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| In [[physics]], the '''parallel axis theorem''', also known as '''Huygens–Steiner 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 (geometry)|parallel]] axis through the object's [[centre of mass]] and the [[perpendicular]] [[distance]] between the axes.
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| ==Mass moment of inertia==
| | In today's community everyone is super busy! Then you are you are one? Because of so many demands made from you, you could be unable to take the time you would probably normally love to on your own physical appearance. Begin to see the ideas presented listed below to give assist you to enhance your accurate elegance!<br><br>If you do not suffer from serious pimples, you need to be certain to use a moisturizing lotion that includes emollients, that can help your skin layer to soak up moisture content from the setting. Other components, for example humectants, may actually bring in moisture content to the epidermis. Dry skin would make use of a cream by using a hefty, rich and creamy uniformity.<br><br>As you get old, exfoliation will become more and more crucial that you the skin. Make use of a glycolic acid-wealthy skin cream, skin wash, or even a retinoid gel to slough away from the best layer pf the dead skin cells as well as to expose the new, radiant new epidermis tissue below. This can be achieved 3 to 4 periods a week for the best outcome.<br><br>Use a handful of tablespoons of sugary almond oils to your cozy bathtub to get a soothing handle that may depart even the roughest, driest pores and skin feeling soft and smooth. You can even use it moderately to extremely dried up and destroyed locks - but only around the in . or more on top of the ends.<br><br>When your eyeliner has a tendency to smear and crease, consider dabbing somewhat of a similar coloured powdered eyeshadow more than the top of the it using a smooth 100 % cotton swab. This will help to keep the eyeliner in position and make your vision makeup products go longer prior to it will require a touch up.<br><br>A fluffy brush as well as a dusting of matte powder are all it takes to freshen up your makeup if you have to change from daytime to night. Then add shimmery natural powder for your cheeks to emphasize your cheek your bones.<br><br>Work with a serious conditioner at least once weekly for additional smooth and healthy hair. Select one time of every week for taking a bath and look at a newspaper or pay attention to audio whilst the strong conditioner soaks into the your hair just before rinsing. A lot of hair merchandise outlines feature a corresponding deeply conditioner.<br><br>Curl eyelashes well before getting mascara on. Your curled eyelashes will not likely only appearance longer than they may be, but the complete area of your vision could be aesthetically lifted and look much brighter. Meticulously press and hold the device on the base of your lashes. Gradually transfer towards the leading of the lashes, in the smooth movement, then press again. This will produce a attractive and organic seeking curve.<br><br>Agree to drinking adequate water every single day. You have to ingest 8 to 10 cups a day for maximum attractiveness gain. Being hydrated will allow you to appear younger, as well as enable you to keep in much better all round health and fitness. Will not ignore yourself and you will probably have great results from your attractiveness program.<br><br>To make your tooth appearance whiter, use lipstick with cool, blue undertones. Lipsticks with warm, orange-based undertones emphasize natural yellowish colour of your the teeth, which makes them look yellower. Lipsticks with amazing, light blue-dependent undertones, on the flip side, will make your teeth appearance brighter. For your very best influence, pick a red lip stick with azure undertones.<br><br>Will not forget your hands must be pampered as well. Hands are often disregarded in elegance treatments. That is why it is stated, in order to know someones age, check out their hands and wrists. Along with everyday treatment with lotion or skin cream, you ought to exfoliate both hands once per week.<br><br>Basic older toilet pieces of paper can be very valuable when you have a shine from doing all of your greatest movements dancing in the group! Remarkably sufficient, the roughness of affordable potty document is very rewarding being an oil blotter. Simply take a sq off and blot to on the greasy spots.<br><br>Hands cream can be quite a fantastic correct for any terrible hair day! Inside the cold of wintertime when stationary has you seeking electrical, rub a tiny amount of cream on your palms and lightly pat hair down. In the moist summer time, perform the identical about the stops of your your hair to tame frizz!<br><br>As you have seen, it is possible to enhance your splendor, regardless of the needs manufactured in your hectic daily life. With these tips, there are actually methods to enhance your beauty throughout the small amount of time you have available. Just take some time on your own! It can make a huge difference!<br><br>If you're ready to learn more information about [http://hdwallpaper.us/profile/vecagle machine tatouer] stop by our own web site. |
| [[File:Steiner.png|thumb|right|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.]]
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| Suppose a body of mass {{math|''m''}} is made to rotate about an axis {{math|''z''}} passing through the body's [[center of mass]]. The body has a moment of inertia {{math|''I''<sub>cm</sub>}} with respect to this axis.
