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C HAP TE R

ROTATIONAL MOTION
Rigid body is defined as a system of particles in which distance between each pair of
particles remains constant (with respect to time) that means the shape and size do not
change, during the motion.
Eg : Fan, Pen, Table, stone and so on.
DY
D BO Type of Motion of rigid body
I
RIG

Pure Translational Pure Rotational Combined Translational

Motion Motion and Rotational Motion

L For Rigid Bodies :

IONA
TAT N Moment of inertia of a rigid body about any axis of
RO OTIO
M rotation. I = ò dm r 2

Moment of Inertia Radius of Gyration (K)

The virtue by which a body revolving about an axis K has no meaning without axis of rotation.
opposes the change in rotational motion is known I = MK2 K is a scalar quantity
as moment of inertia.
Radius of gyaration : K = I
axis M
Perpendicular axis Theorems : Iz = Ix + Iy
m (body lies on the x-y plane)
r
z

o y
· The moment of inertia of a particles with respect
to an axis of rotation is equal to the product of
mass of the particle and square of distance from x
rotational axis. I = mr2
r = perpendicular distance from axis of rotation (Valid only for 2-dimensional body)
Parallel axis Theorem : I = ICM + Md2
· Moment of inertia of system of particle

axis discrete I CM 2
I =ICM+Md
axis
body continuous
r1 m1 dm body CM
r2 r d
m2

r3
m3 I = r2 dm

(for all type of bodies)

I = m1r12 + m2r22 + m3r32 + ..... ICM = moment of inertia about the axis
Passing through the centre of mass
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MOMENT OF INERTIA OF SOME REGULAR BODIES

Shape of Position of the axis Figure Moment of Radius of
the body of rotation Inertia (I) gyration (K)
(1) Circular (a) About an axis A
z
Ring perpendicular to the
plane and passes CM d MR2 R
through the centre
Mass = M B
Radius = R
I
Z
y x'
1 R
(b) About the diametric axis MR2
2 2
x y'

(c) About an axis I1

tangential to the rim 2MR2 2R
and perpendicular to CM R
the plane of the ring M

(d) About an axis

3 3
tangential to the rim O MR2 R
2 2
and lying in the plane R
of ring
(2) Circular Disc (a) About an axis
1 R
passing through the MR2
R 2 2
centre and perpendicular
M = Mass to the plane of disc
R = Radius
y x'

(b) About a diametric MR2 R

axis 4 2
x y'
(c) About an axis
5 5
tangential to the rim CM MR2 R
O I CM 4 2
and lying in the plane R
of the disc
(d) About an axis Iz
3 3
tangential to the rim MR2 R
CM 2 2
& perpendicular to the R
plane of disc M
R

(3) Annular disc (a) About an axis

passing through the M 2 R12 + R22
R2 R2 é R1 + R22 ùû
R1 centre and R1 2ë 2
perpendicular to the M
plane of disc
M = Mass
R1= Internal Radius
R2 = Outer Radius
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P TE R

Shape of Position of the axis Figure Moment of Radius of

the body of rotation Inertia (I) gyration (K)

R2
(b) About a diameteric axis R1 M 2 R12 + R22
é R1 + R22 ùû
M 4ë
2

(4) Solid Sphere (a) About its diametric I

2 2
axis which passes MR2 R
5 5
through its centre
of mass M R

M = Mass
R = Radius
I

(b) About a tangent to

M R 7 7
the Sphere MR2 R
5 5

(5) Hollow (a) About diametric axis I

M 2 2
Sphere passing through centre MR2 R
3 3
(Thin spherical of mass
Shell) R

table tennis ball

M = Mass (b) About a tangent to I

M 5 5
R = Radius the surface MR2 R
3 3
Thickness R
negligible

table tennis ball

(6) Hollow (a) About its geometrical MR2 R
Cylinder axis which is parallel to
its length

M = Mass (b) About an axis which

MR2 ML2 R 2 L2
R = Radius is perpendicular to its + +
L = Length length and passes 2 12 2 12
through its centre of
L
mass

