Moment of Inertia and Centroid MCQ Quiz - Objective Question with Answer for Moment of Inertia and Centroid - Download Free PDF

Explanation:

• The moment of inertia of an area represents how that area is distributed relative to a reference axis. It reflects the section's ability to resist bending — the larger the value, the stiffer the section.

• It can be calculated using integration methods by summing up the contributions of small elemental areas across the entire section. This is particularly useful for irregular or complex shapes where standard formulas are not applicable.

• The unit of moment of inertia depends on the units of both area and distance. If area is measured in square meters (m²), and distance in meters (m), then the unit of moment of inertia becomes meters to the fourth power (m⁴). Thus, units of area and distance both influence the unit of moment of inertia.

• Moment of inertia of an area (second moment of area) is different from mass moment of inertia, which concerns the distribution of mass and is used in rotational dynamics. Concepts like Routh’s Rule apply to mass moment of inertia, not to second moment of area.

 Additional Information Routh's Rule

• Routh’s Rule is a method used in dynamics to determine the mass moment of inertia of a rigid body about an arbitrary axis. It is particularly helpful when the body is unsymmetrical and does not align neatly with standard coordinate axes.

• It is based on the known mass moments of inertia about three mutually perpendicular axes that pass through a common point (usually the center of mass or centroid of the body).

• The rule allows engineers to calculate the moment of inertia about any axis that does not coincide with these principal axes, by combining the known moments of inertia and considering the orientation of the new axis.

• Routh’s Rule is used mainly in mechanical and aerospace engineering, where bodies rotate about complex axes (such as in gyroscopes, propellers, and unsymmetrical machinery), and helps in dynamic analysis and design of rotating systems.

Concept: 

• The centroid is the geometric center of a plane figure or lamina. It is the point at which the entire area of the lamina can be assumed to be concentrated for analysis purposes.
• In uniform materials, the centroid corresponds to the center of mass or center of gravity, assuming uniform density and thickness.

Calculations:

Taking reference as the bottom part of the flange.$A_1=(50\times 10)=500m{m}^2$

$y_1=20+\frac{50}{2}=45mm$

$A_2=(100\times 20)=2000m{m}^2$

$y_2=\frac{20}{2}=10mm$

$\overline{y}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}$

$\overline{y}=\frac{500\times 45+2000\times 45}{2000+500}=\frac{42500}{2500}=17mm$

The section is symmetric about the vertical axis passing through the center of the web.

Therefore, the centroid lies at the centerline of the web horizontally.

$\overline{x}=\frac{100}{2}=50mm$

The centroid is (50 mm, 17 mm).

Concept:

Flange width = 100 mm

Flange thickness = 10 mm

Web height = 80 mm, Web thickness = 10 mm

Total height of section = 10 + 80 + 10 = 100 mm

Centroidal MOI = 449.3 × 10⁴ mm⁴ = 4493000 mm⁴

Calculation:

Area of top flange, A₁ = 100 × 10 = 1000 mm²

Area of bottom flange, A₂ = 100 × 10 = 1000 mm²

Area of web, A₃ = 10 × 80 = 800 mm²

Total area, A = A₁ + A₂ + A₃ = 1000 + 1000 + 800 = 2800 mm²

Distance from centroid to top of section, d = 100 / 2 = 50 mm

$I=I_{centroidal}+A\times d^2$

$I=4493000+2800\times {50}^2=4493000+2800\times 2500$

$I=4493000+7000000=11493000m{m}^4$

Concept:

To calculate the moment of inertia of a flange about the X–X axis of the I-section, we use the Parallel Axis Theorem.

Given:

Flange width, b = 100 mm

Flange thickness, t = 10 mm

Distance between flange centroid and axis X–X, d = 50 mm

Calculation:

Area of flange, A = b × t = 100 × 10 = 1000 mm², d = 50 mm

Using Parallel Axis Theorem,

$I_{X-X}=I_f+A\times d^2$

Substitute values:

$I_{X-X}=I_f+1000\times {50}^2=I_f+1000\times 2500$

$I_{X-X}=I_f+2.5\times {10}^6{\text{mm}}^4$

Explanation:

Moment of Inertia of a Circular Area About a Tangent to the Circle

The moment of inertia (I) of a circular area about a tangent to the circle is calculated using the parallel axis theorem. The theorem states that the moment of inertia of an area about any axis parallel to the centroidal axis is equal to the moment of inertia about the centroidal axis plus the product of the area and the square of the distance between the two axes.

