Download Unit 5: Rotational Motion

Rotational Motion
Unit 5: Rotational Motion
▼ Rotational Motion
Rotational motion of a rigid body is that in which all particles of a body move along
circles in parallel planes. The centre of these circles lie on a fixed line perpendicular to
the parallel planes called axis of rotation.
Rigid body : A rigid body is a body which can rotate with all the parts locked together and
without any change in its shape.
System : A collection of any number of particles interacting with each other.
Internal forces : All the forces exerted by various particles of the system on each other are
called internal forces. Internal forces between two particles are mutual (equal and
opposite).
External forces : To move or stop an object of finite size, we have to apply a force on the
object from outside. This force exerted on a given system is called an external force.
▼ Centre of Mass
Centre of mass of a system (body) is a point where whole mass of the system is
supposed to be concentrated.
Centre of mass of two particle system
Consider a system of two particles A and B of masses mA & mB having position vector
r
r
rA & rB respectively.
r
Let FAB = Internal force on particle A by particle B
r
FBA = Internal force on particle B by particle A
r
FA = External force on particle A
r
FB = External force on particle B
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According to Newton’s second law of motion,
r
r
r
r
d
d
FA =
(mA v A ) & FB =
(mB vB )
dt
dt
1
Total force acting on the system is given by
r
F = Total force acting on particle A + Total force acting on particle B
r
r
r
r
= (FA + FAB ) + (FB + FBA )
r
r
r
r
= (FA + FB ) + (FAB + FBA )
r
r
r
r
r
= (FA + FB )
[Q FAB + FBA = Sum of internal forces = 0]
Rotational Motion
r
r
r
r
r
r
r
d
d
d
\ F = (FA + FB ) =
(mA v A ) +
(mB v B ) =
(mA v A + mB v B )
dt
dt
dt
d é
d r
d r ù
êmA
=
(rA ) + mB
(rB )ú
ê
ú
dt ë dt
dt
û
r
r ù
d éd
ê (mA rA + mB rB )ú
=
ê
ú
dt ëdt
û
r
d2 é r
m r + mB rB ù
2 ë A A
û
dt
Multiply and divide R.H.S by (m A + mB ), we get
r
r
r
d2 éêmA rA + mB rB ù
ú
F = (mA + mB ) 2
.....(1)
dt êë mA + mB ú
û
=
r
Now suppose if point of mass (mA + mB ) having position vector R C.M is acted upon by
r
a force F , then its equation of motion is given by
r
r
d2 r
F = (mA + mB )a = (mA + mB ) 2 éêR C.M ù
.....(2)
ú
û
dt ë
From (1) and (2), we get
r
r
r
mA rA + mB rB
RC.M =
mA + mB
This is the position vector of centre of mass.
Note: The centre of mass of a system of two particles always lies in between the two
particles and on the line joining them.
r
r
r
r + rB
If mA = mB , then R C.M = A
.
2
Centre of mass of n particle system
If a system consists of n particles of masses m1,m2 ,m3 ......mn
r r r
r
whose positions vectors are r1, r2 , r3 ........rn respectively, then
position vector of centre of mass is given by
r
r
r
r
r m1r1 + m2 r2 + m3 r3 + ............ mn rn
r=
m1 + m2 + m3 + ............ mn
y
m1
C.M.
m2
r1
r
r2
m3
r3
x
Important points about centre of mass
 The position of centre of mass is independent of the co-ordinate system chosen.
 The position of centre of mass depends upon the shape of the body and distribution
of mass.
Example : The centre of mass of a circular disc is within the material of the body while
that of a circular ring is outside the material of the body.
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 The centre of mass changes its position only under the translatory motion. There is
no effect of rotatory motion on centre of mass of the body.
2
 In symmetrical bodies in which the distribution of mass is homogenous, the centre
of mass coincides with the geometrical centre or centre of symmetry of the body.
Rotational Motion
r
 If r is a position vector of centre of mass of a system, then velocity of centre of mass
r
r
r
r
ö
r
m1r1 + m2 r2 + m3 r3 + ......÷
dr
dæ
ç
÷
is given by vcm =
=
çç
÷
dt dt èç m1 + m2 + m3 + ...... ÷
ø
r
 If r is a position vector of centre of mass of a system, then acceleration of centre of
r
r
r
r
r
ö
r
m1r1 + m2 r2 + m3 r + .......÷
dvcm d2 r
d2 æ
ç
÷
mass is given by a cm =
=
=
÷
2
2 ç
ç
dt
dt
dt çè m1 + m2 + m3 + ....... ÷
ø
 For an isolated system, external force on the body is zero.
r
r
d r
F = M (vcm ) = 0 i.e v cm = constant .
dt
i.e., centre of mass of an isolated system moves with uniform velocity along a
straight-line path.
▼ Centre of Mass of Rigid Body
Rigid body: A body is said to be rigid body if it does not undergo any change in its shape
and size when very large external force is applied on it.
The centre of mass of a rigid body is a fixed point where whole mass of the body is
supposed to be concentrated. The position of centre of mass depends on two factors:
(1) Geometrical shape of the body
(2) Distribution of mass in the body
Body
Position of centre of mass
(a)
Uniform hollow sphere
Centre of sphere
(b)
Uniform solid sphere
Centre of sphere
(c)
Uniform circular ring
Centre of ring
(d)
Uniform circular disc
Centre of disc
(e)
Uniform rod
Centre of rod
(f)
A plane lamina (Square, Rectangle, Point of intersection of diagonals
Parallelogram)
(g)
Triangular plane lamina
Point of intersection of medians
(h)
Rectangular or cubical block
Points of intersection of diagonals
(i)
Hollow cylinder
Middle point of the axis of cylinder
(j)
Solid cylinder
Middle point of the axis of cylinder
(k)
Cone or pyramid
On the axis of the cone at point
distance 3h/4 from the vertex where h
is the height of cone
Page
S. No.
3
 Position of centre of mass of some rigid bodies:
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Rotational Motion
▼ Law of conservation of linear momentum of system of particles
Consider a system of n particles of masses m1,m2 ,m3 .....,mn . Suppose the forces
F1,F2 ,F3 ,.....,Fn exerted on them produce acceleration a1,a2 ,a3 ,.....,a n respectively.
In the absence of any external force, F total  0
 F1  F 2  F 3  .....  F n  0
 m1 a1  m2 a 2  m3 a 3  .....  mn a n  0
 m1
dv1
dv2
dv3
dv n
 m2
 m3
 .....  mn
0
dt
dt
dt
dt


