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Moments of inertia of rigid bodies (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

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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.

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 formulae 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 modification. Here, mass distribution and distance involve different 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.

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 formulae
• Estimation of MI by inspection
• Part of a rigid body
• Non-uniform mass distribution

Using MI formulae

Example 1

Problem : MIs of a straight wire about a perpendicular through the mid point and a circular frame about its perpendicular central axis are I 1 I 1 and I 2 I 2 respectively. If the lengths of the wires in them are equal, then find the ratio I 1 I 2 I 1 I 2 .

Solution : The MI of the straight wire about a perpendicular through the mid point is :

I 1 = M L 2 12 I 1 = M L 2 12

The MI of the circular frame about its perpendicular central axis is :

I 2 = M R 2 I 2 = M R 2

According to question, the lengths of the wires in them are equal. This means that :

L = 2 π R L = 2 π R

Substituting in the expression of MI of circular frame,

I 2 = M L 2 4 π 2 I 2 = M L 2 4 π 2

Now, the required ratio is :

I 1 I 2 = M L 2 x 4 π 2 12 x M L 2 I 1 I 2 = M L 2 x 4 π 2 12 x M L 2

I 1 I 2 = π 2 3 I 1 I 2 = π 2 3

Example 2

Problem : The moments of inertia of a solid sphere and a ring of same mass about their central axes are same. If R s R s be the radius of solid sphere, then find the radius of the ring.

Solution : Let the radius of the ring is R s R s . Now, the MI of the solid sphere about its central axis is given as :

I s = 2 M R s 2 5 I s = 2 M R s 2 5

The MI of the ring about its central axis is given as :

I r = M R 4 2 I r = M R 4 2

According to question :

M R r 2 = 2 M R s 2 5 M R r 2 = 2 M R s 2 5

R r = ( 2 5 ) x R s R r = ( 2 5 ) x R s

Estimation of MI by inspection

Example 3

Problem : Find the coordinate about which the moment of inertia of a uniform rectangle of dimensions "a" and "2a" is least.

Solution : We can actually determine MI about the given axes by evaluating the integral of MI about each of them 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 such that mass is distributed closest to it.

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 distnace "a"; most distant mass about z-axis is at a distnace √5a/2.

Thus, mass distribution is closest about x-axis.

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 figure.

Solution : We need to look closely about the way MI about an axis is defined. MI of the rigid body about the an axis is given by :

I = M i R i 2 I = M i R i 2

This definition is essentially scalar in nature. The "r" in the expression is perpendicular distance of a particle from the axis of rotation. 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.

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 :

I O = ( 4 M ) R 2 2 I O = ( 4 M ) R 2 2

Now, MI of the quarter plate is 1/4 of the complete disk,

I = 1 4 I O = 1 4 ( 4 M ) R 2 2 = M R 2 2 I = 1 4 I O = 1 4 ( 4 M ) R 2 2 = M R 2 2

QBA (Question based on above) : The MI of a uniform disk is 1.0 kg-m2 about its perpendicular central axis. If a segment, subtending an of 1200 at the center is removed from it, then find the MI of the remaining disk about the axis.

Hint : We see that 120°/360° = 1/3 of the original disk is removed. Therefore, the mass of the remaining part is 1-1/3 = 2/3. Answer is “2/3 kg - m 2 kg - m 2 ”.

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".

Solution : We need to look closely about the way MI about an axis is defined. MI of the rigid body about the an axis is given by :

We shall work out MI from the basic integral expression for elemental mass :

I = r 2 m I = r 2 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’.

From ΔOAB,

r = AB x sin θ r = AB x sin θ

Elemental mass (dm) is, thus, given as :

m = λ x = ( M L ) x m = λ x = ( M L ) x

The MI of elemental mass about inclined axis AA' is :

I = r 2 m = x 2 sin 2 θ ( M L ) x I = r 2 m = x 2 sin 2 θ ( M L ) x

Now, the MI of rod about inclined axis AA' is :

I = I = x 2 sin 2 θ ( M L ) x I = I = x 2 sin 2 θ ( M L ) x

We note here that the angle "θ" is constant for all elemental masses and, thus " sin 2 θ sin 2 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 ) sin 2 θ - L 2 L 2 x 2 x I = ( M L ) sin 2 θ - L 2 L 2 x 2 x

I = ( M L ) sin 2 θ [ x 3 3 ] - L 2 L 2 I = ( M L ) sin 2 θ [ x 3 3 ] - L 2 L 2

I = ( M L ) sin 2 θ [ L 3 24 + L 3 24 ] I = ( M L ) sin 2 θ [ L 3 24 + L 3 24 ]

I = M L 2 12 sin 2 θ I = M L 2 12 sin 2 θ

Note1 : Here, M L 2 12 M L 2 12 is MI of the rod about perpendicular line through COM, denoted by Io. The moment of inertia about an axis making an angle with the rod, therefore can be expressed in terms of "Io" as :

I = I O sin 2 θ I = I O sin 2 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" :

I = ( M L ) sin 2 θ [ x 3 3 ] 0 L I = ( M L ) sin 2 θ [ x 3 3 ] 0 L

I = ( M L ) sin 2 θ x L 3 3 I = ( M L ) sin 2 θ x L 3 3

I = M L 2 3 sin 2 θ I = M L 2 3 sin 2 θ

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”) ?

Solution : Let us now consider a small length “dx” at a distance “x” from the axis of rotation. The elemental mass is,

m = λ x = a x x m = λ x = a x 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 = m = a x x M = m = a x x

Putting the appropriate limit,

M = a [ x 2 3 ] 0 L = a L 2 2 M = a [ x 2 3 ] 0 L = a L 2 2

a = 2 M L 2 a = 2 M L 2

Thus, elemental mass is :

m = ( 2 M L 2 ) x x m = ( 2 M L 2 ) x x

The moment of inertia of the elemental mass about the axis is :

I = r 2 m = 2 M L 2 ) x x x 2 I = r 2 m = 2 M L 2 ) x x x 2

Integrating between the limits “0” and “L”, we have :

I = I = 2 M L 2 ) x x x 2 I = I = 2 M L 2 ) x x x 2

I = I = 2 M L 2 [ x 4 4 ] 0 L I = I = 2 M L 2 [ x 4 4 ] 0 L

I = M L 2 2 I = M L 2 2

Example 7

Problem : A circular disk of radius “R” and mass “M” has non-uniform surface mass density, given by σ = a r 2 σ = a r 2 , 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,

m = σ A = a r 2 2 π r r m = σ A = a r 2 2 π r 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 :

M = m = 2 π a r 3 r M = m = 2 π a r 3 r

Putting the appropriate limit,

M = 2 π a [ r 4 4 ] 0 L = π a R 4 2 M = 2 π a [ r 4 4 ] 0 L = π a R 4 2

a = 2 M π R 4 a = 2 M π R 4

Thus, elemental mass is :

m = ( 2 M π R 4 ) x 2 π r 3 r m = ( 2 M π R 4 ) x 2 π r 3 r

The moment of inertia of the ring of elemental mass about the axis is :

I = ( 2 M π R 4 ) x 2 π r 5 r I = ( 2 M π R 4 ) x 2 π r 5 r

Integrating between the limits “0” and “R”, we have :

I = I = 4 M R 4 r 5 r I = I = 4 M R 4 r 5 r

I = 4 M R 4 [ r 6 6 ] 0 R I = 4 M R 4 [ r 6 6 ] 0 R

I = 4 M R 4 x R 6 6 = 2 M R 2 3 I = 4 M R 4 x R 6 6 = 2 M R 2 3

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