Moment of Inertia formula for different shapes | JEE Main

Moment of inertia or mass moment of inertia is the resistance of a rigid body to change in its angular velocity or we can say, resistance to angular acceleration, when a net external torque acts on it (similar to resistance offered by mass of a particle to acceleration, when a net force acts on it). In this blog we will explore moment of inertia formula for different shapes (ring, disc, hollow cylinder / annular disc, solid cylinder, rod, sphere, hollow sphere, rectangle / rectangular plate, square plate, cube, triangle, cone

For a rigid body made up of discrete particles (for example a pendulum with bob of mass $m$ and massless rod of length $l$), moment of inertia can be calculated using the formula $I = \sum_i m_i r_i^2$

For a continuous rigid body (for example a uniform solid sphere or a uniform rod etc.), $I = \int r^2 dm$

Unit of moment of inertia $I$ is $Kg \ m^2$

Dimensional Formula = $[M^1L^2T^0]$

Now, let’s look at the moment of inertia for different shapes, but before we do that, let’s review the parallel axis and perpendicular axis theorem

Parallel Axis Theorem

Parallel Axis Theorem to find moment of inertia

$I = I_{CM} + Mh^2$

Derivation
Moment of Inertia about $AB$, $I_{CM}$ $=\int r^2dm$

Moment of inertia about $CD$ (parallel to $AB$), $I =\int (h-r)^2 dm$ $=\int (h^2 + r^2 -2hr)dm$ $= \int r^2dm +$ $h^2 \int dm -$ $2h \int rdm$

Now, $\int r dm = M r_{CM} = 0$ as the axis $AB$ is passing through the center of mass (ping us if this didn’t make sense)

Or, $I = I_{CM} + Mh^2$

Perpendicular Axis Theorem (valid ONLY for planar objects)

Perpendicular Axis Theorem to find moment of inertia

$I_z = I_x + I_y$

Derivation
$I_z = \sum_i m_i r_i^2$ $= \sum_i m_i (a_i^2 + b_i^2)$ $=\sum_i m_i a_i^2 + $ $\sum_i m_i + b_i^2$ $=I_x + I_y$ 

Non – Planar Objects:
Note perpendicular axis theorem is NOT valid for non planar or 3 dimensional objects like a cube.

Perpendicular Axis Theorem - Non Planar Object

$I_z \neq I_x + I_y$

Reason
$I_z = \sum_i m_i (a_i^2 + b_i^2$
$I_x = \sum_i m_i (b_i^2 + h_i^2)$
$I_y = \sum_i m_i (b_i^2 + h_i^2)$

Moment of Inertia of Ring

Moment of Inertia of Ring

About the central axis (passing through its center and perpendicular to the plane of the ring):
$I = MR^2$

About the diameter:
$I = \cfrac{1}{2} MR^2$

About the central axis (passing through the center of the ring and perpendicular to the plane of the ring):
$I = \int R^2 dm$ $= R^2 \int dm$ $= MR^2$

About its diameter:
Using perpendicular axis theorem $I_z = I_x + I_y$, we will get $I = \cfrac{1}{2} MR^2$

Top

Moment of Inertia of Disc

Moment of Inertia of Disc

About the central axis (axis passing through the center and perpendicular to the plane of the disc)
$I = \cfrac{1}{2}MR^2$

About its diameter:
$I = \cfrac{1}{4}MR^2$

About its central axis: $dI = r^2 dm$ $dm = \cfrac{M}{\pi R^2} 2 \pi r dr$ $\implies$ $dI = \cfrac{2M}{R^2} r^3 dr$ $I = \int dI = \cfrac{1}{2} MR^2$ About its diameter: Using the perpendicular axis theorem $I_z = I_x + I_y$, we will get $I = \cfrac{1}{4}MR^2$

Top

Moment of Inertia of Hollow Cylinder | Annular Disc

Moment of Inertia of hollow cylinder or annular disc

About the central axis:
$I = \cfrac{1}{2} M (R_1^2 + R_2^2)$

About its central axis:
$dI = r^2 dm$
$dm = \cfrac{M}{\pi (R_2^2 – R_1^2)} 2 \pi r dr$
$\implies$ $dI=$ $\int_{r_1}^{r_2} r^2 dm$ $\cfrac{2 M}{R_2^2 – R_1^2} \int_{r_1}^{r_2} r^3 dr$
Or
$I = \cfrac{1}{2} M (R_1^2 + R_2^2)$

Top

Moment of Inertia of Solid Cylinder

Moment of inertia of solid cylinder

About the axis perpendicular to the length of the rod and passing through its center:
$I = \cfrac{1}{4} MR^2 + \cfrac{1}{12} ML^2$

About the central axis:
$I = \cfrac{1}{2} MR^2$

About the axis perpendicular to its length and passing through its center:
$dI = \cfrac{1}{4} R^2 dm + dm x^2$
$dm = \cfrac{M}{L} dx$
$I = \int dI = \cfrac{1}{4} R^2 \int dm + \cfrac{M}{L} \int_{-L/2}^{L/2} x^2 dx$
$I = \cfrac{1}{4} MR^2 + \cfrac{1}{12} ML^2$

About the central axis:
Same as that of disc about its central axis

Top

Moment of Inertia of a Rod

Moment of Inertia of Rod

About the axis perpendicular to its length and passing through the center:

$I = \cfrac{1}{12} ML^2$

$dI = \int_{-L/2}^{L/2} x^2 dm$ 
$dm = \cfrac{M}{L} dx$
$I = \cfrac{1}{12}ML^2$

Top

Moment of Inertia of Hollow Sphere

Moment of Inertia of Hollow Cylinder

About its diameter:

