5 of the textbook The moment of inertia is related to the rotation of an object about an axi s For a particle with mass the moment of inertia is given by mr where is the distance from the particle to the axis If the density function xy of a lamina ID: 25782 Download Pdf

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5 of the textbook The moment of inertia is related to the rotation of an object about an axi s For a particle with mass the moment of inertia is given by mr where is the distance from the particle to the axis If the density function xy of a lamina

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Winter 2012 Math 255 Moment of Inertia 1 Deﬁnition The moment of inertia is explained in 16.5 of the textbook. The moment of inertia is related to the rotation of an object about an axi s. For a particle with mass , the moment of inertia is given by mr , where is the distance from the particle to the axis. If the density function x,y ) of a lamina occupying a region is given, the moment of inertia of the lamina about the -axis is ZZ x,y dA, and the moment of inertial of the lamina about the -axis is ZZ x,y dA. The moment of inertial about the origin (the polar moment of

i nertia) is ZZ x,y dA. Example Find the moments of inertia for the lamina which occupies the elliptic region that is bounded by ( x/ 2) = 1, and has the density function x,y ) = (= const.). Solution The boundary is expressed as = 1, where x/ 2. Using polar coordinates, we have cos and sin . Note that dA dydx = 2 dydu , and so dA is replaced by 2 rdrd ZZ ρdA ZZ dA = 2 sin θrdrd = 2 cos(2 dr = 2 sin(2

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ZZ ρdA ZZ dA = 8 cos θrdrd = 8 1+cos(2 dr = 8 sin(2 = 2 Note that the mass of the lamina is (2)1. The polar moment of inertia is obtained as ZZ ρdA (2 +1

2 Three Dimensions Let us calculate the moments of inertia of the following elli psoid with con- stant density x,y,z ) = = 1 (1) The moment of inertia about the -axis is calculated as ZZZ dxdydz ρabc sin sin cos sin φdrdθd ρabc 15 (2) Wecansimilarlycalculate and . Notethatthevolume oftheellipsoid is abc . We obtain the moments of inertia as , I , I (3) 3 Spinning Eggs As an application of moment of inertia, let us consider boile d eggs. If we rotate a hard-boiled egg on a table with its axis of symmetry h orizontal, this axis will rise from the horizontal to the vertical.

To analyze this phe- nomenon, we model a boiled egg as an axisymmetric body with ce nter of H. K. Moﬀatt and Y. Shimomura, Spinning eggs – a paradox resolved , Nature 416 (2002) 385–386.

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mass at the origin. Therefore, (4) Suppose the egg is rotating about the vertical axis with angu lar velocity ) = Ω( . Let ) be the angle between the major axis of the egg and the vertical axis. Let ) be the spin. In the homework (Problem 14.4.38), we learned the angular mo mentum ) = ). We can also write = ( mr , where the angular velocity . The angular momentum ) of the egg is

given using the principal axes as ) = Ωsin θ,I d dt ,I (5) and is given in the rotating frame with by ) = Ωcos )sin θ,I d dt ,I Ωsin cos (6) As we studied in Problem 14.4.38, the time derivative of the a ngular mo- mentum is the torque ). In the rotating frame, we have the following Euler’s angular-momentum equation. ∂t (7) Here, ) is determined by the friction between the egg and table. By solving Eq. (7), we can deduce the following relation. h, (8) where is a constant called the Jellett constant. This Jellett constant remains constant in the presence of

friction. (Energy is not conserved in the presence of friction.) The Jellett constant is proportional to , the height of the center of mass, and the angular speed Ω about the vertic al axis. As Ω reduces due to friction, the height will increase because is conserved. This is why a spinning egg will spontaneously ris e up from lying on its side to standing on its end. ToreadthepaperbyMoﬀattandShimomura, abitofphysicsisn eeded. But mathematics in the paper can be understood using the know ledge from Math 255. That is, you can now understand spinning eggs! J. H. Jellett, A

Treatise on the Theory of Friction (Macmillan, London, 1872). For example, see L. D. Landau and E. M. Lifshitz, Mechanics

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