Moment of Inertia of a Cone

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Moment of Inertia of a Cone

The moment of inertia of a cone tells us how the cone’s mass is spread out in relation to the axis it's rotating around. This concept is important in physics and engineering, especially when studying how objects spin. For a solid cone, the moment of inertia depends on its mass, height, base radius, and the specific axis of rotation. Whether the cone is spinning around its central axis or tilted to spin around its base, knowing its moment of inertia helps us understand how much force—or torque—is needed to make it rotate.

1.0Moment of Inertia of A Hollow Cone

Consider a hollow cone (a thin, conical shell with no thickness) that has a total mass M, a base radius R, and a slant height L. The cone is symmetric about its vertical axis, which passes through its apex and the center of the circular base. Our objective is to determine the moment of inertia I of this hollow cone about its central vertical axis. Since the cone is hollow and thin, the mass is distributed uniformly over its curved surface, and we assume negligible thickness.

Radius of the ring at that point $r = \frac{R}{L} \cdot l$

By using similar Triangle, Width of ring dl

Cone Surface Area $A = π R L$

Mass per unit Area, $σ = \frac{M}{π R L}$

Area of the Ring, $d A = 2 π r d l = 2 π (\frac{R}{l} l) d l$

Mass of the Ring, $d m = σ \cdot d A = \frac{M}{π R L} \cdot 2 π (\frac{R}{L} l) d l = \frac{2 Ml}{L ^{2}} d l$

Moment of Inertia of Ring,

$d I = r^{2} d m = (\frac{R}{L} l)^{2} \cdot \frac{2 Ml}{L ^{2}} d l = \frac{2 M R ^{2}}{L ^{4}} l^{3} d l$

$I = \int_{0}^{L} \frac{2 M R ^{2}}{L ^{4}} l^{3} d l = \frac{2 M R ^{2}}{L ^{4}} \int_{0}^{L} l^{3} d l = \frac{2 M R ^{2}}{L ^{4}} [\frac{l ^{4}}{4}]_{0}^{L} = \frac{2 M R ^{2}}{L ^{4}} \cdot \frac{L ^{4}}{4} = \frac{1}{2} M R^{2}$

Moment of inertia of a hollow cone about its central (vertical) axis is 12MR2

2.0Moment of Inertia of A Solid Cone

By similarity of Triangles,

$\frac{r}{x} = \frac{R}{h} \Rightarrow r = (\frac{R}{H} x)$

Mass of Elemental Disc, $d m = \frac{M}{\frac{1}{3} π R ^{2} H} (π r^{2} d x)$

$d m = \frac{3 M}{π R ^{2} H} π (\frac{R ^{2}}{H ^{2}} x^{2}) d x$

$d m = \frac{3 M}{H ^{3}} x^{2} d x$

Moment of Inertia of Disc

$d I = \frac{1}{2} d m r^{2} = \frac{1}{2} (\frac{3 M}{H ^{3}} x^{2} d x) (\frac{R ^{2}}{H ^{2}} x^{2})$

$I = \int d I = \frac{3}{2} \frac{M}{H ^{5}} R^{2} \int_{0}^{H} x^{4} d x = \frac{3}{2} \frac{M}{H ^{5}} (\frac{H ^{5}}{5}) = \frac{3}{10} M R^{2}$

$I = \frac{3}{10} M R^{2}$

Moment of inertia of a solid cone about its central vertical axis is 310MR2

3.0Moment of Inertia of Solid and Hollow Cones (Various Axes)

Type of Cone	Axis of Rotation	Moment of Inertia (I)
Solid Cone	About vertical axis (along height)	I=310MR2
Solid Cone	About base diameter (perpendicular to height)	I=320MR2+4H2
Hollow Cone	About vertical axis (along height)	I=MR2
Hollow Cone	About base diameter (perpendicular to height)	I=12MR2

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Moment of Inertia of a Cone

The moment of inertia of a cone tells us how the cone’s mass is spread out in relation to the axis it's rotating around. This concept is important in physics and engineering, especially when studying how objects spin. For a solid cone, the moment of inertia depends on its mass, height, base radius, and the specific axis of rotation. Whether the cone is spinning around its central axis or tilted to spin around its base, knowing its moment of inertia helps us understand how much force—or torque—is needed to make it rotate.

