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A point mass does not have a moment of inertia around its own axis, but using the parallel axis theorem a moment of inertia around a distant axis of rotation is achieved. d 2 Also, a point mass m at the end of a rod of length r has this same moment of inertia and the value r is called the radius of gyration. y (24.8.1) T rot = 1 2 ( I 1 1 2 + I 2 2 2 + I 3 3 2) These axes, with respect to which the inertia tensor is diagonal, are called the principal axes of inertia, the moments . It is easy to see that the general case about any of the sphere's body axes, x,y or z is taken care of by the inertia tensor with the moments of inertia being, I 0 It should not be confused with the second moment of area, which is used in beam calculations. The square bracket term in $(13.3.9)$ is called the moment of inertia tensor, $\mathbf{I}$, which is usually referred to as the inertia tensor, \[I_{ij} \equiv \sum^{N}_{\alpha} m_{\alpha} \left[ \delta_{ij} \left( \sum^3_k x^2_{\alpha , k} \right) x_{\alpha , i} x_{\alpha , j} \right] \label{13.12}\]. 2 r m 10 The above formula is for the xy plane passing through the center of mass, which coincides with the geometric center of the cylinder. This expression assumes that the rod is an infinitely thin (but rigid) wire. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, it is the rotational analogue to mass (which determines an object's resistance to linear acceleration). 2 r i x 3 , m i y r Furthermore, because of the symmetry of the sphere, each principal moment is the same, so the moment of inertia of the sphere taken about any diameter is . l 2 Variational Principles in Classical Mechanics (Cline), { "13.01:_Introduction_to_Rigid-body_Rotation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13.02:_Rigid-body_Coordinates" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13.03:_Rigid-body_Rotation_about_a_Body-Fixed_Point" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "13.04:_Inertia_Tensor" : "property get [Map 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Tensor, [ "article:topic", "asymmetric top", "spherical top", "authorname:dcline", "license:ccbyncsa", "showtoc:no", "symmetric top", "licenseversion:40", "source@http://classicalmechanics.lib.rochester.edu" ], https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FClassical_Mechanics%2FVariational_Principles_in_Classical_Mechanics_(Cline)%2F13%253A_Rigid-body_Rotation%2F13.10%253A_General_Properties_of_the_Inertia_Tensor, $ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}$ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$, 13.9: Perpendicular-axis Theorem for Plane Laminae, 13.11: Angular Momentum and Angular Velocity Vectors, Asymmetric Top: $I_1 \neq I_2 \neq I_3$, source@http://classicalmechanics.lib.rochester.edu. 4 Accessibility StatementFor more information contact us atinfo@libretexts.org. 13 I r 0 I I ) A uniform sphere, or a uniform cube, rotating about a point displaced from the center-of-mass also behave inertially like a symmetric top. d = I [ ( 1 0 . m ( ( For the moment of inertia about the z axis, we found, I I am looking for exact or even approximate formulas for the moment of inertia of a hollow spheroid (oblate and prolate.) = ( 2 r 3 Students have to keep in mind that we are talking about the moment of inertia of a solid sphere about its central axis above. 10 Details about the moment of inertia of a sphere. h 2 i ) y As usual, the Lagrangian L = T V where the potential energy V is a function of six variables in general, the center of mass location and the . About an axis passing through the tip: 2 Therefore, i , The mass moment of inertia is often also known as the rotational inertia, and sometimes as the angular mass. I d = A quick inspection of (1) reveals that the inertia tensor is symmetric, so the the products of inertia are, I i i As a result, the inertial properties of any body about a body-fixed point are equivalent to that of an ellipsoid that has the same three principal moments of inertia. {\displaystyle I={\begin{bmatrix}{\frac {1}{3}}ml^{2}&0&0\\0&0&0\\0&0&{\frac {1}{3}}ml^{2}\end{bmatrix}}}, I r h + 150 Diagonalization may be accomplished by an appropriate rotation of the axes in the body. When calculating moments of inertia, it is useful to remember that it is an additive function and exploit the parallel axis and perpendicular axis theorems. 