d
m1 v1  m2 v 2  m3 v 3  .....  mn v n  0
dt
 m1 v1  m2 v2  m3 v 3  .....  mn v n  cons tan t

 p1  p2  p3  .....  pn  cons tan t
Hence, if no external force acts on a system, the total linear momentum of the system
is conserved. This is the law of conservation of linear momentum.
Motion of the CM of fire crackers exploding in air: Initially, a fire cracker moves along a parabolic
path. It explodes in flight. Each fragment will follow its own parabolic path. The
explosion is caused by internal forces only. They do not influence the motion of CM.
The total external force (force of gravity) is same before and after the explosion. So the
centre of mass of all the fragments will continue to move along the same parabolic
path of the cracker as before explosion.
Earth-moon system: The moon moves around the earth in a circular orbit and the earth
moves around the sun in an elliptical orbit. It will be more correct to say that the
centre of mass of earth-moon system moves around the sun in an elliptical orbit, not
the earth and the moon themselves. Here the mutual forces of gravitation between the
earth and moon are internal forces while the sun’s attraction of both earth and moon
are the external forces acting on the centre of mass of the earth moon system.
▼ Moment of Inertia
Moment of inertia is the property of a body due to which it opposes any change in its
state of rest or of uniform rotation.
Moment of inertia of a particle:
The moment of inertia of a particle about a given axis of rotation is the product of mass of
particle (m) and the square of the perpendicular distance(r) of particle from the axis of
rotation.
\ I = mr2
Moment of inertia of a body made up of
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r
Rotational Motion
of particles:
The moment of inertia of a body about a given axis of rotation is the sum of the product of
masses of the particles and the square of their perpendicular distances from the axis of
rotation.
\ I = m1r12 + m2r22 + m3r32 + ......
Important points about moment of inertia (M.O.I)
 Dimensional formula of M.O.I = [ML2 T0 ]
 S.I. unit of M.O.I : kgm2
 Moment of inertia depends on following factors:
(1) mass of the body
(2) distribution of mass about the axis of rotation
(3) position and orientation of axis of rotation.
 It is not a vector as direction (clockwise or anti-clockwise) is not to be specified and
also not a scalar as it has different values in different directions. Actually it is a
tensor quantity.
Physical significance of M.O.I
M.O.I of a body about an axis of rotation always opposes any change in its state of rest
or of uniform rotational motion. The greater the M.O.I of a body, the greater is the
torque required to change its state of rotation. The M.O.I of a body plays the same role
in rotational motion as mass plays in linear motion.
▼ Radius of Gyration
Radius of gyration of a body about a given axis is the perpendicular distance of a point
from the axis of rotation where whole mass of the body is supposed to be concentrated
so that moment of inertia remains the same. It is denoted by K.
If mass of the body is M, then moment of inertia of the body
about the given axis is given by
2
I = MK
…..(1)
m
We know, I = m1r12 + m2r22 + m3r32 + ....... + mnrn2
m
m
r2
r3
r5
r1
m
r4
m
If m1 = m2 = m3 = ....... = m , then
I = m(r12 + r22 + r32 + ..........rn2 )
.....(2)
From (1) and (2),
Mk 2 = m(r12 + r22 + r32 + ............ + rn2 )
nmk 2 = m(r12 + r22 + r32 + .......... + rn2 )
[Q M = nm]
M
Hence 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.
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r12 + r22 + r32 + ........... + rn2
n
Page
\ K=
k
Rotational Motion
Important points about Radius of gyration
 Radius of gyration depends on following factors:
(1) shape and size of the body
(2) position and orientation of axis of rotation
(3) distribution of mass of the body w.r.t. the axis of rotation.
 Radius of gyration does not depend on the mass of body.
 Dimensional formula of K = [M0L1T0 ] .
 S.I. unit: metre (m)
 Physical significance of radius of gyration:
Through this concept, a real body (particularly irregular) is replaced by a point mass
for dealing its rotational motion.
Example : In case of a disc rotating about an axis through its centre of mass and
perpendicular to its plane
k=
I
=
M
(1 2)MR2
R
=
M
2
So instead of disc we can assume a point mass M at a distance (R / 2) from the
axis of rotation for dealing the rotational motion of the disc.
▼ Theorem of parallel axes
According to this theorem, moment of inertia of a body (I) about a given axis is equal to
the sum of moment of inertia of the body about an axis parallel to given axis and
passing through centre of mass of the body Ig and Ma 2 where M is the mass of the
body and a is the perpendicular distance between the two axes.
I = Ig + Ma 2
IG
I
a
G
Example: Moment of inertia of a disc about an axis through its centre and
1
perpendicular to the plane is MR 2 , so moment of inertia about an axis through its
2
tangent and perpendicular to the plane will be
I = Ig + Ma 2
IG
I
1
MR 2 + MR 2
2
3
\ I = MR 2
2
R
G
Page
6
I=
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Rotational Motion
▼ Theorem of perpendicular axes
According to this theorem, the sum of moment of inertia of a plane lamina about two
mutually perpendicular axes lying in its plane is equal to its moment of inertia about
an axis perpendicular to the plane of lamina and passing through the point of
intersection of first two axes.
Z
Iz = Ix + Iy
X
Y
Example : Moment of inertia of a disc about an axis through its centre of mass and
1
perpendicular to its plane is MR 2 , so if the disc is in x–y plane, then
2
Iz = Ix + Iy
Z
1
Þ MR2 = 2ID [As ring is symmetrical body so Ix = Iy = ID ]
2
1
\ ID = MR2
4
ID
ID
X
O
Y
▼ M.O.I of some standard bodies
Body
Axis of rotation
Moment of Inertia
Ring
About an axis passing
through C.G. and
perpendicular to its plane
MR2
About its diameter
Ring
About a tangential axis
perpendicular to its own
plane
3
MR 2
2
MR2
Page
Ring
About a tangential axis in
its own plane
1
MR 2
2
7
Ring
Figure
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Rotational Motion
About an axis passing
through C.G. and
perpendicular to its plane
1
MR 2
2
About its Diameter
1
MR 2
4
Disc
About a tangential axis in
its own plane
5
MR 2
4
Disc
About a tangential axis
perpendicular to its own
plane
3
MR 2
2
Disc
Disc
R
Solid cylinder
About its own axis
Solid Sphere
About its diametric axis
Passing through the centre
of mass and perpendicular
to the plane
2
MR 2
5
1
ML2
12
M
[
12
2
L
 b2 ]
Page
Rectangular
lamina of
length l and
breadth b
About on axis passing
through its centre of mass
and perpendicular to the
rod.
L
8
Long thin rod
1
MR 2
2
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Rotational Motion
▼ Torque or moment of force
The turning effect of the force about the axis of rotation is called moment of force or
torque. It is measured as the product of the magnitude of the force and the
perpendicular distance between the line of action of the force and the axis of rotation.
Rotation
Rotation
P
F
r
O
F
r
O
(B)
(A)
r
If the particle is rotating in XY-plane about the origin under the effect of force F and
r
the position vector of the particle is r , then
Torque,   r  F
In scalar form,   r F sin 
r
r
[where  is the angle between r and F ]
Important points about torque
 SI unit of torque : newton-metre (Nm) and CGS unit : dyne-cm
 Dimensional formula of torque : [ML2 T2 ] .
 Torque is an axial vector. Its direction is always perpendicular to the plane
r
r
containing vector r and F in accordance with right hand screw rule.
 Maximum and minimum value of torque:
(a) Maximum value of torque, max  rF when   900
i.e. when r  F
(b) Minimum value of torque, min  0 when   00 or 1800 i.e. when r F
 For a given force(F) and angle( q ), magnitude of torque depends on r. The more is
the value of r, the more will be the torque and easier to rotate the body.
Example : (i) Handles are provided near the free edge of the door.
(ii) The handle of screw driver is taken thick.
(iii) The handle of hand-pump is kept long.
(iv) The arm of wrench used for opening the tap is kept long.

