User:Prof McCarthy/parallel axis theorem

In physics, the parallel axis theorem or Huygens–Steiner theorem can be used to determine the second moment of area or the mass moment of inertia of a rigid body about any axis, given the body's moment of inertia about a parallel axis through the object's centre of mass and the perpendicular distance (r) between the axes.

Mass moment of inertia for planar dynamics

A rigid body undergoing planar movement is constrained so that all points trajectories line in planes parallel to a reference plane. The planar dynamics of a rigid body is defined by Newton's second law,

$\mathbf {F} =M\mathbf {A} ,\quad \mathbf {T} =I_{R}\alpha ,$

where A is the acceleration of the center of mass R and I_R is the mass moment of inertia of the body about the center of mass.

Recall that the center of mass R of the body is defined by the requirement that the relative position coordinates r-R of points in the volume V of the body weighted by density ρ(r""") sum to zero, that is,

$\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV=0.$

The moment of inertia I_S about an arbitrary reference point S in the body is defined by

$I_{S}=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {S} )\cdot (\mathbf {r} -\mathbf {S} )dV.$

$I_{z}=I_{cm}+mr^{2},\,$

where:

$I_{cm}\!$ is the moment of inertia of the object about an axis passing through its centre of mass;

$m\!$ is the object's mass;

$r\!$ is the perpendicular distance between the axis of rotation and the axis that would pass through the centre of mass.

This rule can be applied with the stretch rule and perpendicular axis theorem to find moments of inertia for a variety of shapes.

Parallel axes rule for area moment of inertia

The parallel axes rule also applies to the second moment of area (area moment of inertia) for a plane region D:

$I_{z}=I_{x}+Ar^{2},\,$

where:

$I_{z}\!$ is the area moment of inertia of D relative to the parallel axis;

$I_{x}\!$ is the area moment of inertia of D relative to its centroid;

$A\!$ is the area of the plane region D;

$r\!$ is the distance from the new axis z to the centroid of the plane region D.

Note: The centroid of D coincides with the centre of gravity (CG) of a physical plate with the same shape that has uniform density.

Proof

We may assume, without loss of generality, that in a Cartesian coordinate system the perpendicular distance between the axes lies along the x-axis and that the centre of mass lies at the origin. The moment of inertia relative to the z-axis, passing through the centre of mass, is:

$I_{cm}=\int {(x^{2}+y^{2})}dm$

The moment of inertia relative to the new axis, perpendicular distance r along the x-axis from the centre of mass, is:

$I_{z}=\int {((x-r)^{2}+y^{2})}dm$

If we expand the brackets, we get:

$I_{z}=\int {(x^{2}+y^{2})}dm+r^{2}\int dm-2r\int {x}dm$

The first term is I_cm, the second term becomes mr², and the final term is zero since the origin is at the centre of mass. So, this expression becomes:

$I_{z}=I_{cm}+mr^{2}\,$

In classical mechanics

In classical mechanics, the Parallel axis theorem (also known as Huygens-Steiner theorem) can be generalized to calculate a new inertia tensor J_ij from an inertia tensor about a centre of mass I_ij when the pivot point is a displacement a from the centre of mass:

$J_{ij}=I_{ij}+m(\|{\boldsymbol {a}}\|^{2}\delta _{ij}-a_{i}a_{j})\!$

where

${\boldsymbol {a}}=a_{1}{\boldsymbol {\hat {x}}}+a_{2}{\boldsymbol {\hat {y}}}+a_{3}{\boldsymbol {\hat {z}}}\!$

is the displacement vector from the centre of mass to the new axis, and

$\delta _{ij}\!$

is the Kronecker delta.

We can see that, for diagonal elements (when i = j), displacements perpendicular to the axis of rotation results in the above simplified version of the parallel axis theorem.

See also

• Perpendicular axis theorem
• Stretch rule
• Jakob Steiner
• Christiaan Huygens

References

• Parallel axis theorem