2 0 + {\displaystyle I_{x}=I_{y}={\frac {1}{12}}m\left(3\left(r_{2}^{2}+r_{1}^{2}\right)+4h^{2}\right)}, I ( Components of $\mathbf{L}$ and $\boldsymbol{\omega}$ can be taken along the three body-fixed axes denoted by $i$. 12 0 ) The principal moments (eigenvalues) and principal axes (eigenvectors) are obtained as roots of the secular determinant and are real. Additionally, if we talk about the moment of inertia of the sphere about its axis on the surface it is expressed as; I = (7/5) MR 2 Moment Of Inertia Of Sphere Derivation y Summarizing the above discussion, the inertia tensor has the following properties. F.3 Spherical Coordinates Refer to Figures F-3 and F-4 x = r sin 8 cos , *g = r sin Spherical coordinates: dv =r sin d d d sin, X3 = r cos e rsin = tan* r= Vx} + x3 + x}, = cos - 1 Then, the diagonal moments of inertia of the inertia tensor are, \[\begin{align} s z = For the example of a solid sphere, we found that the moment of inertia of a Solid Sphere was, I + {\displaystyle I_{13}=I_{31}} {\displaystyle I_{x,\mathrm {solid} }=I_{y,\mathrm {solid} }=I_{z,\mathrm {solid} }={\frac {\phi ^{2}}{10}}ms^{2}\,\!} {\displaystyle I_{x}=I_{y}=m\left({\frac {3}{20}}r^{2}+{\frac {3}{80}}h^{2}\right)\,\! 20 It is important to understand this distinction and the more general case about an arbitrary axis is handled by the inertia tensor. 3 2 T = 1 2MV2 + 1 2Iikik. s m i l 3 I (spherical shell) = kg m 2. When r1 = r2, h {\displaystyle I_{11}=I_{22}={\frac {1}{4}}MR^{2}+{\frac {1}{12}}ML^{2}}, If you imagine a very dense pencil, where R is small and L is large, you might have an inertia tensor that looks like, I r m o Therefore subtracting these equations gives, \[\sum_i I_{mm}\omega_{im}\omega_{in} \sum_k I_{nn}\omega_{km}\omega_{kn} = 0 \], \[(I_{mm} I_{nn}) \sum_k \omega_{km} \omega_{kn} = 0 \], \[(I_{mm} I_{nn}) \boldsymbol{\omega}_m \cdot \boldsymbol{\omega}_n = 0 \label{13.53}\], \[\boldsymbol{\omega}_m \cdot \boldsymbol{\omega}_n = 0 \label{13.54}\]. It is essentially a reorientation of the orthogonal axis system. (moment of inertia tensor of solid of hemisphere)/ (semialgebraic description of hemisphere) Since we have chosen z as our axis of rotation, then z in formula (2) is the distance from dm (dV) to the z axis. 0 m z A spherical top is a body having three degenerate principal moments of inertia. r 12 0 {\displaystyle I_{x}=I_{y}={\frac {1}{4}}m(r_{1}^{2}+r_{2}^{2})}, I The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. r r {\displaystyle I_{11}=I_{22}=I_{33}=I}. I 0 I Li = 3 j Iijj. ] Suppose now that your body be symmetric with respect the plane y O z, then we can make an orthogonal change of variables ( x, y, z) ( x, y, z) : 0 3 r + 0 4 2 This page was last edited on 13 September 2020, at 02:35. The inertial properties of a body for rotation about a specific body-fixed location is defined completely by only three principal moments of inertia irrespective of the detailed shape of the body. 1 2 = We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 2 ( 3 = The moment of inertia tensor for a sphere is $$\mathbf I = \begin{pmatrix}\frac25MR^2&0&0 \\ 0&\frac25MR^2&0 \\ 0&0&\frac25MR^2\end{pmatrix}.$$ After rotation, due to the symmetry of the sphere, the MOI tensor is unchanged: $\mathbf I' = \mathbf I.$ In particular it still has three non-zero entries on the main diagonal. {\displaystyle I_{x}=I_{y}={\frac {\pi \rho h}{12}}\left(3(r_{2}^{4}-r_{1}^{4})+h^{2}(r_{2}^{2}-r_{1}^{2})\right)}, I For example, the inertia tensor for a cube is very different when the fixed point is at the center of mass compared with when the fixed point is at a corner of the cube. m {\displaystyle [kg/m^{2}]} 1 M x 3 (1) and the moment of inertia tensor is. 3 y 2 = z Typically this occurs when the mass density is constant, but in some cases the density can vary throughout the object as well. Figure 13.2. s To obtain the scalar moments of inertia I above, the tensor moment of inertia I is projected along some axis defined by a unit vector n according to the formula: where the dots indicate tensor contraction and the Einstein summation convention is used. 4.2: Inertia Tensor. , {\displaystyle I=\left[{\begin{matrix}10&0&0\\0&10&0\\0&0&100\end{matrix}}\right]}, https://en.wikiversity.org/w/index.php?title=PlanetPhysics/Inertia_Tensor&oldid=2206060, Creative Commons Attribution-ShareAlike License. 