Parallel axis theorem

File:Steiner.png

Suppose a body of mass {{math|m}} is rotated about an axis {{math|z}} passing through the body's center of mass. The body has a moment of inertia {{math|I_cm}} with respect to this axis.

The parallel axis theorem states that if the body is made to rotate instead about a new axis {{math|z′}}, which is parallel to the first axis and displaced from it by a distance {{math|d}}, then the moment of inertia {{math|I}} with respect to the new axis is related to {{math|I_cm}} by

:$I\; =\; I\_\backslash mathrm\{cm\}\; +\; md^2.$

Explicitly, {{math|d}} is the perpendicular distance between the axes {{math|z}} and {{math|z′}}.

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

Image:Parallelaxes-1.png

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 center of mass lies at the origin. The moment of inertia relative to the z-axis is then

:$I\_\backslash mathrm\{cm\}\; =\; \backslash int\; (x^2\; +\; y^2)\; \backslash ,\; dm.$

The moment of inertia relative to the axis {{math|z′}}, which is at a distance {{math|D}} from the center of mass along the x-axis, is

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

Expanding the brackets yields

:$I\; =\; \backslash int\; (x^2\; +\; y^2)\; \backslash ,\; dm\; +\; D^2\; \backslash int\; dm\; -\; 2D\backslash int\; x\backslash ,\; dm.$

The first term is {{math|I_cm}} and the second term becomes {{math|MD²}}. The integral in the final term is a multiple of the x-coordinate of the center of mass{{snd}}which is zero since the center of mass lies at the origin. So, the equation becomes:

:$I\; =\; I\_\backslash mathrm\{cm\}\; +\; MD^2.$

The parallel axis theorem can be generalized to calculations involving the inertia tensor.{{citation

| last = Abdulghany | first = A. R.

| date = October 2017

| doi = 10.1119/1.4994835

| issue = 10

| journal = American Journal of Physics

| pages = 791–795

| title = Generalization of parallel axis theorem for rotational inertia

| volume = 85| doi-access = free

| bibcode = 2017AmJPh..85..791A

}} Let {{math|I_ij}} denote the inertia tensor of a body as calculated at the center of mass. Then the inertia tensor {{math|J_ij}} as calculated relative to a new point is

:$J\_\{ij\}=I\_\{ij\}\; +\; m\backslash left(|\backslash mathbf\{R\}|^2\; \backslash delta\_\{ij\}-R\_i\; R\_j\backslash right),$

where $\backslash mathbf\{R\}=R\_1\backslash mathbf\{\backslash hat\{x\}\}+R\_2\backslash mathbf\{\backslash hat\{y\}\}+R\_3\backslash mathbf\{\backslash hat\{z\}\}\backslash !$ is the displacement vector from the center of mass to the new point, and {{math|δ_ij}} is the Kronecker delta.

For diagonal elements (when {{math|i {{=}} 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 of coordinate-free notation as

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

where E₃ is the {{nobr|3 × 3}} identity matrix and $\backslash otimes$ is the outer 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. In this generalization, the inertia tensor can be moved from being reckoned about any reference point $\backslash mathbf\{R\}\_\{ref\}$ to some final reference point $\backslash mathbf\{R\}\_F$ via the relational matrix $M$ as:

:$I\_\{F\}\; =\; I\_\backslash mathrm\{ref\}\; +\; m(M[\backslash mathbf\{R\},\backslash mathbf\{R\}]\; -\; 2M[\backslash mathbf\{R\},\backslash mathbf\{C\}])$

where $\backslash mathbf\{C\}$ is the vector from the initial reference point to the object's center of mass and $\backslash mathbf\{R\}$ is the vector from the initial reference point to the final reference point ($\backslash mathbf\{R\}\_F\; =\; \backslash mathbf\{R\}\_\{ref\}\; +\; \backslash mathbf\{R\}$). The relational matrix is given by

:

-1/2(r_x c_y + r_y c_x) & (r_x c_x + r_z c_z) & -1/2(r_y c_z + r_z c_y) \\

-1/2(r_x c_z + r_z c_x) & -1/2(r_y c_z + r_y c_z) & (r_x c_x + r_y c_y) \end{array}\right]

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 {{math|I_z}} is the area moment of inertia of D relative to the parallel axis, {{math|I_x}} is the area moment of inertia of D relative to its centroid, {{math|A}} is the area of the plane region D, and {{math|r}} is the distance from the new axis {{math|z}} to the centroid of the plane region D. The centroid of D coincides with the centre of gravity of a physical plate with the same shape that has uniform density.

