# User:Prof McCarthy/parallel axis theorem

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In physics, the parallel axis theorem or HuygensSteiner 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.

## Polar moment of inertia for planar dynamics

The mass properties of a rigid body that is constrained to move parallel to a plane are defined by its center of mass R=(x, y) in this plane, and its polar moment of inertia IR about this point. The parallel axis theorem provides a convenient relationship between the moment of inertia IS and the moment of inertia about the center of mass IR.

Recall that the center of mass R has the property

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

where r is integrated over the volume V of the body. The polar moment of inertia of a body undergoing planar movement can be computed relative to any reference point S,

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

where S is constant and r is integrated over the volume V.

In order to obtain the moment of inertia IS in terms of the moment of inertia IR, introduce the vector d from S to the center of mass R,

${\displaystyle I_{S}=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} +\mathbf {d} )\cdot (\mathbf {r} -\mathbf {R} +\mathbf {d} )dV=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )\cdot (\mathbf {r} -\mathbf {R} )dV+2\mathbf {d} \cdot (\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV)+(\int _{V}\rho (\mathbf {r} )dV)\mathbf {d} \cdot \mathbf {d} ,}$

The first term is the moment of inertia IR, the second term is zero by definition of the center of mass, and the last term is the total mass of the body times the square magnitude of the vector d. Thus,

${\displaystyle I_{S}=I_{R}+Md^{2},}$

which is known as the parallel axis theorem.

Let a rigid assembly of points Pi be constrained to move in trajectories parallel to a reference plane.

If the movement of a rigid body is constrained so that all if its point trajectories line in planes parallel to a reference plane, the body is said to undergo planar movement. The dynamics of a rigid body in planar movement is defined by Newton's second laws restricted to the plane,

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

where the resultant force has two components, F=(Fx, Fy), and the resultant toque T is a scalar. Similarly, A=(ax, ay) is the acceleration of the center of mass R=(x, y) and the scalar IR 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,

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

For planar movement of a rigid body, the moment of inertia IS about an arbitrary reference point S is defined by

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

where S is constant and r is integrated over the volume V.

The parallel axis theorem provides a convenient relationship between the moment of inertia IS and the moment of inertia about the center of mass IR. Let d be the vector from S to the center of mass, so S=R-d, then the moment of inertia about S becomes

${\displaystyle I_{S}=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} +\mathbf {d} )\cdot (\mathbf {r} -\mathbf {R} +\mathbf {d} )dV=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )\cdot (\mathbf {r} -\mathbf {R} )dV+2\mathbf {d} \cdot (\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV)+(\int _{V}\rho (\mathbf {r} )dV)\mathbf {d} \cdot \mathbf {d} ,}$

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

where:

${\displaystyle I_{cm}\!}$ is the moment of inertia of the object about an axis passing through its centre of mass;
${\displaystyle m\!}$ is the object's mass;
${\displaystyle r\!}$ 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:

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

where:

${\displaystyle I_{z}\!}$ is the area moment of inertia of D relative to the parallel axis;
${\displaystyle I_{x}\!}$ is the area moment of inertia of D relative to its centroid;
${\displaystyle A\!}$ is the area of the plane region D;
${\displaystyle r\!}$ 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:

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

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

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

If we expand the brackets, we get:

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

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

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

## 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 Jij from an inertia tensor about a centre of mass Iij when the pivot point is a displacement a from the centre of mass:

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

where

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

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

${\displaystyle \delta _{ij}\!}$

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.