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| 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|''r''}}, then the moment of inertia {{math|''I''}} with respect to the new axis is related to {{math|''I''<sub>cm</sub>}} by
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| :<math> I = I_\mathrm{cm} + mr^2.</math>
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| Explicitly, {{math|''r''}} is the perpendicular distance between the axes {{math|''z''}} and {{math|''z′''}}.
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| 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.
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| [[Image:Parallelaxes-1.png|thumb|right|Parallel axes rule for area moment of inertia]]
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| ===Proof===
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| 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 is
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| :<math>I_\mathrm{cm} = \int (x^2 + y^2) \, dm.</math>
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| The moment of inertia relative to the axis {{math|''z′''}}, which is a perpendicular distance {{math|''r''}} along the ''x''-axis from the centre of mass, is
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| :<math>I = \int \left[(x - r)^2 + y^2\right] \, dm</math>
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| Expanding the brackets yields
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| :<math>I = \int (x^2 + y^2) \, dm + r^2 \int dm - 2r\int x\, dm.</math>
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| The first term is {{math|''I''<sub>cm</sub>}}, the second term becomes {{math|''mr''<sup>2</sup>}}, and the final term is zero since the origin of the coordinates is at the centre of mass. So, the equation becomes:
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| :<math> I = I_\mathrm{cm} + mr^2.</math>
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| === Tensor generalization ===
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| The parallel axis theorem can be generalized to calculations involving the [[Moment of inertia#The inertia tensor|inertia tensor]]. Let {{math|''I<sub>ij</sub>''}} denote the inertia tensor of a body as calculated at the center of mass. Then the inertia tensor {{math|''J<sub>ij</sub>''}} as calculated relative to a new point is
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| :<math>J_{ij}=I_{ij} + m\left(|\mathbf{R}|^2 \delta_{ij}-R_i R_j\right),</math>
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| where <math>\mathbf{R}=R_1\mathbf{\hat{x}}+R_2\mathbf{\hat{y}}+R_3\mathbf{\hat{z}}\!</math> is the displacement vector from the centre of mass to the new point, and {{math|δ<sub>''ij''</sub>}} is the [[Kronecker delta]].
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| 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.
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| The generalized version of the parallel axis theorem can be expressed in [[Component-free treatment of tensors|coordinate-free notation]] as
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| :<math> \mathbf{J} = \mathbf{I} + m \left[\left(\mathbf{R} \cdot \mathbf{R}\right) \mathbf{E}_{3} - \mathbf{R} \otimes \mathbf{R} \right],</math>
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| where '''E'''<sub>3</sub> is the {{nobr|3 × 3}} [[identity matrix]] and <math>\otimes</math> is the [[outer product]].
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| ==Area moment of inertia==
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| The parallel axes rule also applies to the [[second moment of area]] (area moment of inertia) for a plane region ''D'':
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| :<math>I_z = I_x + Ar^2,</math>
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| where {{math|''I<sub>z</sub>''}} is the area moment of inertia of ''D'' relative to the parallel axis, {{math|''I<sub>x</sub>''}} 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.
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| ==Polar moment of inertia for planar dynamics==
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| [[File:Steiners sats.PNG|thumb|right|Polar moment of inertia of a body around a point can be determined from its polar moment of inertia around the center of mass.]]
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| 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''<sub>''R''</sub> 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<sub>S</sub> around an arbitrary point '''S''' and the moment of inertia I<sub>R</sub> about the center of mass '''R'''.
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| Recall that the center of mass '''R''' has the property
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| :<math> \int_V \rho(\mathbf{r}) (\mathbf{r}-\mathbf{R}) \, dV=0, </math>
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| 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''',
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| : <math> I_S = \int_V \rho(\mathbf{r}) (\mathbf{r}-\mathbf{S})\cdot (\mathbf{r}-\mathbf{S}) \, dV,</math>
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| where '''S''' is constant and '''r''' is integrated over the volume ''V''.
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| In order to obtain the moment of inertia ''I''<sub>''S''</sub> in terms of the moment of inertia ''I''<sub>''R''</sub>, introduce the vector '''d''' from '''S''' to the center of mass '''R''',
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| : <math>
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| \begin{align}
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| I_S & = \int_V \rho(\mathbf{r}) (\mathbf{r}-\mathbf{R}+\mathbf{d})\cdot (\mathbf{r}-\mathbf{R}+\mathbf{d}) \, dV \\
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| & = \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}.