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Shape of Position of the axis Figure Moment of Radius of

the body of rotation Inertia (I) gyration (K)
(c) About an axis
MR2 ML2 R2 L2
perpendicular to its + +
2 3 2 3
length and passing
through one end of the
cylinder L

(7) Solid (a) About its geometrical I R

MR2
Cylinder axis, which is along
R M 2 2
M = Mass its length
R = Radius
L = Length

(b) About an axis I

3 3
tangential to the M
MR2 R
R 2 2
cylinderical surface and
parallel to its
geometrical axis

(c) About an axis

ML2 MR2 L2 R2
passing through the + +
12 4 12 4
centre of mass and
perpendicular to its
length
L

(8) Thin Rod (a) About an axis L

ML2
passing through centre
12 12
of mass and
Thickness is perpendicular to its L
negligible length
w.r.t. length
Mass = M (b) About an axis L
ML2
Length = L passing through one
3 3
end and perpendicular
to length of the rod L

(9) Rectangular (a) About an axis passing b

Mb 2
Plate through centre of mass
b 12 2 3
and perpendicular to
side b in its plane a

M = Mass (b) About an axis a

Ma 2
a = Length passing through centre
12 2 3
b = Breadth of mass and b
perpendicular to side a
a in its plane.

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Shape of Position of the axis Figure Moment of Radius of

the body of rotation Inertia (I) gyration (K)

(c) About an axis

passing throught centre
(
M a2 + b 2 ) a2 + b 2
of mass and 12 12
b
perpendicular to plane a

(10) Cube About an axis passes Ma2 a

through centre of mass
6 6
and perpendicular to
face

a
Mass = M
Side a
(11) About an axis
Uniform thin rod Passing through I C = mr 2 r
r
bent into shape of center and
C
an arc of mass m perpendicular to the
plane containing
the arc
(12) About an axis
mr 2
Sector of a Passing through r IC = r/ 2
2
uniform disk center and C

of mass m perpendicular to the

plane containing
the sector.

TORQUE
r
Torque about point : rt = rr ´ F Line of action
of force
Magnitude of torque = Force × perpendicular F
P
distance of line of action q
of force from the axis of rsinq r

rotation. O
t=rFsinq
Direction of torque can be determined by using right hand thumb rule.
ROTATIONAL EQUILIBRIUM
If a rigid body is in rotational equilibrium under the action of several coplanar forces, the Fn
resultant torque of all the forces about any axis perpendicular to the plane containing the Fi
F1
forces must be zero.
In the figure a body is shown under the action of several external
F2
coplanar forces F1, F2, … …Fi, and Fn.
v
åtP = 0 T1
y

x
T2
Here P is a point in the plane of the forces about which we calculate
D C
torque of all the external forces acting on the body. The flexibility available A B
in selection of the point P provides us with advantages that we can select
l /4 l/2
such a point about which torques of several unknown forces will become
200
zero or we can make as many number of equations as desired by selecting 400

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several different points. The first situation yields to a simpler equation to be solved and second situation though does not give
independent equation, which can be used to determine additional unknowns yet may be used to check the solution.
The above condition reveals that a body cannot be in rotational equilibrium under the action of a single force unless
the line of action passes through the mass center of the body.
A case of particular interest arises where only three coplanar forces are involved and the body is in rotational
equilibrium. It can be shown that if a body is in rotational equilibrium under the action of three forces, the lines
of action of the three forces must be either concurrent or parallel. This condition provides us with a graphical
technique to analyze rotational equilibrium.
Equilibrium of Rigid Bodies
A rigid body is said to be in equilibrium, if it is in translational as well as rotational equilibrium both. To analyze
such problems conditions for both the equilibriums must be applied.
Rotation about fixed axis not passing through mass center
In this kind of rotation the axis of rotation remains fixed and does not passes through the mass center.
Rotation of door is a common example of this category. Doors are hinged about their edges; therefore their
axis of rotation does not pass through the mass center. In this kind of rotation motion the mass center
executes circular motion about the axis of rotation.
In the figure, free body diagram and kinetic diagram of a body r
R r
rotating about a fixed axis through point P is shown. It is easy to conceive r P maC
P Fi
that as the body rotates its mass center moves on a circular path of radius r
r I pa
rP / C . The mass center of the body is in translation motion with acceleration C C
r r r
aC on circular path of radius rP / C . To deal with this kind of motion, we have F1 Fn
to make use of both the force and the torque equations.
r r r r r
Translation of mass center SFi = MaC = M a ´ rC / P - M w 2 rC / P
r r
Centroidal Rotation St C = I C a
r r r r
Making use of parallel axis theorem ( )
I P = MrP2 / C + I C and aC / P = a ´ rC / P - w 2 rC / P
we can write the following equation also.
r r
Pure Rotation about P St P = I P a