2 y Note that every fixed point in a body has a specific inertia tensor. ( 0 r {\displaystyle I={\begin{bmatrix}{\frac {1}{12}}m(3r^{2}+h^{2})&0&0\\0&{\frac {1}{12}}m(3r^{2}+h^{2})&0\\0&0&{\frac {1}{2}}mr^{2}\end{bmatrix}}}, I 3 / [ 10 That is, try: {\displaystyle I_{x}=I_{y}={\frac {I_{z}}{2}}\,} z The Rotational Inertia or moment of inertia of a solid sphere rotating about a diameter is . 1 offset by I 21 2 [4] Classical Mechanics - Moment of inertia of a uniform hollow cylinder, "Moment of Inertia J Calculation Formula", Tutorial on deriving moment of inertia for common shapes, https://en.wikipedia.org/w/index.php?title=List_of_moments_of_inertia&oldid=1152625557, A uniform annulus (disk with a concentric hole) of mass, Thick-walled cylindrical tube with open ends, of inner radius. 0 i m For a spherical top with three identical principal moment of inertia, the principal axes system can have any orientation with respect to the origin. M I semialgebraic description of hemisphere. I r 3 The components of the inertia tensor at a specified point depend on the orientation of the coordinate frame whose origin is located at the specified fixed point. I This page titled 13.10: General Properties of the Inertia Tensor is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. It doesn't look right. I For using this approach, the first thing we need to calculate is the kinetic energy of the body in an . 1 The moment of inertia of a sphere of uniform density and radius R is. i {\displaystyle I_{x}=I_{y}={\frac {1}{12}}m\left(3\left(r_{2}^{2}+r_{1}^{2}\right)+h^{2}\right)} ( s 2 i 2 I m i = m r y ) 22 A Consider a uniform solid cylinder of mass M, radius R, height h. The density is then. 2 ) i x \notag \end{align}\], while the off-diagonal products of inertia are, \[\begin{align} I_{yx} & = I_{xy} \equiv - \sum^N_{\alpha} m_{\alpha} [x_{\alpha} y_{\alpha}] \\[4pt] \notag I_{zx} & = I_{xz} \equiv - \sum^N_{\alpha} m_{\alpha} [x_{\alpha} z_{\alpha}] \\[4pt] \notag I_{zy} & = I_{yz} \equiv - \sum^N_{\alpha} m_{\alpha} [y_{\alpha} z_{\alpha}] \end{align}\], Note that the products of inertia are symmetric in that, The above notation for the inertia tensor allows the angular momentum \ref{13.12} to be written as, \[\begin{align} L_x & = I_{xx} \omega_x + I_{xy} \omega_y + I_{xz} \omega_z \\[4pt] \notag L_y & = I_{yx} \omega_x + I_{yy} \omega_y + I_{yz} \omega_z \\[4pt] \notag L_z & = I_{zx} \omega_x + I_{zy} \omega_y + I_{zz} \omega_z \end{align}\]. M 1 and the products of inertia equal to 0. ( + r = = The moment of inertia, otherwise known as the mass moment of inertia, angular mass, second moment of mass, or most accurately, rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis, akin to how mass determines the force needed for a desired acceleration. 3 i For a symmetric top with two identical principal moments of inertia, any orientation of two orthogonal axes perpendicular to the symmetry axis are satisfactory eigenvectors. 1 This page titled 13.4: Inertia Tensor is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. {\displaystyle I_{\mathrm {hollow} }={\frac {1}{12}}ms^{2}\,\!} 12 {\displaystyle I_{11}} Variational Principles in Classical Mechanics (Cline), { 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( + 100 t 3 #1 lilphy 28 0 Homework Statement hello, i want to calculate the inertia tensor of the combination of a point mass and a sphere in the object's frame, the center of mass is at the origin. Recitation 6 Notes: Moment of Inertia and Imbalance, Rotating Slender Rod Moment of Inertia Recall that Frame Axyz , the object coordinate system, is attached to (and rotates with) the rotating rigid body A is chosen to be EITHER xed in space (v. A = 0) OR the rigid body's center of mass (v. A = v. G) so. r {\displaystyle {\frac {r_{2}^{5}-r_{1}^{5}}{r_{2}^{3}-r_{1}^{3}}}={\frac {5}{3}}r_{2}^{2}} About an axis passing through the base: m 0 h Examples of an oblate spheroid are an orange, or a frisbee. 0 w 4 I {\displaystyle I_{22}} , and 4 2 y 11 , = 2 r + = x x 2 A less obvious consequence of the spherical symmetry is that any orientation of three mutually perpendicular axes about the center of mass of a uniform cube is an equally good principal axis system. 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