Parallel axis theorem

"Steiner's theorem" redirects here. Not to be confused withSteiner's theorem (geometry).

Theparallel axis theorem,also known asHuygens–Steiner theorem,or just asSteiner's theorem,named afterChristiaan HuygensandJakob Steiner,can be used to determine themoment of inertiaor thesecond moment of areaof arigid bodyabout any axis, given the body's moment of inertia about aparallelaxis through the object'scenter of gravityand theperpendiculardistancebetween the axes.

Mass moment of inertia

The mass moment of inertia of a body around an axis can be determined from the mass moment of inertia around a parallel axis through the center of mass.

Suppose a body of massmis rotated about an axiszpassing through the body'scenter of mass.The body has a moment of inertiaI_cmwith respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axisz′,which is parallel to the first axis and displaced from it by a distanced,then the moment of inertiaIwith respect to the new axis is related toI_cmby

$I=I_{\mathrm {cm} }+md^{2}.$

Explicitly,dis the perpendicular distance between the axeszandz′.

The parallel axis theorem can be applied with thestretch ruleandperpendicular axis theoremto find moments of inertia for a variety of shapes.

Parallel axes rule for area moment of inertia

Derivation

We may assume, without loss of generality, that in aCartesian coordinate systemthe perpendicular distance between the axes lies along thex-axis and that the center of mass lies at the origin. The moment of inertia relative to thez-axis is then

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

The moment of inertia relative to the axisz′,which is at a distanceDfrom the center of mass along thex-axis, is

$I=\int \left[(x-D)^{2}+y^{2}\right]\,dm.$

Expanding the brackets yields

$I=\int (x^{2}+y^{2})\,dm+D^{2}\int dm-2D\int x\,dm.$

The first term isI_cmand the second term becomesMD².The integral in the final term is a multiple of the x-coordinate of thecenter of mass– which is zero since the center of mass lies at the origin. So, the equation becomes:

$I=I_{\mathrm {cm} }+MD^{2}.$

Tensor generalization

The parallel axis theorem can be generalized to calculations involving theinertia tensor.LetI_ijdenote the inertia tensor of a body as calculated at the center of mass. Then the inertia tensorJ_ijas calculated relative to a new point is

$J_{ij}=I_{ij}+m\left(|\mathbf {R} |^{2}\delta _{ij}-R_{i}R_{j}\right),$

where$\mathbf {R} =R_{1}\mathbf {\hat {x}} +R_{2}\mathbf {\hat {y}} +R_{3}\mathbf {\hat {z}} \!$is the displacement vector from the center of mass to the new point, andδ_ijis theKronecker delta.

For diagonal elements (wheni=j), displacements perpendicular to the axis of rotation results in the above simplified version of the parallel axis theorem.

The generalized version of the parallel axis theorem can be expressed in the form ofcoordinate-free notationas

$\mathbf {J} =\mathbf {I} +m\left[\left(\mathbf {R} \cdot \mathbf {R} \right)\mathbf {E} _{3}-\mathbf {R} \otimes \mathbf {R} \right],$

whereE₃is the3 × 3identity matrixand$\otimes$is theouter product.

Further generalization of the parallel axis theorem gives the inertia tensor about any set of orthogonal axes parallel to the reference set of axes x, y and z, associated with the reference inertia tensor, whether or not they pass through the center of mass.

Second moment of area

The parallel axes rule also applies to thesecond moment of area(area moment of inertia) for a plane regionD:

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

whereI_zis the area moment of inertia ofDrelative to the parallel axis,I_xis the area moment of inertia ofDrelative to itscentroid,Ais the area of the plane regionD,andris the distance from the new axiszto thecentroidof the plane regionD.ThecentroidofDcoincides with thecentre of gravityof a physical plate with the same shape that has uniform density.

Polar moment of inertia for planar dynamics

Polar moment of inertia of a body around a point can be determined from its polar moment of inertia around the center of mass.

The mass properties of a rigid body that is constrained to move parallel to a plane are defined by its center of massR= (x,y) in this plane, and its polar moment of inertiaI_Raround an axis throughRthat is perpendicular to the plane. The parallel axis theorem provides a convenient relationship between the moment of inertia I_Saround an arbitrary pointSand the moment of inertia I_Rabout the center of massR.

Recall that the center of massRhas the property

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

whereris integrated over the volumeVof the body. The polar moment of inertia of a body undergoing planar movement can be computed relative to any reference pointS,

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

whereSis constant andris integrated over the volumeV.

