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 I 1 I 1 and I 2 I 2 respectively. If the composition of wires are same and 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. As the composition of the wires same, the mass of two entities are same. Also :

L = 2 π R ⇒ R = L 2 π L = 2 π R ⇒ R = L 2 π

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 : 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

Let the radius of the ring is R r R r . The MI of the ring about its central axis is given as :

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

According to question, the moments of inertia of the solid sphere and the ring about their central axes are same :

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

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.

Figure 1: MI of rectangular plate about an axis.

Rectangular plate

Solution : 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.

Figure 2: MI of rectangular plate about z-axis.

Rectangular plate

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.

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.

Figure 3: MI of quarter plate about z-axis.

Quarter plate

Solution : We need to look closely 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.

Figure 4: MI of complete circular plate.

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 :

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

Note : 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 kg - m 2 kg - m 2 about its perpendicular central axis. If a segment, subtending an angle of 120° 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 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 - m 2 kg - m 2 ”.

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 figure.

Figure 5: MI of a rod about an axis making an angle with the rod

MI of a rod

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

đ I = r 2 đ m đ I = r 2 đ 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’.

Figure 6: MI of a rod about an axis making an angle with the rod

MI of a rod

From ΔOAB,

r = x sin θ r = x sin θ

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 :

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

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

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

The MI of rod about inclined axis AA', therefore, is :

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

We note here that the angle "θ" is constant for all elemental masses along the rod 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 đ x ⇒ I = ( M L ) sin 2 θ ∫ - L 2 L 2 x 2 đ 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 Ic. The moment of inertia about an axis making an angle with the rod, therefore, can be expressed in terms of "Ic" as :

⇒ I = I C sin 2 θ ⇒ I = I C 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" :

Figure 7: MI of a rod about an axis making an angle with the rod and passing through one of the edges

MI of a rod

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

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”) ?

Figure 8: Mass per unit length of the rod varies from one end to another.

MI of a rod

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

đ m = λ đ x = a x đ x đ m = λ đ x = a x đ 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 = ∫ đ m = ∫ a x đ x M = ∫ đ m = ∫ a x đ 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 :

⇒ đ m = a x đ x = ( 2 M L 2 ) x đ x ⇒ đ m = a x đ x = ( 2 M L 2 ) x đ x

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

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

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

⇒ I = ∫ đ I = ∫ ( 2 M L 2 ) x 3 đ x ⇒ I = ∫ đ I = ∫ ( 2 M L 2 ) x 3 đ x

⇒ I = 2 M L 2 [ x 4 4 ] 0 L ⇒ 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,

Figure 9: Mass per unit areas of the circular plate varies in radial direction.

MI of a circular plate

đ m = σ đ A = a r 2 2 π r đ r đ m = σ đ A = a r 2 2 π r đ 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 = ∫ đ m = 2 π a ∫ r 3 đ r M = ∫ đ m = 2 π a ∫ r 3 đ 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 :

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

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

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

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

⇒ I = ∫ đ I = 4 M R 4 ∫ r 5 đ r ⇒ I = ∫ đ I = 4 M R 4 ∫ r 5 đ 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