Centroidal Moment of Inertia:

The moment of inertia of a circular area about its centroidal axis in the plane of the lamina is:

I_centroid = (π × r⁴) / 4

Parallel Axis Theorem:

Using the parallel axis theorem, the moment of inertia about a tangent to the circle is:

I_tangent = I_centroid + A × d²

Where:

• I_centroid = Moment of inertia about the centroidal axis
• A = Area of the circular lamina
• d = Distance between the centroidal axis and the tangent axis (equal to the radius r of the circle)

Calculation:

The area (A) of the circular lamina is given by:

A = π × r²

Substituting the values:

I_tangent = I_centroid + (π × r²) × r²

I_tangent = (π × r⁴) / 4 + π × r⁴

I_tangent = π × r⁴ × (1/4 + 1)

I_tangent = π × r⁴ × (5/4)

I_tangent = (5π × r⁴) / 4

Concept:

The CG of a semicircular plate of  r radius, from its base, is

$\overline{y}=\frac{4r}{3\pi }$

Calculation:

Given:

r = 33 cm

$\overline{y}=\frac{4r}{3\pi }=\frac{4\times 33}{3\times \frac{22}{7}}$

y̅ = 14 cm

∴ the C.G. of a semicircular plate of 66 cm diameter, from its base, is 14 cm.

Additional Information

C.G. of the various plain lamina are shown below in the table. Here x̅  & y̅  represent the distance of C.G. from x and y-axis respectively.

Circle
Semicircle
Triangle
Cone
Rectangle
Quarter Circle
Solid hemisphere

CONCEPT:

Moment of inertia:

• The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
• The moment of inertia of a particle  is

⇒ I = mr2

Where r = the perpendicular distance of the particle from the rotational axis.

• Moment of inertia of a body made up of a number of particles (discrete distribution)

⇒ I = m1r12 + m2r22 + m3r32 + m4r42 + -------

Rotational kinetic energy: 

• The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
• A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
• Mathematically rotational kinetic energy can be written as -

⇒ KE $=\frac{1}{2}I{\omega }^2$

Where I = moment of inertia and ω = angular velocity.

EXPLANATION:

• The moment of inertia of the ring about an axis passing through the center and perpendicular to its plane is given by

⇒ I_ring = MR²

• Moment of inertia of the disc about an axis passing through center and perpendicular to its plane is given by -

$\Rightarrow I_{disc}=\frac{1}{2}M{R}^2$

• As we know that mathematically rotational kinetic energy can be written as

$\Rightarrow KE=\frac{1}{2}I{\omega }^2$

• According to the question, the angular velocity of a thin disc and a thin ring is the same. Therefore, the kinetic energy depends on the moment of inertia.
• Therefore, a body having more moments of inertia will have more kinetic energy and vice - versa.
• So, from the equation, it is clear that,

⇒ Iring > Idisc

∴ Kring > Kdisc

• The ring has higher kinetic energy.

Centre of gravity of various plane areas

1) Triangle

2) Semi circle of radius ‘R’

3) This Hollow Cone (Right angled)

4) Trape zoid

$\overline{y}=\frac{ga+6}{\left(a+6\right)}\left(\frac{4}{3}\right)$

5) sine wave

6) 4^th degree curve

$\overline{x}=\left(\frac{6\left(N+1\right)}{2\left(N+2\right)}\right)$

$\overline{y}=\left(\frac{hN}{ZN+1}\right)$

Important Points

1. Centre of gravity of a thin hollow cone = 1/3
2. Centre of gravity of a solid cone = 1/4

Concept:

Moment of inertia of circular plate,

${\mathrm{I}}_{\mathrm{x}\mathrm{x}}=\frac{\pi }{64}{\mathrm{d}}^4$

Moment of inertia of Square plate,

${\mathrm{I}}_{\mathrm{x}\mathrm{x}}=\frac{\mathrm{d}\times {\mathrm{d}}^3}{12}$

Calculation:

The ratio of the moment of inertia of a circular plate to that of a square plate is, Which is less than 1.

Important Points 

The following table shows the Second moment of inertia of different shapes

Shape	Figure	Moment of Inertia
Rectangle		${\mathrm{I}}_{\mathrm{x}\mathrm{x}}=\frac{\mathrm{b}{\mathrm{d}}^3}{12}$ ${\mathrm{I}}_{\mathrm{y}\mathrm{y}}=\frac{\mathrm{d}{\mathrm{b}}^3}{12}$
Triangle		${\mathrm{I}}_{\mathrm{x}\mathrm{x}}=\frac{\mathrm{b}{\mathrm{h}}^3}{36}$ ${\mathrm{I}}_{\mathrm{y}\mathrm{y}}=\frac{\mathrm{h}{\mathrm{b}}^3}{36}$
Circle		${\mathrm{I}}_{\mathrm{x}\mathrm{x}}=\frac{\pi }{64}{\mathrm{d}}^4$ ${\mathrm{I}}_{\mathrm{y}\mathrm{y}}=\frac{\pi }{64}{\mathrm{d}}^4$
Semicircle		${\mathrm{I}}_{\mathrm{x}\mathrm{x}}=0.11{\mathrm{R}}^4$ ${\mathrm{I}}_{\mathrm{y}\mathrm{y}}=\frac{\pi }{8}{\mathrm{R}}^4$
Quarter circle		${\mathrm{I}}_{\mathrm{x}\mathrm{x}}=0.055{\mathrm{R}}^4$ ${\mathrm{I}}_{\mathrm{y}\mathrm{y}}=0.055{\mathrm{R}}^4$