In order to obtain the moment of inertiaI_Sin terms of the moment of inertiaI_R,introduce the vectordfromSto the center of massR,

${\begin{aligned}I_{S}&=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} +\mathbf {d} )\cdot (\mathbf {r} -\mathbf {R} +\mathbf {d} )\,dV\\&=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )\cdot (\mathbf {r} -\mathbf {R} )dV+2\mathbf {d} \cdot \left(\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )\,dV\right)+\left(\int _{V}\rho (\mathbf {r} )\,dV\right)\mathbf {d} \cdot \mathbf {d}.\end{aligned}}$

The first term is the moment of inertiaI_R,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 vectord.Thus,

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

which is known as the parallel axis theorem.

Moment of inertia matrix

The inertia matrix of a rigid system of particles depends on the choice of the reference point.There is a useful relationship between the inertia matrix relative to the center of massRand the inertia matrix relative to another pointS.This relationship is called the parallel axis theorem.

Consider the inertia matrix [I_S] obtained for a rigid system of particles measured relative to a reference pointS,given by

$[I_{S}]=-\sum _{i=1}^{n}m_{i}[r_{i}-S][r_{i}-S],$

wherer_idefines the position of particleP_i,i= 1,...,n.Recall that [r_i−S] is the skew-symmetric matrix that performs the cross product,

$[r_{i}-S]\mathbf {y} =(\mathbf {r} _{i}-\mathbf {S} )\times \mathbf {y},$

for an arbitrary vectory.

LetRbe the center of mass of the rigid system, then

$\mathbf {R} =(\mathbf {R} -\mathbf {S} )+\mathbf {S} =\mathbf {d} +\mathbf {S},$

wheredis the vector from the reference pointSto the center of massR.Use this equation to compute the inertia matrix,

$[I_{S}]=-\sum _{i=1}^{n}m_{i}[r_{i}-R+d][r_{i}-R+d].$

Expand this equation to obtain

$[I_{S}]=\left(-\sum _{i=1}^{n}m_{i}[r_{i}-R][r_{i}-R]\right)+\left(-\sum _{i=1}^{n}m_{i}[r_{i}-R]\right)[d]+[d]\left(-\sum _{i=1}^{n}m_{i}[r_{i}-R]\right)+\left(-\sum _{i=1}^{n}m_{i}\right)[d][d].$

The first term is the inertia matrix [I_R] relative to the center of mass. The second and third terms are zero by definition of the center of massR,

$\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )=0.$

And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix [d] constructed fromd.

The result is the parallel axis theorem,

$[I_{S}]=[I_{R}]-M[d]^{2},$

wheredis the vector from the reference pointSto the center of massR.

Identities for a skew-symmetric matrix

In order to compare formulations of the parallel axis theorem using skew-symmetric matrices and the tensor formulation, the following identities are useful.

Let [R] be the skew symmetric matrix associated with the position vectorR= (x,y,z), then the product in the inertia matrix becomes

$-[R][R]=-{\begin{bmatrix}0&-z&y\\z&0&-x\\-y&x&0\end{bmatrix}}^{2}={\begin{bmatrix}y^{2}+z^{2}&-xy&-xz\\-yx&x^{2}+z^{2}&-yz\\-zx&-zy&x^{2}+y^{2}\end{bmatrix}}.$

This product can be computed using the matrix formed by the outer product [RR^T] using the identity

$-[R]^{2}=|\mathbf {R} |^{2}[E_{3}]-[\mathbf {R} \mathbf {R} ^{T}]={\begin{bmatrix}x^{2}+y^{2}+z^{2}&0&0\\0&x^{2}+y^{2}+z^{2}&0\\0&0&x^{2}+y^{2}+z^{2}\end{bmatrix}}-{\begin{bmatrix}x^{2}&xy&xz\\yx&y^{2}&yz\\zx&zy&z^{2}\end{bmatrix}},$

where [E₃] is the 3 × 3 identity matrix.

Also notice, that

$|\mathbf {R} |^{2}=\mathbf {R} \cdot \mathbf {R} =\operatorname {tr} [\mathbf {R} \mathbf {R} ^{T}],$

where tr denotes the sum of the diagonal elements of the outer product matrix, known as its trace.

See also

• Christiaan Huygens
• Jakob Steiner
• Moment of inertia
• Perpendicular axis theorem
• Rigid body dynamics
• Stretch rule

External links

• Parallel axis theorem
• Moment of inertia tensor
• Video about the inertia tensor