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| \end{align}
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| </math>
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| The first term is the moment of inertia ''I''<sub>''R''</sub>, 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,
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| :<math> I_S = I_R + Md^2, \, </math>
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| which is known as the parallel axis theorem.<ref>{{Citation |first=Burton |last=Paul |year=1979 |title=Kinematics and Dynamics of Planar Machinery |publisher=Prentice Hall |isbn=978-0-13-516062-6 |doi= }}</ref>
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| ==Moment of inertia matrix==
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| 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.
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| Consider the inertia matrix [I<sub>S</sub>] obtained for a rigid system of particles measured relative to a reference point '''S''', given by
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| :<math> [I_S] = -\sum_{i=1}^n m_i[r_i-S][r_i-S],</math>
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| where '''r'''<sub>''i''</sub> defines the position of particle ''P''<sub>''i''</sub>, ''i'' = 1, ..., ''n''. Recall that [''r''<sub>''i''</sub> − ''S''] is the skew-symmetric matrix that performs the cross product,
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| :<math> [r_i -S]\mathbf{y} = (\mathbf{r}_i - \mathbf{S})\times \mathbf{y},</math>
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| for an arbitrary vector '''y'''.
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| Let '''R''' be the center of mass of the rigid system, then
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| :<math> \mathbf{R} = (\mathbf{R}-\mathbf{S}) + \mathbf{S} = \mathbf{d} + \mathbf{S},</math>
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| where '''d''' is the vector from the reference point '''S''' to the center of mass '''R'''. Use this equation to compute the inertia matrix,
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| :<math> [I_S] = -\sum_{i=1}^n m_i[r_i- R + d][r_i - R+ d].</math>
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| Expand this equation to obtain
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| : <math> [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].</math>
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| The first term is the inertia matrix [''I''<sub>''R''</sub>] relative to the center of mass. The second and third terms are zero by definition of the center of mass '''R''',
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| :<math> \sum_{i=1}^n m_i(\mathbf{r}_i -\mathbf{R}) = 0.</math>
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| And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix [''d''] constructed from '''d'''.
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| The result is the parallel axis theorem,
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| :<math> [I_S] = [I_R] - M[d]^2,</math>
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| where '''d''' is the vector from the reference point '''S''' to the center of mass '''R'''.<ref>T. R. Kane and D. A. Levinson, [http://www.amazon.com/Dynamics-Theory-Applications-Mechanical-Engineering/dp/0070378460 Dynamics, Theory and Applications], McGraw-Hill, NY, 2005.</ref>
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| ===Identities for a skew-symmetric matrix===
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| In order to compare formulations of the parallel axis theorem using skew-symmetric matrices and the tensor formulation, the following identities are useful.
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| Let [''R''] be the skew symmetric matrix associated with the position vector '''R''' = (''x'', ''y'', ''z''), then the product in the inertia matrix becomes
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| :<math> -[R][R]= -\begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}^2 = \begin{bmatrix}
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| y^2+z^2 & -xy & -xz \\ -y x & x^2+z^2 & -yz \\ -zx & -zy & x^2+y^2 \end{bmatrix}.</math>
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| This product can be computed using the matrix formed by the outer product ['''R''' '''R'''<sup>T</sup>] using the identify
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| :<math> -[R]^2 = |\mathbf{R}|^2[E_3] -[\mathbf{R}\mathbf{R}^T]=
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| \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},</math>
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| where [''E''<sub>3</sub>] is the 3 × 3 identify matrix.
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| Also notice, that
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| :<math> |\mathbf{R}|^2 = \mathbf{R}\cdot\mathbf{R} =\operatorname{tr}[\mathbf{R}\mathbf{R}^T],</math>
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| where tr denotes the sum of the diagonal elements of the outer product matrix, known as its trace.
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| ==See also==
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| * [[Moment of inertia]]
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| * [[Perpendicular axis theorem]]
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| * [[Stretch rule]]
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| * [[Jakob Steiner]]
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| * [[Christiaan Huygens]]
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| * [[Rigid body dynamics]]
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| ==References==
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| {{reflist}}
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| ==External links==
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| *[http://scienceworld.wolfram.com/physics/ParallelAxisTheorem.html Parallel axis theorem]
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| [[Category:Mechanics]]
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| [[Category:Physics theorems]]
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| [[fr:Moment d'inertie#Théorème de transport (ou théorème d'Huygens ou théorème de Steiner)]]
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