A body is said to be in rotational equilibrium if resultant torque acting on it is zero
i.e.   0 .
In case of beam balance, the system will be in rotational equilibrium if 1  2  0
 F2
1
 F1
 F2
1
2
0
R
2
However, if 1  2 , L.H.S. will move downwards
and if if 1  2 , R.H.S. will move downward.
l1
l2
F1
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
Rotational Motion
 Physical significance of torque: Torque is the cause of rotatory motion and in rotational
motion, it plays same role as force plays in translatory motion.
F
 Couple: A couple is a combination of two equal
but opposite forces not acting along the same
line of action. The effect of couple is given by its
r
F
torque,   r  F .
▼ Expression for torque
 Expression for torque in Cartesian coordinates:
Consider a particle of mass m rotating in a plane XY about the origin. Let P be the
position of the particle at any instant where OP  r and XOP  . Let the rotation
occur under the action of force F applied at P along PA .
Y
Fy
In small time dt, let the particle at P reach Q where


F
OQ  r  dr and POQ  d.
dy
r+dr
From OPQ, OP  PQ  OQ

A
Q
dx
Fx
P

dθ
 PQ  OQ  OP  r  dr  r  dr
O
r
θ
X
Small amount of work done in rotating the particle from P to Q is dW  F.dr .
If Fx ,Fy are rectangular components of force F and dx, dy are rectangular components
of displacement dr , then F  ˆiFx  ˆjFy & dr  ˆidx  ˆjdy.