ANGULAR MOMENTUM (MOMENT OF LINEAR MOMENTUM)

Angular momentum of a particle about a given axis is the l According to Newtons Second Law’s for rotatory
product of its linear momentum and perpendicular distance r
of line of action of linear momentum vector from the axis r dL r
r r r motion t = = Ia .
of rotation, L = r ´ p dt

l Angular Impulse = Change in angular momentum.

l If a large torque acts on a body for a small time
mv

q r
p=

then, angular impulse= tdt

r
q
Conservation of Angular Momentum
b=
rs i Angular momentum of a particle or a system
nq
remains constant if t ext = 0 about that point or axis
of rotation.
Magnitude of Angular momentum
= Linear momentum × Perpendicular distance DL
If t = 0 then = 0 Þ L= constant
of line of action of momentum from the Dt
axis of rotation
L = mv × r sinq Þ Lf = Li or I1w1 = I2w2
Direction of angular momentum can be used by using
right hand thumb rule.
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Examples of Conservation of Angular Momentum

l If a person skating on ice folds his arms then his M.I. l A diver jumping from a height folds his arms and legs
decreases and 'w' increases. (I decrease) in order to increase no. of rotation in air
by increasing 'w'.

Ii
wi If
wf

Ii>If
w i< w f

l If a person moves towards the centre of rotating platform then 'I' decrease and 'w' increase.

ROTATIONAL KINETIC ENERGY

1 2
Kinetic Energy of Rotation KE R = Iw
2
1 2 L2 1
l Other forms K= Iw = = Lw
2 2I 2
1
l If external torque acting on a body is equal to zero (t = 0), L=constant Kµ , K µw
I
q2
l Rotational Work : Wr = tq (If torque is constant) Wr = ò tdq (If torque is variable)
q1

1 2 1 2 1
l The work done by torque = Change in kinetic energy of rotation. W = Iw - Iw = I(w 22 - w12 )
2 2 2 1 2
dW dq DW
l Instantaneous power = =t = tw Average power Pav =
dt dt Dt
COMBINED TRANSLATIONAL AND ROTATIONAL MOTION OF A RIGID BODY

When a body perform translatory motion as well as rotatory Pure rolling

motion then it is known as rolling. vCM =Rw CM
In Pure Rolling R
w CM
vCM
(i) If the velocity of point of contact with respect to the surface
is zero then it is known as pure rolling.
Contact point

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(ii) If a body is performing rolling then the velocity of any point of the body with respect to the surface
r r r r
is given by v = v CM + wCM ´ R

A VCM A wR A VCM+w R=2VCM

VCM VCM wR VCM=wR
2VCM
VCM VCM E VCM VCM=0 VCM VCM
F C E
F F VCM
C C
VCM VCM 2VCM
C
wR
VCM VCM=wR
VCM wR
B B P VCM-wR=0
Only T.M. Only R.M. Pure rolling motion

Only Translatory motion + Only Rotatory Motion = Rolling motion. For pure rolling above body
VA = 2VCM VE = 2 VCM VF = 2 VCM VB = 0

