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ROTATIONAL INERTIA
(MOMENT OF INERTIA)
ROTATIONAL INERTIA (MOMENT OF INERTIA)
ROTATIONAL INERTIA (MOMENT OF INERTIA)
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The rotational inertia of an object is a measure of the resistance of
the object to changes in its rotational motion
For a system of particles of masses mi at distances ri from an axis
passing through a point P the rotational inertia of the system about
the axis is given by:
I  m1r12  m2 r22  ...   mi ri 2
i
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Definition of
moment of inertia
SI unit of moment of inertia is the kgm2
For a solid object the rotational inertia is found by evaluating an integral
as we will see later
In a rigid body the distances ri are constant, and I is independent of
how the body is rotating around a given axis. The rotational inertia of
some common shapes about some of their symmetry axes is given in
Table 9.2 of your textbook

Greater a body’s moment of inertia, the harder it is to start the
body rotating if it’s at rest and the harder it is to stop its rotation
if it’s already rotating
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A.
B.
C.
PARALLEL-AXIS THEOREM. EXAMPLE
Moments of inertia for different rotation axes
One-piece machine part consists of three heavy connectors linked by light
molded struts.
What is the moment of inertia of this body about an axis through point A,  to
the plane of the slide?

What is the moment of inertia of this body about an axis coinciding the rod BC?
If the body rotates about an axis through A  to the plane of the slide with
angular speed 4.0 rad/s, what is its kinetic energy?

A part of a mechanical linkage has a mass of 3.6 kg. We measure its
moment of inertia about an axis 0.15 m from its center of mass to be
Ip=0.132 kg·m2.
What is the moment of inertia Icm about a parallel axis through the
center of mass?
I cm  I p  Md 2 
 0.132 kg  m 2  (3.6kg )(0.15m) 2  0.051 kg  m 2
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PARALLEL-AXIS THEOREM
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INERTIA CALCULATIONS
This theorem gives the rotational inertia of an object of mass M
about an axis, P, that is parallel to and a distance d away from an
axis that passes through the object's center of mass.
For a continuous distribution of mass the sum of the masses times the
square of the distances to the axis of rotation which defines the moment
of inertia become an integral.

If the object is divided into small mass elements dm in such a manner
that all of the points in a particular mass element are the same
perpendicular distance r from the axis of rotation then the moment of
inertia is given by
2
.

I p  I cm  Md
Result show that Icm is less than Ip. This is as it should be: the moment
of inertia for an axis through the center of mass is lower than for any
other parallel axis.
2
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Parallel-Axis Theorem
I   r dm
To evaluate the integral, you need to represent r and dm in terms of the
same integration variable. 1-D object, slender rod: use coordinate x
along the length and relate dm to an increment dx. 3-D object: express
dm in terms of element of volume dV and density .
dm
  const I    r 2 dV
, I   r 2  (V )dV
dV
of integral are determined by the shape and
dV  dx dy dz  Limits
dimensions of the body

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INERTIA CALCULATIONS
INERTIA CALCULATIONS
Uniform thin rod, axis  to length
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Slender uniform rod with mass M and
length L.
I
Compute its moment of inertia about an
axis through O, at an arbitrary distance h
from the end.
Choose as an element of mass a short section of rod with length dx at a
distance x from O. The ratio of the mass dm of this element to the total
mass M is equal to the ratio of its length dx to the total length L:
dm dx

M
L
M
 x dm  L
2
M
dm 
dx
L

 M  x 3 
1
2
2
h x dx   L  3   3 M ( L  3Lh  3h )

 h
Lh
Lh
2
Evaluate this general expression about an axis through the left
end; the right end; through the center. Compare with Table 9.2.
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Hollow or solid cylinder, rotating about
axis of symmetry
Hollow, uniform cylinder with length L, inner radius
R1, outer radius R2. Compute its moment of inertia
about the axis of symmetry.
Choose as a volume element a thin cylindrical shell
of radius r, thickness dr, and length L. All parts of
this element are at very nearly the same distance
from the axis. The volume of this element:
dm  dV   (2rLdr )
2L 4
4
 r dm  R r  (2rLdr )  2L R r dr  4 ( R2  R1 ) 
1
1
1
L 2
V  L( R22  R12 ) I  M ( R22  R12 )

( R2  R12 )( R22  R12 )
2
2
2
R2
2
R2
3
1
M ( R22  R12 )
2
I
1
MR 2
2
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If cylinder is solid, R1=0, R2=R:
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If cylinder has a very thin wall, R1 and R2 are very
nearly equal:
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INERTIA CALCULATIONS
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Hollow or solid cylinder, rotating about axis
of symmetry
I  MR 2
Note: moment of inertia of a cylinder about an axis of
symmetry depends on its mass and radii, but not on
its length!
INERTIA CALCULATIONS
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Uniform sphere, axis through center
Uniform sphere with radius R. the axis is through
its center. Find the moment of inertia about the
axis is through the center of this sphere.
Divide sphere into thin disks of thickness dx,
whose moment of inertia we already know. The
radius r of the disk is
The volume is
The mass is
r  R2  x2
dV  r 2 dx   ( R 2  x 2 ) dx
dm  dV  r 2 dx   ( R 2  x 2 )dx
R

The moment of inertia for the disk of radius r and mass dm is
dI 
1 2
1
r dm 
2
2
2
 x 2 [ ( R 2  x 2 )dx] 
2

2
( R 2  x 2 ) 2 dx
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Density is 200 kg/m3. Find Iz.
INERTIA CALCULATIONS
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Uniform sphere, axis through center
Integrating from x=0 to x=R gives the moment
of inertia of the right hemisphere.
From symmetry, the total I for the entire sphere
is just twice this:
I  ( 2)
I

2
 (R
R
0
8 5
R
15
I
2
MR 2
5
2
 x 2 ) 2 dx
Volume of the sphere
V
The mass M of the sphere

4 R 3
3
M  V 
4 3
R
3
Note: moment of inertia of a solid sphere is less than the
moment of inertia of a solid cylinder of the same mass and
radius! (Reason is that more of the sphere’s mass is located
close to the axis)
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Rod: 3 kg/m
Plate: 12 kg/m2
Find I.
Rods mass density 3 kg/m.
Find IA.
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