 dW  F.dr  ˆiFx  ˆjFy . ˆidx  ˆjdy  Fx dx  Fy dy
.....(1)
Let the coordinates of the point P be (x,y).
 x  r cos  & y  r sin 
dx
dy
 r sin    y &
 r cos   x
d
d
 dx   yd & dy  xd
Also,
Equation (1) becomes
dW  Fx ( yd)  Fy (xd)  (x Fy  yFx )d   d
where   x Fy  yFx
Therefore,   x Fy  yFx is an expression for torque in cartesian co-ordinates.
 Expression for torque in Polar coordinates:
Suppose the line of action of force F makes an angle  with X-axis.
 Fx  F cos  & Fy  F sin 
the
point
P
where
OP  r
and
   x Fy  yFx  (rcos )Fsin   (r sin )Fcos   rF(sin  cos   cos  sin )  rF sin(  )
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If x, y are
the
coordinates
XOP  , then x  rcos  & y  r sin .

F
Rotational Motion
Y
F sin 
Let  be the angle which the line of action of F makes
with the position vector OP  r. From diagram, it is
clear that      or     
Therefore,   rF sin 
is an expression for torque in
O
polar coordinates.

α
θ
d

P

r
N
▼ Angular Momentum
F
F cos 
X
The turning momentum of particle about the axis of rotation is called the angular
momentum of the particle.
If p is the linear momentum of particle and r is its position vector from the point of
rotation, then angular momentum ( L ) is given by
L  rp
In scalar form, L  r psin 
r
r
[where  is the angle between r and p ]
Important points about angular momentum
 SI unit of angular momentum : kgm2s–1 or Js.
 Dimensional formula : [ML2T-1 ]
 Angular momentum is an axial vector. Its direction is always perpendicular to the
plane of rotation and along the axis of rotation.
 Maximum and minimum value of angular momentum:
(a) Maximum value of angular momentum, Lmax  rp when   900 i.e. when r  p
(b) Minimum value of angular momentum Lmin  0 when   00 or1800 i.e. when r p
ˆ and p  p ˆi  p ˆj  p k
ˆ
 In cartesian co-ordinates if r  xiˆ  yjˆ  zk
x
y
z
ˆi
Then L  r  p  x
px
ˆj
y
py
ˆ
k
z =(yp z  zp y )iˆ  (xp z  zp x ) ˆj  (xp y  yp x )kˆ
pz
 In case of circular motion, L  r  p  m(r  v)  mvr sin 
 L  mvr  mr2
[As r  v and v  r ]
Also, L  I
[As mr2 = I]
In vector form, L  I
 We know, L  I 
dL
d
I
 I  
dt
dt
Page
11
It means the rate of change of angular momentum is equal to the net torque acting
on the particle. [Rotational analogue of Newton's second law]
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Rotational Motion
▼ Expression for angular momentum
 Expression for angular momentum in Cartesian coordinates:
Consider a particle of mass m rotating in a plane XY about the origin. Let P be the
position of the particle at any instant where OP  r and XOP  .
If px  mv x & py  mv y are the x and y components of linear momentum of the body,
then according to Newton’s 2nd law of motion, we have
dp y
dv y
dp x
dv
d
d
Fx 

(mv x )  m x & Fy 

(mv y )  m
dt
dt
dt
dt
dt
dt
Put these values in   x Fy  yFx
dv y
dv x
dt
dt
 dv y
dv 
   m x
y x 
dt 
 dt
  xm
 ym
.....(1)
dv y
dv
d
dx
dy
(x v y  yv x )  x
 vy
 y x  vx
dt
dt
dt
dt
dt
dv y
dv
x
 vy vx  y x  vx vy
dt
dt
dv y
dv
x
y x
dt
dt
Put this value in (1),
Now,
 dv y
dv
  m x
y x
dt
 dt
dL
But  
dt
 L  xp y  yp x

d
d
d
  m (x v y  yv x )  (xm v y  ymv x )  (xp y  yp x )
dt
dt
dt

Therefore, L  x py  ypx
is an expression for angular momentum in cartesian
coordinates.
 Expression for angular momentum in Polar coordinates:
Let P(x,y) be position of a particle of mass m and linear momentum p. Let OP  r and
XOP  , then x  rcos  & y  r sin .