1.0Moment of Inertia of A Hollow Cone

Consider a hollow cone (a thin, conical shell with no thickness) that has a total mass M, a base radius R, and a slant height L. The cone is symmetric about its vertical axis, which passes through its apex and the center of the circular base. Our objective is to determine the moment of inertia I of this hollow cone about its central vertical axis. Since the cone is hollow and thin, the mass is distributed uniformly over its curved surface, and we assume negligible thickness.

Radius of the ring at that point $r = \frac{R}{L} \cdot l$

By using similar Triangle, Width of ring dl

Cone Surface Area $A = π R L$

Mass per unit Area, $σ = \frac{M}{π R L}$

Area of the Ring, $d A = 2 π r d l = 2 π (\frac{R}{l} l) d l$

Mass of the Ring, $d m = σ \cdot d A = \frac{M}{π R L} \cdot 2 π (\frac{R}{L} l) d l = \frac{2 Ml}{L ^{2}} d l$

Moment of Inertia of Ring,

$d I = r^{2} d m = (\frac{R}{L} l)^{2} \cdot \frac{2 Ml}{L ^{2}} d l = \frac{2 M R ^{2}}{L ^{4}} l^{3} d l$

$I = \int_{0}^{L} \frac{2 M R ^{2}}{L ^{4}} l^{3} d l = \frac{2 M R ^{2}}{L ^{4}} \int_{0}^{L} l^{3} d l = \frac{2 M R ^{2}}{L ^{4}} [\frac{l ^{4}}{4}]_{0}^{L} = \frac{2 M R ^{2}}{L ^{4}} \cdot \frac{L ^{4}}{4} = \frac{1}{2} M R^{2}$

Moment of inertia of a hollow cone about its central (vertical) axis is 12MR2

2.0Moment of Inertia of A Solid Cone

By similarity of Triangles,

$\frac{r}{x} = \frac{R}{h} \Rightarrow r = (\frac{R}{H} x)$

Mass of Elemental Disc, $d m = \frac{M}{\frac{1}{3} π R ^{2} H} (π r^{2} d x)$

$d m = \frac{3 M}{π R ^{2} H} π (\frac{R ^{2}}{H ^{2}} x^{2}) d x$

$d m = \frac{3 M}{H ^{3}} x^{2} d x$

Moment of Inertia of Disc

$d I = \frac{1}{2} d m r^{2} = \frac{1}{2} (\frac{3 M}{H ^{3}} x^{2} d x) (\frac{R ^{2}}{H ^{2}} x^{2})$

$I = \int d I = \frac{3}{2} \frac{M}{H ^{5}} R^{2} \int_{0}^{H} x^{4} d x = \frac{3}{2} \frac{M}{H ^{5}} (\frac{H ^{5}}{5}) = \frac{3}{10} M R^{2}$

$I = \frac{3}{10} M R^{2}$

Moment of inertia of a solid cone about its central vertical axis is 310MR2

3.0Moment of Inertia of Solid and Hollow Cones (Various Axes)

Type of Cone	Axis of Rotation	Moment of Inertia (I)
Solid Cone	About vertical axis (along height)	I=310MR2
Solid Cone	About base diameter (perpendicular to height)	I=320MR2+4H2
Hollow Cone	About vertical axis (along height)	I=MR2
Hollow Cone	About base diameter (perpendicular to height)	I=12MR2

Frequently Asked Questions

You have two cones of identical mass and radius — one solid and one hollow. Which one offers more resistance to rotation about its vertical axis and why?

A solid cone is spun about a horizontal axis passing through its base (perpendicular to the height). If a hollow cone of the same size and mass is spun the same way, will their moments of inertia be the same?

Why is it easier to topple a hollow cone than a solid one resting on the same base?

Frequently Asked Questions

You have two cones of identical mass and radius — one solid and one hollow. Which one offers more resistance to rotation about its vertical axis and why?

A solid cone is spun about a horizontal axis passing through its base (perpendicular to the height). If a hollow cone of the same size and mass is spun the same way, will their moments of inertia be the same?

Why is it easier to topple a hollow cone than a solid one resting on the same base?

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Moment of Inertia of a Cone

1.0Moment of Inertia of A Hollow Cone

2.0Moment of Inertia of A Solid Cone

3.0Moment of Inertia of Solid and Hollow Cones (Various Axes)

Table of Contents

Join ALLEN!

Moment of Inertia of a Cone

1.0Moment of Inertia of A Hollow Cone

2.0Moment of Inertia of A Solid Cone

3.0Moment of Inertia of Solid and Hollow Cones (Various Axes)

Table of Contents