### Parallel axis theorem (20705 views - Calculations (Mech&Elec))

In physics, the parallel axis theorem, also known as Huygens–Steiner theorem, or just as Steiner's theorem, after Christiaan Huygens and Jakob Steiner, can be used to determine the mass moment of inertia or the second moment of area of a rigid body about any axis, given the body's moment of inertia about a parallel axis through the object's center of gravity and the perpendicular distance between the axes.
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## Parallel axis theorem

### Parallel axis theorem

In physics, the parallel axis theorem, also known as Huygens–Steiner theorem, or just as Steiner's theorem, after Christiaan Huygens and Jakob Steiner, can be used to determine the mass moment of inertia or the second moment of area of a rigid body about any axis, given the body's moment of inertia about a parallel axis through the object's center of gravity and the perpendicular distance between the axes.

## Mass moment of inertia

Suppose a body of mass m is made to rotate about an axis z passing through the body's centre of gravity. The body has a moment of inertia Icm with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis z′ which is parallel to the first axis and displaced from it by a distance d, then the moment of inertia I with respect to the new axis is related to Icm by

$I=I_{\mathrm {cm} }+md^{2}.$ Explicitly, d is the perpendicular distance between the axes z and 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.

### Derivation

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

$I_{\mathrm {cm} }=\int (x^{2}+y^{2})\,dm.$ The moment of inertia relative to the axis z′, which is a perpendicular distance d along the x-axis from the centre of mass, 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 is Icm and the second term becomes md2. The integral in the final term is the x-coordinate of the centre of mass, which is zero by construction. So, the equation becomes:

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

The parallel axis theorem can be generalized to calculations involving the inertia tensor. Let Iij denote the inertia tensor of a body as calculated at the centre of mass. Then the inertia tensor Jij as 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 centre of mass to the new point, and δij is the Kronecker delta.

For diagonal elements (when 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

$\mathbf {J} =\mathbf {I} +m\left[\left(\mathbf {R} \cdot \mathbf {R} \right)\mathbf {E} _{3}-\mathbf {R} \otimes \mathbf {R} \right],$ where E3 is the 3 × 3 identity matrix and $\otimes$ 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. 

## 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 Iz is the area moment of inertia of D relative to the parallel axis, Ix is the area moment of inertia of D relative to its centroid, A is the area of the plane region D, and r is the distance from the new axis 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.

## 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 = (xy) in this plane, and its polar moment of inertia IR around an axis through R that is perpendicular to the plane. The parallel axis theorem provides a convenient relationship between the moment of inertia IS around an arbitrary point S and the moment of inertia IR about the center of mass R.

Recall that the center of mass R has the property

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

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

{\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 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,

$I_{S}=I_{R}+Md^{2},\,$ 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 mass R and the inertia matrix relative to another point S. This relationship is called the parallel axis theorem.

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

$[I_{S}]=-\sum _{i=1}^{n}m_{i}[r_{i}-S][r_{i}-S],$ where ri defines the position of particle Pi, i = 1, ..., n. Recall that [ri − 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 vector y.

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

$\mathbf {R} =(\mathbf {R} -\mathbf {S} )+\mathbf {S} =\mathbf {d} +\mathbf {S} ,$ where d is the vector from the reference point S to the center of mass R. Use this equation to compute the inertia matrix,

$[I_{S}]=-\sum _{i=1}^{n}m_{i}[r_{i}-R+d][r_{i}-R+d].$ 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 [IR] relative to the center of mass. The second and third terms are zero by definition of the center of mass R,

$\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )=0.$ And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix [d] constructed 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.

### 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 vector R = (xyz), 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 [R RT] using the identify

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