Velocity at a point on rim of sphere Rolling Kinetic Energy under pure rolling

v net = v 2 + R 2w 2 + 2vRw cos q Rolling body

VCM
v
q
q
Rw
surface
Rolling Kinetic Energy
1 1 1 2 1 æ v2 ö
E = mv 2 + Iw2 = mv + mK2 ç 2 ÷
2 2 2 2 èR ø
For pure rolling v = Rw 1 2æ K2 ö
Rolling Kinetic Energy E = mv ç 1 + 2 ÷
2 è R ø
q
v net = 2v cos
2 Etranslation : Erotation : ETotal = 1 :
K2 K2
: 1 +
R2 R2

K2 E trans 1 E trans 1 E rotation K

= K2
2

Body = = R K2
R2 E rotation R2
2
E total 1 + KR2 E total 1 + R2

Ring 1 1 1/2 1/2

Disc 1/2 2 2/3 1/3
Solid sphere 2/5 5/2 5/7 2/7
Spherical shell 2/3 3/2 3/5 2/5
Solid cylinder 1/2 2 2/3 1/3
Hollow cylinder 1 1 1/2 1/2

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General Plane Motion: Rotation about axis in translation motion

r
Rotation of bodies about an axis in translation motion can be dealt with either F1
as superposition of translation of mass center and centroidal rotation or r
Fi
assuming pure rotation about the instantaneous axis of rotation. In the r
I Ca r
figure is shown the free body diagram and kinetic diagram of a body in MaC
general plane motion. C
r r
n r
r F2 Fn
Translation of mass center å Fi = MaC
i =0
n
r r
Centroidal Rotation åt
i -1
C = IC a
This kind of situation can also be dealt with considering it rotation about IAR. It gives sometimes quick solutions,
especially when IAR is known and forces if acting at the IAR are not required to be found.

Rolling Motion on an inclined plane

w m
Rolling body

VCM
height

Inclined plane q

Applying Conservation of energy l Linear accleration on reaching the lowest point a

g sin q
1 2 1 2 =
mgh = mv + Iw 1 + K2 / R2
2 2
l Time taken to reach the lowest point of the plane is

2æ v ö
2
1 2 1 2s(1 + K 2 / R 2 )
mgh = 2 mv + 2 mK ç 2 ÷ t=
èR ø g sin q

K2
1 2æ K ö 2 l Least, will reach first
R2
mgh = 2 mv ç 1 + R2 ÷ ...(1)
è ø
K2
Maximum, will reach last
h = s sinq ...(2) R2

from (1) & (2) K2

equal, will reach together
R2
l When ring, disc, hollows sphere, solid sphere rolls
2gh 2gs sin q on same inclined plane then
VRolling = =
K2
1+ 2 K2 vS > vD > vH > vR aS > aD > aH > aR
R 1+
R2 tS < tD < tH < tR

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For a pure rolling body after one full rotation

A3
A2
A4

A1 A5
2p R

displacement of lowermost point =2pR

distance = 8R

Angular Momentum in general plane motion

Angular momentum of a body in plane motion can also be written similar y
to torque equation or kinetic energy as sum of angular momentum about the r
vC
w
axis due to translation of mass center and angular momentum of centroidal
C
rotation about centroidal axis parallel to the original axis.
Consider a rigid body of mass M in plane motion. At the instant shown its mass r
rC
r r O x
center has velocity v and it is rotating with angular velocity w about an axis
r
perpendicular to the plane of the figure. It angular momentum Lo about an axis passing though the origin
and parallel to the original is expressed by the following equation.
r r r r
Lo = rC ´ ( Mv C ) + I C w
The first term of the above equation represent angular momentum due to translation of the mass center and
the second term represents angular momentum in centroidal rotation.
Angular momentum in rotation about fixed axis y
Consider a body of mass M rotating with angular velocity w about a
fixed axis perpendicular to plane of the figure passing through point P r r
vC
rC / P
P. Making use of the parallel axis theorem I P = MrC2 / P + I C and w

