angular acceleration about it.
We may rewrite equation (9) in vector form as
τ =Iα
This equation is called Fundamental equation of
rotation or law of rotation.This corresponds to F = m α,
which is the fundamental equation of linear motion.
Principle of conservation of Angular momentum:
According to this principle, when no external torque
acts on a system of particles,then the total angular
momentum of the system remains always a constant.
Some examples of conservation of Angular momentum
1.An ice skater or a ballet dancer can increases her angular
velocity by folding her arms and bringing the stretched
legs close to other leg.
2.The driver jumping from the spring board sometimes
exhibits somersaults in air before touching the water layer.
Theorem of Perpendicular axis:- It state that the
momentum of inertia of a plane lamina about an axis
perpendicular to its plane is equal to the sum of moments
of inertia of lamina about any two mutually perpendicular
axis to its plane and intersecting each other at the point
where the perpendicular axis passing through it.
Theorem of parallel axis:- It state that the moment
of inertia of a rigid body about any axis is equal to its
moment of inertia about a parallel axis through its centre
of mass and the product of mass of the body and the
square of the perpendicular distance between the axis.
Moment of inertia of a thin circular ring
suppose M is the mass of a thin circular ring of radius R
with centre O.we have to calculate moment of inertia of
the ring about an axis YOY’,perpendicular to the plane
of the ring and passing through its centre,Fig.5(b).12.
Y
O
Y’
R
dx
Length of the ring = circumference = 2πR
Mass per unit length of the ring = M/2πR
Consider a small element of the ring of length dx
Mass of this element = M dx
2πR
Moment of inertia of this element about YOY’
= [(M/2πR)*dx] R2 =( MR/2π)*dx
As the small element can lie anywhere over the entire
length of the ring i.e., from x = 0 to x = 2πR, therefore,
Moment of inertia of the entire circular ring about
YOY’
I = MR2 = mass(radius)2
Moment of inertia of thin uniform Rod
Suppose M is mass of a thin uniform rod of length l. We
have to calculate moment of inertia of the rod about an
axis YY` passing through the centre O of the rod and
perpendicular to its length,
Mass per unit length of the rod = M/l
Consider a small element of length = dx, of the rod at a
distance x from the centre.
Mass of the element = M/l * dx
Moment of inertia of this element about the given axis
= mass x (distance)2
= M/l * dx (x 2)
= M/l .x2 .dx
As the element chosen can lie anywhere from left end of
the rod (x = -l/2) to the right end of the rod (x = + /2),
therefore,
Moment of inertia of the uniform rod about YOY` is
I = 1/12 Ml2
BODY
1. Uniform rod of length l.
2. Uniform rectangular
lamina of length l and
breadth b.
3. Uniform circular ring
of radius of radius r.
4. Uniform circular disc
of radius r.
5.Hollow cylinder of
radius r.
6. Solid cylinder of
radius r.
7.Hollow sphere of
radius r.
8. Solid sphere of radius r.
AXIS
1. Perpendicular to rod
through its centre.
2.Perpendicular to lamina
and through its centre.
M.I.
1/12Ml
M(l2
+b2)/12
3. Perpendicular to its
MR2
plane through the centre
4.Perpendicular to its plane
2
and through the centre. ½ MR
5. Axis of cylinder
6. Axis of cylinder
7. Diameter
MR2
½ MR2
2/3
8. Diameter
MR2
2/5 MR2
Linear Motion
1. Distance /displacement (s)
Rotational Motion
1. Angle or angular
displacement(θ)
2. Linear velocity, v = ds/ dt
2. Angular velocity ,ω = d θ/dt
3. Linear acceleration,a = dv/dt 3. Angular acceleration,a = dω/dt
4. Mass (m)
4. Moment of inertia (I)
5. Linear momentum p = mv
5. Angular momentum L = I ω
6. Force F = ma
6. Torque, τ = I ω
7. Also, force F = dp/dt
7. Also, torque τ = dL/dt
8. Translational K.E. = ½ mv2
8. Rotational K.E. = ½ I ω2
9. Work done, W = Fs
9. Work done, W = τθ
10. Power P = Fv
10. Power = τω
11. Linear momentum of a
11. Angular momentum of a system is
system is conserved
conserved when no external
when no external force acts
torque acts on the system.
on the system. ( Principle of
(Principle of conservation of
conservation of linear
momentum)
angular momentum )
12. Equation of translation
motion
(i) v = u + at
(ii) S = ut +1/2 at2
(iii) v - u = 2as,where the
symbols have their usual
meaning.
13. Distance travelled in nth
second ,snth = u+a/2(2n-1)
12. Equation of rotation motion
(i) ω2 = ω1 + αt
(ii) θ = ω1t + ½αt2
(iii) ω22 - ω12 = 2α θ,
Where the symbols have
their usual meaning
13 Angle traced in nth second.
θnth = ω1 + α/2(2n –1)