Statics/Moment of Inertia (contents)

The moment of inertia can be defined as the second moment about an axis and is usually designated the symbol I. The moment of inertia is very useful in solving a number of problems in mechanics. For example, the moment of inertia can be used to calculate angular momentum, and angular energy. Moment of inertia is also important in beam design.

Shape moment of inertia for flat shapes

The area moment of inertia takes only shape into account, not mass.

It can be used to calculate the moment of inertia of a flat shape about the x or y axis when I is only important at one cross-section.

$I_{x}=\int y^{2}\,dA$

$I_{y}=\int x^{2}\,dA$

$I_{z}=\int r^{2}\,dA$

Because, for flat shapes, $r^{2}=x^{2}+y^{2}$ the following is true $I_{z}=I_{x}+I_{y}$

The shape moment of inertia of the cross-section of a beam is used in Structural Engineering in order to find the stress and deflection of the beam.

Shape moment of inertia for 3D shapes

The moment of inertia I=∫r²dm for a hoop, disk, cylinder, box, plate, rod, and spherical shell or solid can be found from this figure.

For more shapes see Mass Moments Of Inertia Of Common Geometric Shapes.

Mass moment of inertia

The mass moment of inertia takes mass into account. The mass moment of inertia of a point mass about a reference axis is equal to mass multiplied by the square of the distance from that point mass to the reference axis:

$I_{pointmass}=r^{2}m\,$

The metric units are kg*m^2.

The mass moment of inertia of any body of mass rotating around any axis is equal to the sum of the mass moment of inertia of each of the particles of that body:

$I_{m}=\sum r_{i}^{2}m_{i}$

Rather than adding up each particle individually, sometimes we can take a mathematical shortcut by integrating over all the particles:

$I_{m}=\int r^{2}\,dm$

Radius of gyration

The radius of gyration is the radius at which you could concentrate the entire mass to make the moment of inertia equal to the actual moment of inertia. If the mass of an object was 2kg, and the moment of inertia was $18kg*m^{2}$, then the radius of gyration would be 3m. In other words, if all of the mass was concentrated at a distance of 2m from the axis, then the moment of inertia would still be $18kg*m^{2}$. Radius of gyration is represented with a k.

$k={\sqrt {I/m}}\,$

The formula for the area radius of gyration replaces the mass with area.

$k={\sqrt {I/A}}\,$

Parallel axis theorem

x runs through the center of mass, while x' is parallel to x

If the moment of inertia is known about an axis that runs through the center of mass, then the moment of inertia about any parallel axis is given by,

$I_{parallel-axis}=I_{center\,of\,mass}+md^{2}$

where d is the distance between the two axis of rotation.

further reading

• Mechanics/Rigid Bodies