F
Let  be the angle which the line of action of p makes with the position vector OP  r.
psin 
P

r
O

α
θ
d

N
p
pcos 
X
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 px  pcos  & py  psin 
Rotational Motion
Put these values in L  xpy  ypx
 L  xpy  ypx  (rcos )psin   (r sin )pcos   rp(sin  cos   cos  sin )  rF sin(  )
From diagram, it is clear that      or     
Therefore, L  rpsin  is an expression for angular momentum in polar coordinates.
▼ Law of conservation of angular momentum
Statement: Angular momentum of a system (may be particle or body) remains constant if
resultant torque acting on it zero.
According to Newton’s second law for rotational motion,  
dL
dt
So if the net external torque on a particle (or system) is zero, then
dL
0
dt
i.e. L  L1  L 2  L 3  ....... = constant.
As L  I so if   0 then I  constant  I 
1

Since angular momentum ( I) remains constant so when I decreases, angular velocity 
increases and vice-versa.
Examples of law of conservation of angular momentum :
(1) The angular velocity of revolution of a planet around the sun in an elliptical orbit
increases when the planet come closer to the sun and vice-versa because when planet
comes closer to the sun, it's moment of inertia I decreases, therefore  increases.
(2) A circus acrobat performs feats involving spin by bringing his arms and legs closer
to his body or vice-versa. On bringing the arms and legs closer to body, his moment of
inertia I decreases. Hence  increases.
(3) A person-carrying heavy weight in his
hands and standing on a rotating platform
can change the speed of platform. When the
person suddenly folds his arms. Its moment
of inertia decreases and in accordance the
angular speed increases.
(4) Effect of change in radius of earth on its time period:
L  I  constant

T  R2
2
2
MR2 
 constant
5
T
[if M remains constant]
If R becomes half then time period will become one-fourth i.e.
24
 6hrs.
4
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L 
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Angular momentum of the earth,
Rotational Motion
▼ Equilibrium of rigid bodies
A rigid body is said to be in equilibrium if the net external force acting on it does not
change the translational and rotational states of the body.
Forces are equal
and act along the
same line.
Forces are equal
and does not act
along the same
line.
Forces are unequal
and act along the
same line.
Forces are unequal
and does not act
along the same
line.
 F  0 &   0
Body will remain
stationary if initially it
was at rest.
 F  0 &   0
Rotation i.e. spinning.
 F  0 &   0
Translation i.e. slipping or
skidding.
 F  0 &   0
Rotation and translation
both i.e. rolling.
F
F
F
F
F2
F1
F1
F2
▼ Equations of linear motion & rotational motion
Linear Motion
Rotational Motion
(1)
If linear acceleration is 0, u = constant If
angular
acceleration
  cons tant and   t
and s = u t.
(2)
If linear acceleration a = constant, If angular acceleration  = constant
then
then
(i)
v  u  at
(ii) s  ut 
(i)
1 2
at
2
(iv) snth  u 
0,
2  1  t
(ii)   1t 
(iii) v2  u2  2as
is
1 2
t
2
(iii) 22  12  2
1
a(2n  1)
2
(iv) nth  1  (2n  1)

2
If acceleration is not constant, the If acceleration is not constant, the
above equation will not be applicable. above equation will not be applicable.
In this case
In this case
(ii) a 
dx
dt
(i)
dv d2 x

dt dt2
(ii)  

d
dt
d d2

dt dt2
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v
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(i)
Rotational Motion
▼ Analogy between translatory motion & rotatory motion
Translatory motion
Rotatory motion
Mass
m
Moment of Inertia
I
Displacement
s
Angular displacement