r r r r C
equation vC = w ´ rC / P we can express the angular momentum LP of
the body about the fixed rotational axis.
r r O x
LP = I P w
The above equation reveals that the angular momentum of a rigid body in plane motion can also be expressed
in a single term due to rotation about the instantaneous axis of rotation.
Angular momentum in pure centroidal rotation
In pure centroidal rotation, mass center remains at rest, therefore
w
angular momentum due to translation of the mass center vanishes.
r r C
LC = I C w
Eccentric Impact
In eccentric impact the line of impact which is the common normal drawn at the point of impact does not
passes through mass center of at least one of the colliding bodies. It involves change in state of rotation
motion of either or both the bodies.
Consider impact of two A and B such that the mass center CB of B does
CB Line of
not lie on the line of impact as shown in figure. If we assume bodies to Impact
be frictionless their mutual forces must act along the line of impact.
A
The reaction force of A on B does not passes through the mass center CA
B

of B as a result state of rotation motion of B changes during the impact.

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Problems of Eccentric Impact

Problems of eccentric impact can be divided into two categories. In one category both the bodies under going
eccentric impact are free to move. No external force act on either of them. There mutual forces are responsible
for change in their momentum and angular momentum. In another category either or both of the bodies are
hinged.
Eccentric Impact of bodies free to move
Since no external force acts on the two body system, we can use principle of conservation of linear momentum,
principle of conservation of angular momentum about any point and concept of coefficient of restitution.
The coefficient of restitution is defined for components of velocities of points of contacts of the bodies along
the line of impact.
While applying principle of conservation of angular momentum care must be taken in selecting the point about
which we write the equation. The point about which we write angular momentum must be at rest relative to the
selected inertial reference frame and as far as possible its location should be selected on line of velocity of the
mass center in order to make zero the first term involving moment of momentum of mass center.
Eccentric Impact of hinged bodies
When either or both of the bodies are hinged the reaction of the hinge during the impact act as
external force on the two body system, therefore linear momentum no longer remain conserved and
we cannot apply principle of conservation of linear momentum. When both the bodies are hinged we
cannot also apply conservation of angular momentum, and we have to use impulse momentum principle
on both the bodies separately in addition to making use of coefficient of restitution. But when one of
the bodies is hinged and other one is free to move, we can apply conservation of angular momentum
about the hinge.
Ex. A uniform rod of mass m and length l is suspended from a fixed support and can O

rotate freely in the vertical plane. A small ball of mass m moving horizontally with velocity
vo strikes elastically the lower end of the rod as shown in the figure. Find the angular
velocity of the rod and velocity of the ball immediately after the impact. v o

Sol. The rod is hinged and the ball is free to move. External forces acting on
the rod ball system are their weights and reaction from the hinge. Weight of O w' O

the ball as well as the rod are finite and contribute negligible impulse during
the impact, but impulse of reaction of the hinge during impact is considerable
and cannot be neglected. Obviously linear momentum of the system is not
conserved. The angular impulse of the reaction of hinge about the hinge is vo v'
zero. Therefore angular momentum of the system about the hinge is
Before the impact Immediately after
conserved. Let velocity of the ball after the impact becomes v'B and angular the impact
velocity of the rod becomes w'.
We denote angular momentum of the ball and the rod about the hinge before the impact by L and L and
B1 R1
after the impact by L and L .
B2 R2
Applying conservation of angular momentum about the hinge, we have
r r r r
LB 1 + LR1 = LB 2 + LR 2 ® mv o l + 0 = mv B¢ l + I o w ¢
Substituting 1
3 M l2 for Io, we have
3mv B¢ + M lw ¢ = 3mv o (1)
The velocity of the lower end of the rod before the impact was zero and immediately after the impact it
becomes lw' towards right. Employing these facts we can express the coefficient of restitution according to eq.
¢ - v Pn
v Qn ¢
e= lw ¢ - v B¢ = ev o
v pn - v Qn ® (2)

From eq. (1) and (2), we have

Velocity of the ball immediately after the impact v B¢ =

(3m - eM ) v o Ans.
3m + M
3 (1 + e ) mv o
Angular velocity of the rod immediately after the impact w¢ = Ans.
(3m + M ) l
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