Connexions module: m14365
1
Moments of inertia of rigid bodies
(application)
∗
Sunil Kumar Singh
This work is produced by The Connexions Project and licensed under the
†
Creative Commons Attribution License
Abstract
Solving problems is an essential part of the understanding process.
Questions and their answers are presented here in the module text format as if it were an extension of the
theoretical treatment of the topic. The idea is to provide a verbose explanation of the solution, detailing the
application of theory. Solution presented here, therefore, is treated as the part of the understanding process
not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely
absorbed unless they are put to real time situation.
1 Hints on solving problems
•
Moment of inertia is a scalar quantity without any directional property. This has important implication
in solving problems, which involve removal of a part of the rigid body from the whole. We need to
simply use the MI formula for a given shape with changed mass of the remaining body, ensuring that
the pattern of mass distribution about the axis has not changed.
•
Moment of inertia about an oblique axis involves MI integration with certain modication. Here, mass
distribution and distance involve dierent variables. We are required to express integral in terms of
one variable so that the same can be integrated by single integration process.
•
In determining relative MIs, we should look how closely or how distantly mass is distributed.
This
enables us to compare MIs without actual calculation in some cases.
•
So far, we have studied calculation of MI for uniform objects. However, we can also evaluate MI integral,
if variation in mass follows certain pattern and the same can be expressed in terms of mathematical
expression involving variable.
2 Representative problems and their solutions
We discuss problems, which highlight certain aspects of the calculation of moment of inertia of regularly
shaped rigid bodies.
For this reason, questions are categorized in terms of the characterizing features
pertaining to the questions as :
•
•
Using MI formula
Estimation of MI by inspection
∗ Version
1.5: Nov 14, 2009 8:00 am US/Central
† http://creativecommons.org/licenses/by/2.0/
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Connexions module: m14365
•
•
2
Part of a rigid body
Non-uniform mass distribution
2.1 Using MI formula
Example 1
Problem :
The moment of inertia of a straight wire about its perpendicular bisector and moment of
inertia of a circular frame about its perpendicular central axis are
I1
and
I2
respectively. If the composition
of wires are same and lengths of the wires in them are equal, then nd the ratio
Solution :
I1 I2
.
The MI of the straight wire about a perpendicular through the mid point is :
M L2
12
I1 =
The MI of the circular frame about its perpendicular central axis is :
I2 = M R 2
According to question, the lengths of the wires in them are equal. As the composition of the wires same,
the mass of two entities are same. Also :
L = 2πR
⇒ R =
L
2π
Substituting in the expression of MI of circular frame,
M L2
4π 2
I2 =
Now, the required ratio is :
I1
I2
Example 2
Problem :
are same. If
Rs
Solution :
=
M L2 x 4π 2
12 x M L2
I1
I2
=
π2
3
The moments of inertia of a solid sphere and a ring of same mass about their central axes
be the radius of solid sphere, then nd the radius of the ring.
The MI of the solid sphere about its central axis is given as :
Is =
Let the radius of the ring is
Rr
2M Rs 2
5
. The MI of the ring about its central axis is given as :
Ir = M Rr 2
According to question, the moments of inertia of the solid sphere and the ring about their central axes
are same :
M Rr 2 =
⇒ Rr =
http://cnx.org/content/m14365/1.5/
√
2M Rs 2
5
2
5
x Rs
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3
2.2 Estimation of MI by inspection
Example 3
Problem :
Find the axis of the coordinate system about which the moment of inertia of a uniform
rectangle of dimensions "a" and "2a" is least.
Rectangular plate
Figure 1:
Solution :
MI of rectangular plate about an axis.
We can actually determine MI about the given axes by evaluating the integral of MI about
each of the axes and then compare. However, we can as well estimate least MI as mass of the rectangle is
uniformly distributed. MI is least for the axis about which the mass is distributed closest to it.
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Rectangular plate
Figure 2:
MI of rectangular plate about z-axis.
We can see here that the most distant mass about x-axis is at a distance "a/2"; most distant mass about
y-axis is at a distance "a"; most distant mass about z-axis at a distance between "a" and "
√
5a/2".
Here, mass distribution is closest about x-axis. Thus, moment of inertia about x-axis is least.
2.3 Part of a rigid body
Example 4
Problem :
Find the the MI of one quarter of a circular plate of mass "M" and radius "R" about z-axis,
as shown in the gure.
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Quarter plate
Figure 3:
Solution :
MI of quarter plate about z-axis.
We need to look closely the way MI about an axis is dened. MI of the rigid body about
the an axis is given by :
I =
This denition is essentially scalar in nature.
a particle from the axis of rotation.
P
Mi Ri 2
The "R" in the expression is perpendicular distance of
Does it matter whether the particle lie on left or right of the axis?
Obviously no. It means that MI has no directional attribute. Thus, we can conclude that MI of the quarter
plate is arithmetic quarter (1/4) of the MI of a complete disk, whose mass is 4 times greater than that of
quarter plate.
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Circular plate
Figure 4:
MI of complete circular plate.
Hence, mass of the corresponding complete disk is 4M and radius is same R as that of quarter plate. The
MI of the complete disk about perpendicular axis to its surface is :
IO = ( 4M )
R2
2
Now, MI of the quarter plate is 1/4 of the complete disk,
I =
Note :
1
4 IO
=
1
4
( 4M )
R2
2
=
M R2
2
We notice that the expression of MI has not changed for the quarter plate. We must, however,
be aware that the symbol "M" represents mass of the quarter plate - not that of the complete circular plate.