Velocity
v
Angular velocity

Acceleration
a
Angular acceleration

Linear momentum
p  mv
Angular Momentum
L  I
Force
F  ma 
Torque
  I 
Work done
W  F.s
Work done
W  . 
Linear K.E.
1
mv2
2
Rotational K.E.
1 2
I
2
Power
P  F.v
Power
W  . 
d
(mv)
dt
d
(I)
dt
Conceptual Questions
CQ.1.
What is centre of gravity?
Ans.
Centre of gravity of a body is a point where whole weight of the body is
supposed to be concentrated.
CQ.2.
Ans.
What is the significance of the centre of mass of the system of particles?
Centre of mass of the system of particles helps in describing the behavior of a
macroscopic body in terms of the laws developed for microscopic particles.
CQ.3.
Ans.
Can geometrical centre and centre of mass of a body coincide
? If yes, give examples.
Yes, the geometrical centre and centre of mass of spheres, circular rings,
circular discs, cubes, cylinders etc. coincide.
CQ.4. Explain why is it difficult to open a door by pushing or pulling it near the hinge?
Ans.
Since the door rotates about the hinge, so the distance of the point of
application (r) of force is small when the door is pushed near the hinge. Due to
this, small torque will be produced which will not be enough to rotate the door.
CQ.5. When there is no external torque acting on a rotating body, which of the following quantities
change: (i) Angular acceleration (ii) Angular momentum, (iii) Angular speed.
Ans.
(i) The relation between torque and angular acceleration is   I
Hence, angular acceleration will not change, when external torque acting the
body is zero.
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0
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Since   0
can
Rotational Motion
(ii) The relation between torque and angular momentum is given by  
dL
dt
dL
 0  L  Cons tant
dt
Hence, angular momentum of the body does not change, if external torque
acting on the body is zero.
(iii) Since L  I  cons tant if   0 , then L is constant.
When   0

Therefore, angular speed will change, if M.I. of the body changes.
CQ.6.
Ans.
In a fly wheel, most of the mass is concentrated at the rim. Explain why?
Most of the mass of a flywheel is concentrated at the rim so as to have large
moment of inertia for the same mass. This is due to the fact that the moment of
inertia varies as the square of the distance from the axis of rotation.
CQ.7.
Ans.
If all the ice on the polar caps of the earth melts, how will it effect the duration of the day?
When ice melts, the water will go much farther away from the axis of rotation. It
will increase the M.I. of earth. According to the law of conservation of angular
momentum (I  cons tant) , angular velocity of the earth will decrease.
2
2
.
or T 
T

As  decreases, so T will increase. Thus, duration of the day will become longer.
Now,  
CQ.8. A cat is able to land on its feet after fall. Explain why?
Ans.
According to law of conservation of angular momentum, I  cons tant or  
1
I
When the cat lands after falling, it stretches its tail. Due to this M.I. of the cat
increases and hence its angular speed decreases. Thus, the cat lands safely
after the fall.
CQ.9. Is the angular velocity of rotation of hour hand of a watch greater or smaller than the angular
of earth’s rotation about its axis?
2
Ans.
We know,  
T
2 2
 For hour hand, 1 

[ T1  12hrs]
T1 12
For earth, 2 

2 2

T2 24
1 T2 24


2
2 T1 12
velocity
[ T2  24 hrs]
 1  22
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CQ.10. If a body is rotating, is it necessarily being acted upon by an external torque?
Ans.
No. Torque is required only for producing angular acceleration.
16
So angular velocity of rotation of hour hand is double than that of earth’s
rotation.
Rotational Motion
CQ.11. Why two spheres of the same mass and same radius, one solid and the other hollow have
unequal
M.I. ?
Ans.
M.I. of a body depends upon the distribution of mass of the body about the axis
of rotation. Mass of solid sphere is distributed uniformly about the axis of
rotation. On the other hand, mass of hollow sphere is distributed away from the
axis of rotation. Hence M.I. of hollow sphere is larger than the M.I. of solid
sphere.
CQ.12. Does M.I. of a rigid body change with the speed of rotation? Explain.
Ans.
Moment of inertia of a rigid body depends only on the distribution of mass of
the body about the axis of rotation and is independent of the speed of rotation.
Hence M.I. of a rigid body does not change with the speed of rotation.
Numerical Problems
Problem 1.
Two particles of masses 100g and 400g have position vectors
ˆ and 6 ˆi  3 ˆj  2k
ˆ cm respectively at a given time. Find the
2 ˆi  4 ˆj  12k
position vector at the centre of mass.
Problem 2.
Locate the centre of mass of a system of three particles of masses 1kg,
2kg and 3kg placed at the corners of an equilateral triangle of 1 m side.
Problem 3.
Four identical spheres each of radius a are placed on a horizontal table
touching one another so that their centres lies at the corners of a square
of side 2a. Find the position of their centre of mass.
Problem 4.
ˆ
Two bodies of masses 5 kg and 1 kg are moving with velocities ˆi  5 ˆj  2k
ˆ ms 1 respectively. Find the velocity of the centre of
and 5 ˆi  20 ˆj  2k
mass.
Problem 5.
Calculate the moment of inertia of a system of masses about AB axis
passing through the midpoint of the base of the triangle as shown in the
figure.
C
m
a
x
m
m
A
B
Four spheres each of diameter 2r and mass M are placed with their
Page
Problem 6.
a
17
a
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Rotational Motion
centres on the four corners of a square of side R. Calculate the M.I of the
system about one side of the square taken as the axis.
r
R
O
Problem 7.
The moment of inertia of circular ring about the tangent to the ring
which is parallel to the diameter of the ring is 15 g cm2 . If the mass of the
ring is 2g, what is its radius?
Problem 8.
Calculate the K.E. of the earth due to its rotation about its own axis.
Mass of the earth is 6 1024 kg and radius of the earth is 6400 km.
Problem 9.
Energy of 484J is spent in increasing the speed of a flywheel from 60
rpm to 360 rpm. Find the M.I. of the flywheel.
Problem 10.
A constant torque of 500Nm turns a wheel of M.I. 100 kg m2 about an
axis through its centre. Find the gain in angular velocity in 2 seconds.
Problem 11.
ˆ N acts at a point (3iˆ  2jˆ  4k)
ˆ m from the origin.
A force of (2iˆ  4ˆj  2k)
What is the magnitude of torque?
Problem 12.
The resultant of the system in the figure is a force of 8N parallel to the
given force through R . What is the value of PR ?
R
Q
P
Problem 13.
A flywheel of mass 1000 kg and radius 1m is rotating at the rate of 420
rpm. Find the constant retarding torque required to stop the wheel in 14
rotations. Assuming mass to be concentrated at the rim.
Problem 14.
A disc of mass 5kg and radius 40cm rotates about an axis passing
through its centre and perpendicular to its plane. It makes 10
revolutions in one second. Calculate its (i) rotational kinetic energy and
(ii) angular momentum.
Problem 15.
What will be the duration of the day, if the earth suddenly shrinks to
1/64 of its original volume, mass remains unchanged?
Two discs of same mass and thickness are made of materials having
different densities. Which one of them will have larger M.I.?
Problem 16.
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3N
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5N
Rotational Motion
Problem 17.
If the angular momentum of a rotating body is increased by 200%, then
how much % of its kinetic energy of rotation will increase?
Problem 18.
The angular velocity of a body is

ˆ
  2iˆ  3ˆj  4k
and a torque

ˆ acts on it. What is the rotational power?
  ˆi  2ˆj  3k
Problem 19.
The wheel of a car is rotating at the rate of 1200 revolutions per minute.
On pressing the accelerator for 10sec, it starts rotating at 4500
revolutions per minute. What is the angular acceleration of the wheel?
Problem 20.
A wheel is at rest. Its angular velocity increases uniformly and becomes
60 rad/sec after 5 sec. What is the total angular displacement?
Answers
ˆ
1. 4.4 ˆi  3.2jˆ  4k
5.
3
ma 2
4
 7
3
2.  ,

 12 4 
6.
2
M(2r2  5R2 )
5
4.
7. 2.2cm
8. 2.6 1029 J
9. 0.7 kg m2
10. 10 rads1
13. 11 rads1
14. (i) 788.8J (ii) 25.12 Js
11. 24.4Nm 12. PR 
16. M.I. of disc with less density will be larger
19. 1980 degree /sec2
3
RQ
5
15. 1.5 hour
17. 800%
20. 150 rad
Page
19
18. 20W
1
ˆ ms 1
(5 ˆj  8 k)
6
3. (a,a)
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