$I = \cfrac{2}{3} MR^2$

Let’s consider the differential ring of mass $dm$ as shown: 

$dm = \cfrac{M}{4 \pi R^2}2\pi R \sin \theta . R d\theta$ $=\cfrac{1}{2} M \sin \theta d \theta$

$dI = R^2 \sin^2 \theta dm$ $=\cfrac{1}{2} MR^2 \sin^3 \theta d\theta$

$I = \int dI$ $= \cfrac{1}{2} MR^2 \int_0^{\pi} \sin^3 \theta d\theta$

$I = \cfrac{2}{3} MR^2$

Top

Moment of Inertia of Solid Sphere

Moment of Inertia of Solid Cylinder

About its diameter:

$I = \cfrac{2}{5} MR^2$

Consider a spherical shell or radius $r$ and thickness $dr$

For this differential strip,
$dm = \cfrac{M}{\cfrac{4}{3} \pi R^3} 4 \pi r^2 dr$ $=\cfrac{3M}{R^3} r^2dr$
$dI = \cfrac{2}{3} dm r^2$ $=\cfrac{2M}{R^3} r^4 dr$
$I = \cfrac{2M}{R^3} \int_0^R r^4 dr$ $=\cfrac{2}{5} MR^2$

Top

Moment of Inertia of Rectangle | Rectangular Plate

Moment of Inertia of Rectangle or Rectangular Plate

About the axis passing through its center and perpendicular to its plane,

$I = \cfrac{M}{12} (a^2 + b^2)$

Let’s consider differential rod like element of width $dx$,

$dm = \cfrac{M}{ab}adx$ $=\cfrac{M}{b} dx$

$dI = \cfrac{1}{12} dm a^2 + dm (\cfrac{b}{2} – x)^2$

$I = \int dI$ $=\cfrac{1}{12} \cfrac{Ma^2}{b} \int_0^b dx + \cfrac{M}{b} \int_0^b (\cfrac{b^2}{4} + x^2 – bx)dx$

$I = \cfrac{1}{12} M (a^2 + b^2)$

Top

Moment of Inertia of Square Plate

Moment of Inertia of Square

About the axis through its center and perpendicular to its plane:

$I = \cfrac{1}{6} Ma^2$

About the axis passing through its center and in the plane of the square:

$I = \cfrac{1}{12} Ma^2$

About the axis through its center and perpendicular to its plane:
Derivation is same as that for rectangular plate

About the axis passing through its center and in plane of the square ($I_x$ or $I_y$):
From perpendicular axis theorem: $I_z = I_x + I_y$
From Symmetry $I_x = I_y$

$\implies$ $I_x = I_y = \cfrac{1}{12} Ma^2$

Top

Moment of Inertia of Cube

Moment of Inertia of Cube

About an axis passing through its center and perpendicular to any two faces,

$I = \cfrac{1}{6} Ma^2$

Consider the differential square plate element of thickness $dx$ 

$dm = \cfrac{M}{a^3} a^2 dx$ $= \cfrac{M}{a} dx$

$dI = \cfrac{1}{6} a^2 dm$

$I = \int dI = \cfrac{1}{6} Ma \int_0^a dx$

$I = \cfrac{1}{6} Ma^2$

Top

Moment of Inertia of Triangle

1) Moment of Inertia of Equilateral Triangle about centroid

Moment of Inertia of Equilateral Triangle About Centroid

Moment of inertia about an axis passing through the centroid and perpendicular to the triangle, $I = \cfrac{1}{12} Ma^2$

Let’s find the moment of inertia of an equilateral triangle of side $a$ indirectly by dividing the equilateral triangle into 4 equilateral triangles of slide $\cfrac{a}{2}$ as shown.

Now, the moment of inertia is proportional to mass and square of the side lengths, 

So moment of inertia of smaller equilateral triangles, $i$, (about perpendicular axis passing through their centroids) would be $\cfrac{1}{16}$ of the moment of inertia $I$ of the full triangle

So using parallel axis theorem, we can write

$I = 4 \times \cfrac{I}{16} + 3 \times \cfrac{M}{4} \times (\cfrac{a}{2 \sqrt{3}})^2$

Or, $I = \cfrac{1}{12} Ma^2$

2) Moment of Inertia of Triangle (of base $a$ and height $h$) about centroid (with axis in-plane of triangle)

Moment of Inertia of Triangle about axis passing through centroid but in plane of triangle

Moment of inertia about the axis passing through centroid and parallel to base:

$I = \cfrac{\rho ah^3}{36} = \cfrac{Mh^2}{18}$

Consider the differential element $dm$ as shown $dm = \cfrac{M}{\cfrac{1}{2}ah} \cfrac{a}{h} x dx$ $dI = dm (\cfrac{2}{3}h – x)^2$ $I = \int_0^h dI$

3) Moment of Inertia of Triangle (of base $a$ and height $h$) about its base

Moment of Inertia about base

Moment of inertia about the base:

$I = \cfrac{\rho ah^3}{12} = \cfrac{Mh^2}{6}$

Using parallel axis theorem: 

$I_{base} = I_{centroid} + M(\cfrac{h}{3})^2$ $=\cfrac{Mh^2}{6}$

Top

Moment of Inertia of Solid Cone

Moment of Inertia of Solid Cone

Moment of inertia of solid cone about its central axis, $I$ $=\cfrac{3}{10} MR^2$

Consider the differential disc element as shown

$dm = \cfrac{M}{\cfrac{1}{3} \pi R^2 h} \pi (\cfrac{R}{h} x)^2 dx$
$dI = \cfrac{1}{2} (\cfrac{R}{h} x)^2 dm$

$I = \int_0^h dI$

Top

Leave a Reply