File:Steiners sats.PNG

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 I_R around an axis through R that is perpendicular to the plane. The parallel axis theorem provides a convenient relationship between the moment of inertia I_S around an arbitrary point S and the moment of inertia I_R about the center of mass R.

Recall that the center of mass R has the property

:$\backslash int\_V\; \backslash rho(\backslash mathbf\{r\})\; (\backslash mathbf\{r\}-\backslash mathbf\{R\})\; \backslash ,\; dV=0,$

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,

: $I\_S\; =\; \backslash int\_V\; \backslash rho(\backslash mathbf\{r\})\; (\backslash mathbf\{r\}-\backslash mathbf\{S\})\backslash cdot\; (\backslash mathbf\{r\}-\backslash mathbf\{S\})\; \backslash ,\; dV,$

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

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

: $$

\begin{align}

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{align}

The first term is the moment of inertia I_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 vector d. Thus,

:$I\_S\; =\; I\_R\; +\; Md^2,\; \backslash ,$

which is known as the parallel axis theorem.{{Citation |first=Burton |last=Paul |year=1979 |title=Kinematics and Dynamics of Planar Machinery |publisher=Prentice Hall |isbn=978-0-13-516062-6 }}

The inertia matrix of a rigid system of particles depends on the choice of the reference point.{{citation|first1=T. R.|last1=Kane|first2=D. A.|last2=Levinson|title=Dynamics, Theory and Applications|publisher= McGraw-Hill, New York|year= 2005}} There is a useful relationship between the inertia matrix relative to the center of mass R and the inertia matrix relative to another point S. 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 point S, given by

:$[I\_S]\; =\; -\backslash sum\_\{i=1\}^n\; m\_i[r\_i-S][r\_i-S],$

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

:$[r\_i\; -S]\backslash mathbf\{y\}\; =\; (\backslash mathbf\{r\}\_i\; -\; \backslash mathbf\{S\})\backslash times\; \backslash mathbf\{y\},$

for an arbitrary vector y.

Let R be the center of mass of the rigid system, then

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

where d is the vector from the reference point S to the center of mass R. Use this equation to compute the inertia matrix,

:$[I\_S]\; =\; -\backslash sum\_\{i=1\}^n\; m\_i[r\_i-\; R\; +\; d][r\_i\; -\; R+\; d].$

Expand this equation to obtain

: $[I\_S]\; =\; \backslash left(-\backslash sum\_\{i=1\}^n\; m\_i\; [r\_i\; -\; R][r\_i\; -\; R]\backslash right)\; +\; \backslash left(-\backslash sum\_\{i=1\}^n\; m\_i[r\_i\; -\; R]\backslash right)[d]\; +\; [d]\backslash left(-\backslash sum\_\{i=1\}^n\; m\_i[r\_i\; -\; R]\backslash right)\; +\; \backslash left(-\backslash sum\_\{i=1\}^n\; m\_i\backslash right)[d][d].$

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 mass R,

:$\backslash sum\_\{i=1\}^n\; m\_i(\backslash mathbf\{r\}\_i\; -\backslash mathbf\{R\})\; =\; 0.$

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

The result is the parallel axis theorem,

:$[I\_S]\; =\; [I\_R]\; -\; M[d]^2,$

where d is the vector from the reference point S to the center of mass R.

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 vector R = (x, y, z), then the product in the inertia matrix becomes

:

y^2+z^2 & -xy & -xz \\ -y x & 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 [R R^T] using the identity

:$-[R]^2\; =\; |\backslash mathbf\{R\}|^2[E\_3]\; -[\backslash mathbf\{R\}\backslash mathbf\{R\}^T]=$

\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

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

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

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

{{reflist}}

{{commons category|Steiner's parallel axis theorem}}

• [http://scienceworld.wolfram.com/physics/ParallelAxisTheorem.html Parallel axis theorem]
• [http://mathworld.wolfram.com/MomentofInertia.html Moment of inertia tensor]
• [https://www.youtube.com/watch?v=mFVz7iCc45I Video about the inertia tensor]

Category:Mechanics

Category:Physics theorems

Category:Moment (physics)

fr:Moment d'inertie#Théorème de transport (ou théorème d'Huygens-Steiner)