QBA (Question based on above) :
The MI of a uniform disk is 1.0
central axis. If a segment, subtending an angle of 120
◦
kg − m2
about its perpendicular
at the center, is removed from it, then nd the MI
of the remaining disk about the axis.
Hint :
◦
We see that 120 /360
◦
= 1/3 part of the complete disk is removed. Therefore, the mass of the
remaining part is 1-1/3 = 2/3. Answer is 2/3
kg − m2
.
2.4 Oblique axis
Example 5
Problem :
Determine the moment of inertia of a uniform rod of length "L" and mass "m" about an
axis passing through its center and inclined at an angle "θ " as shown in the gure.
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MI of a rod
Figure 5:
Solution :
MI of a rod about an axis making an angle with the rod
We shall work out MI from the basic integral expression for elemental mass :
I = r2 m
Let us consider an small element "dx" along the length, which is situated at a linear distance "x" from
the axis. The elemental mass, however, is at a perpendicular distance AB from the axis of rotation not x'.
MI of a rod
Figure 6:
From
MI of a rod about an axis making an angle with the rod
∆OAB,
r = xsinθ
The above expression allows us to convert all distances involved in the calculation of MI in terms of a
single variable "x". Now, the elemental mass (dm) is, given as :
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m = λx =
M
L
x
The MI of elemental mass about inclined axis AA' is :
I = r2 m = x2 sin2 θ
M
L
x
The MI of rod about inclined axis AA', therefore, is :
I =
R
x2 sin2 θ
R
I =
M
L
x
We note here that the angle "θ " is constant for all elemental masses along the rod and, thus "
sin2 θ
"
together with "M/L" can be taken out of the integral sign. The appropriate limits of integration covering
the length of the rod are "-L/2" and "L/2" :
⇒ I =
M
L
sin2 θ
R
⇒ I =
M
L
sin2 θ
h
⇒ I =
M
L
⇒ I =
sin2 θ
h
L
2
− L
2
x3
3
L3
24
i
+
x2 x
L
2
− L
2
L3
24
i
2
M L2
12 sin θ
M L2
12 is MI of the rod about perpendicular line through COM, denoted by Ic. The moment
of inertia about an axis making an angle with the rod, therefore, can be expressed in terms of "Ic" as :
Note1 :
Here,
⇒ I = IC sin2 θ
Note2 :
The MI of the rod at an inclined axis passing through one of its edge is obtained by integrating
between limits "0" and "L" :
MI of a rod
Figure 7:
MI of a rod about an axis making an angle with the rod and passing through one of the
edges
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M
L
I =
sin2 θ
⇒ I =
M
L
⇒ I =
h
x3
3
i
sin2 θ x
L
0
L3
3
2
M L2
3 sin θ
2.5 Non-uniform mass distribution
Example 6
Problem :
The mass per unit length of a non-uniform rod of mass M and length L is given as
ax, where x is measured from left end and a is a constant. What is its moment of inertia about an axis
perpendicular to the rod and passing through its left end ( point O) ?
MI of a rod
Figure 8:
Solution :
Mass per unit length of the rod varies from one end to another.
Let us now consider a small length dx at a distance x from the axis of rotation. The
elemental mass is,
m = λ x = ax x
In order to evaluate the constant a, we integrate the elemental mass for the whole length of the rod and
equate the same to its mass as :
M =
R
m =
R
ax x
h
L
0
=
Putting the appropriate limit,
⇒ M = a
x2
3
⇒ a =
Thus, elemental mass is :
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i
2M
L2
aL2
2
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⇒
2M
L2
m = ax x =
x x
The moment of inertia of the elemental mass about the axis is :
I = r 2 m = x2
2M
L2
x x
Integrating between the limits 0 and L, we have :
⇒ I =
R
I =
⇒ I =
2M
L2
⇒ I =
Example 7
Problem :
σ = ar2
2M
L2
R
h
x4
4
i
x3 x
L
0
M L2
2
A circular disk of radius R and mass M has non-uniform surface mass density, given by
, where r is the distance from the center of the disk and a is a constant. What is its MI about
the perpendicular central axis ?
Solution :
Let us now consider a small concentric area of thickness dr at a distance r from the axis
of rotation. The elemental mass is,
MI of a circular plate
Figure 9:
Mass per unit areas of the circular plate varies in radial direction.
m = σ A = ar2 2πr r
In order to evaluate the constant a, we integrate the elemental mass for the whole of the circular disk
and equate the same to its mass as :
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M =
R
m = 2πa
r3 r
R
Putting the appropriate limit,
⇒ M = 2πa
h
i
r4
4
L
0
πaR4
2
=
2M
πR4
⇒ a =
Thus, elemental mass is :
⇒
m =
2M
πR4
x2πr3 r
The moment of inertia of the ring of elemental mass about the axis is :
2M
πR4
I = r2 m =
x2πr5 r
Integrating between the limits 0 and R, we have :
⇒ I =
R
⇒ I =
⇒ I =
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4M
R4
4M
R4
I =
4M
R4
x
h
R6
6
r6
6
=
R
i
r5 r
R
0
2M R2
3