CONCEPT:

• Parallel axis theorem: Moment of inertia of a body about a given axis I is equal to the sum of moment of inertia of the body about an axis parallel to given axis and passing through centre of mass of the body Io and Ma2, where ‘M’ is the mass of the body and ‘a’ is the perpendicular distance between the two axes.

⇒ I = I_o + Ma²

EXPLANATION:

• For a uniform rod with negligible thickness, the moment of inertia about its centre of mass is:

$I_{cm}=\frac{1}{12}M{L}^2$

Where M = mass of the rod and L = length of the rod

∴ The moment of inertia about the end of the rod is

$\Rightarrow I_{end}=I_{cm}+M{d}^2$

$\Rightarrow I_{end}=\frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2=\frac{1}{3}M{L}^2$

Concept:

Area moment of inertia is given by, I = A × k²

where A is an area of section and k is radius of gyration of the section.

For circular section, k = D/4

Calculation:

$\mathrm{A}=\frac{\pi }{4}{\mathrm{D}}^2$

$\mathrm{k}=\frac{\mathrm{D}}{4}$

$\therefore \mathrm{I}=\mathrm{A}\times {\mathrm{k}}^2=\frac{\pi }{64}{\mathrm{D}}^4=\frac{\pi }{4}{R}^4$

here, the area moment of inertia for the quadrant is ${\mathrm{I}}_{\mathrm{q}\mathrm{u}\mathrm{a}}=\frac{1}{4}I=\frac{\pi }{16}{R}^4$

Explanation:

Moment of inertia:

Moment of inertia is a measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element’s distance from the axis.

I = ∑( m₁r₁² + m₂r₂² + m₃r₃² +m₄r₄² + …….. + m_nr_n²)

Moment of inertia of a thin spherical shell of mass M and radius R about its diameter.

$\mathrm{I}=\frac{2}{3}\mathrm{M}{\mathrm{R}}^2$

Additional Information

Moment of inertia of some important shapes:

Concept:

Area Moment of Inertia:

• It is a geometrical property of an area which reflects how its points are distributed with regards to an arbitrary axis.
• It is also known as 2^nd moment of area or 2^nd Moment of Inertia.
• Its SI unit is ‘m⁴’
• Mathematically, it is represented as

$I_x=\int \int y^{2}dxdyand{I}_y=\int \int x^{2}dxdy$

Calculation:

Given:

width(b) = 3 cm, height(h)= 4 cm

For the rectangular section, the Moment of Inertia is given by

$I_{xx}=\frac{b{h}^3}{12}=\frac{3\times 4^3}{12}=16c{m}^4$

Mass Moment of Inertia:

It is a measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element’s distance from the axis.

It’s SI unit is kg-m²

Mathematically, $I=\sum _{i=1}^n{m}_i{r}_i^2$

MOI of Some Standard Shapes:

Explanation:

Moment of inertia of different section:

For circular cross-section,

$I_{xx}=I_{yy}=\frac{\pi D^4}{64}$

According to perpendicular axis theorem,

I_zz = I_xx + I_yy

∴ I_zz = 2 × I_xx

∴ I_xx is always less than I_zz

Concept:

Moment of inertia:

A moment of inertia is a measure of the resistance of a body to angular acceleration about a given axis that is equal to the sum of the products of each element of mass in the body and the square of the element’s distance from the axis.

I = ∑( m1r12 + m2r22 + m3r32 +m4r42 + …….. + mnrn2)​

Moment of inertia of a thin spherical shell of mass M and radius R about its diameter.

$\mathrm{I}=\frac{2}{3}\mathrm{M}{\mathrm{R}}^2$

• For a rigid body system, the moment of inertia is the sum of the moments of inertia of all its particles taken about the same axis.

$I=\sum m_i{r_i}^2$

where I is the Moment of Inertia, m is point mass, r is the perpendicular distance from the axis of rotation.

The moment of inertia of different bodies is given in the below table: