Paradigms in Physics: Rotational Motion solutions to TRANSLATING TENSORS Worksheet 3 by Philip J. Siemens TOPICS page A. Checking previous exercises 2 1. Brick 2. Cage B. Translating the CageX inertial tensor 4 1. From corner of cage to center of mass 2. From center of cage to center of mass C. Numerical values © 2003 Philip J. Siemens 9 version February 11, 2003 Oregon State University Worksheet Solutions to Translating Tensors page 2 A. CHECKING PREVIOUS EXERCISES We can use formula C.7 in the chapter on Rigid Bodies to compare the tensors we calculated in the worksheet Inertial Integrals, in the center of mass and another coordinate system. (This equation is also 11.49 in Marion and Thornton.) The inertial tensor in the center-of-mass coordinate system is given by I'ij = Iij – Tij (C.7) where Tij is the inertia of a point mass M at the center-of-mass's position R : Tij = M (R 2δij – R i R j ) . A.1. Brick X Center of mass: Y Z = (C.7a) z 12 A 1B 2 12 C A y C x The translation tensor is M Tij = M (R 2δij – R i R j ) = 4 B2+C2 –AB –AC –AB A 2+C2 –AC B –BC . A 2+B 2 –AC The tensors for the brick, calculated in the Worksheet Inertial Integrals , are origin at corner, Iij = 4(B2+C2) M –3AB 12 –3AC –3AB 4(A 2+C2) –3BC center of mass, I'ij = –3BC 4(A 2+B 2) –3AC M , 12 2 2 0 B +C A 2+C2 0 0 0 These 3 matrices actually work in eq. (C.7)! © 2003 Philip J. Siemens Paradigms in Physics: Rotational Motion 0 0 . A 2+B 2 √ Oregon State University Worksheet Solutions to Translating Tensors A.2. Cage X Y Z Center of mass: = 12 A 1B 2 12 C page 3 z A y C B x The translation tensor is the same as for the brick. The tensors calculated in the Worksheet on Inertial Integrals are origin at corner, Iij = 2C+8B 3(A+C) 12AB +12ABC2+8C3(A+B) σ –6AB(AB+BC+AC) 12 –6AC(AB+BC+AC) –6AB(AB+BC+AC) 12A 2BC+8A 3(B+C) +12ABC 2+8C3(A+B) –6BC(AB+BC+AC) –6BC(AB+BC+AC) 12A 2BC+8A 3(B+C) +12AB 2C+8B 3(A+C) –6AC(AB+BC+AC) center of mass, I'ij = σ 6 2C B+C3(A+C)+3AB 0 3(A+B)+3ABC2 0 The difference is Iij – I'ij M = 4 © 2003 Philip J. Siemens 0 0 A 3(B+C)+3A 2BC +B 3(A+C)+3AB 2C 0 A 3(B+C)+3A 2BC +C3(A+B)+3ABC 2 0 B2+C2 –AB –AC –AB A 2+C2 –AC –BC . A 2+B 2 –AC Paradigms in Physics: Rotational Motion check comparison: the difference is equal to T i j Oregon State University Worksheet Solutions to Translating Tensors page 4 B. TRANSLATING THE CAGEX INERTIAL TENSOR Use formula (C.7) in the chapter on Rigid Bodies to compare the tensors you calculated in the worksheet Inertial Integrals, in the center of mass and another coordinate system. The inertial tensor in the center-of-mass coordinate system is given by I'ij = Iij – Tij (C.7) where Tij is the inertia of a point mass M at the center-of-mass's position R : Tij = M (R 2δij – R i R j ) . (C.7a) B.1. From corner of cage to center of mass The CageX apparatus consists of a hollow rectangular cage with two square faces and four narrow rectangular faces. z L Its length in the x and y directions is L, and the length in the z direction is h . The walls of the cage have a uniform surface mass density, and its total mass is M . In addition there is a clay sphere of mass m fastened to the corner of the cage at the origin. The radius of this sphere is r . L h x y mass m radius r The expression for the center of mass of the cage-ball system in terms of M , m , L , h , and r is: M L L h R = M+m 2 , 2 , 2 The numerical values from the Inertial Integrals Worksheet are © 2003 Philip J. Siemens X Y Z ________ = ________ . ________ Paradigms in Physics: Rotational Motion Oregon State University Worksheet Solutions to Translating Tensors page 5 B.1. From corner of cage to center of mass (continued) The translation tensor for the cage-ball apparatus expressed in terms of M , m , L , and h is M2 1 MT (R 2δij – R i R j ) = Tij = M+m 4 L2+h2 –L2 –Lh –L2 L2+h2 –Lh –Lh 2L2 –Lh The inertial tensor for the cage-ball apparatus in the original coordinate system from section C of the Worksheet on Inertial Integrals expressed in terms of M , m , L , and h is 4+20L3h –6L3(L+2h) –6L2h(L+2h) 8L 2 2 3 +12L h +16Lh M 8L4+20L3h –6L2h(L+2h) Iij = 24(L2+2Lh)–6L3(L+2h) +12L2h2+16Lh3 2 –6L h(L+2h) –6L2h(L+2h) 16L4+40L3h _________________________________ The center-of-mass inertial tensor calculated from these results is I'ij = Mm 1 M+m 4 L2+h2 –L2 –Lh –L2 L2+h2 –Lh –Lh 2L2 –Lh + M 12(L2+2Lh) © 2003 Philip J. Siemens L4+L33h +8Lh 0 0 Paradigms in Physics: Rotational Motion 0 0 2L4+8L3h 0 L4+L3h +8Lh3 0 Oregon State University Worksheet Solutions to Translating Tensors page 6 B.2. From center of cage to center of mass This time we work in a coordinate system with origin at the center of the cage. z L The expression for the center of mass of the cage-ball system in terms of M , m , L , and h is: L y h m L L h R = – M+m 2 , 2 , 2 . The numerical values from the Inertial Integrals Worksheet are x X Y Z ________ = ________ . ________ The translation tensor for the cage-ball apparatus expressed in terms of M, m , L , and h is MT (R 2δij – R i R j ) = © 2003 Philip J. Siemens mass m radius r m2 1 Tij = M+m 4 L2+h2 –L2 –Lh Paradigms in Physics: Rotational Motion –L2 L2+h2 –Lh –Lh 2L2 –Lh Oregon State University Worksheet Solutions to Translating Tensors page 7 B.2. From center of cage to center of mass (continued) The inertial tensor for the ball only in the cage-centered coordinate system of section B of the Worksheet on Inertial Integrals may be approximately expressed in terms of m , L , and h as Iij = L2+h2 m –L2 4 –Lh –L2 L2+h2 –Lh –Lh 2L2 –Lh The inertial tensor ,for the cage only in the cage-centered coordinate system found in section B of the Worksheet on Inertial Integrals, expressed in terms of M , m , L , h , and r is M Iij = 12(L2+2Lh) 3h L4+4L +3L2h2+2Lh3 0 0 4 3 L +4L h 0 +3L2h2+2Lh3 0 2L4+8L3h 0 0 ––––––––––––––––––––––––––––––––– The inertial tensor for the cage plus ball in the cage-centered coordinate system of section B of the Worksheet on Inertial Integrals, expressed in terms of m , L , and h is therefor L2+h2 –L2 –Lh m –L2 2 2 L +h –Lh + Itot = = Iball + Icage = 4 –Lh –Lh 2L2 M 12(L2+2Lh) © 2003 Philip J. Siemens 3h L4+4L 2 2 +3L h +2Lh3 0 0 . 4 3 L +4L h 0 2 2 3 +3L h +2Lh 0 2L4+8L3h 0 Paradigms in Physics: Rotational Motion 0 Oregon State University Worksheet Solutions to Translating Tensors page 8 B.2. From center of cage to center of mass (concluded) The translation tensor for the cage-ball apparatus, expressed in terms of M , m , L , and h , found on page 6, is m2 1 Tij = M+m 4 L2+h2 –L2 –Lh –L2 L2+h2 –Lh –Lh 2L2 –Lh The inertial tensor for the cage plus ball in the cage-centered coordinate system, calculated on the previous page, expressed in terms of m , L , and h is m It = 4 L2+h2 –L2 –Lh –L2 –Lh 2L2 –Lh L2+h2 –Lh + M 12(L2+2Lh) 3h L4+4L +3L2h2+2Lh3 0 0 . 4 3 L +4L h 0 +3L2h2+2Lh3 0 2L4+8L3h 0 0 ––––––––––––––––––––––––––––––––– The center-of-mass inertial tensor calculated from these results is I'ij = Mm 1 M+m 4 L2+h2 –L2 –Lh –L2 L2+h2 –Lh –Lh 2L2 –Lh + M 12(L2+2Lh) L4+L33h +8Lh 0 0 Comparing this result to that of part B.1, we observe that © 2003 Philip J. Siemens 0 0 2L4+8L3h 0 L4+L3h +8Lh3 0 they are the same . Paradigms in Physics: Rotational Motion Oregon State University Worksheet Solutions to Translating Tensors page 9 C. Numerical values The inertial tensor for the cage-ball apparatus in the original coordinate system, found in section C.3 of Worksheet 3 Inertial Integrals, is Iij = numerically expressed in units of____ . The measured numerical values of the center-of-mass coordinates, from page 5 of the CageX1 Workbook, are X Y Z ________ = ________ . ________ The corresponding translation tensor for the cage-ball apparatus, evaluated numerically from the expression Tij = M (R 2δij – R i R j ) isTij = The center-of-mass inertial tensor calculated from these results, is I'ij = numerically expressed in units of ____. © 2003 Philip J. Siemens Paradigms in Physics: Rotational Motion Oregon State University Worksheet Solutions to Translating Tensors page 10 B.3. Numerical values continued We can compare some of the elements of the inertial tensor to the measured results from the CageX1 Workbook, page 10: Cases Observed moment I'obs (units © 2003 Philip J. Siemens Corresponding Calculated element I'?? value above ) (units Paradigms in Physics: Rotational Motion Comparison ) Oregon State University InertiaCalcs_cage.nb 1 CageX Inertia Tensor in Center-of-Mass frame (correction to Translating Tensors worksheet solution, p. 5 and p. 8, to Rotating Tensors worksheet solution, p. 4, and to Principal Axes worksheet solution, p. 6) Written as one piece: M H3 h L H2 m+ML+2 h H4 m+ML+L H4 m+ML+2 h L H5 m+2 MLL i ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ j 12 H2 h+LL Hm+ML j j j j 2 j L mM j - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ j 4 Hm+ML j j j j j hLmM - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ 4 Hm+ML k 2 3 3 L mM - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ 4 Hm+ML 2 2 M H3 h2 L H2 m+ML+2 h3 H4 m+ML+L3 H4 m+ML+2 h L2 H5 m+2 MLL ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 12 H2 h+LL Hm+ML ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ hLmM - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ 4 Hm+ML y z z z z z hLmM z z - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ z 4 Hm+ML z z z z L2 M H10 h m+4 L m+4 h M+L ML z ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅ Å ÅÅÅÅÅ 6 H2 h+LL Hm+ML { hLmM - ÅÅÅÅÅÅÅÅ ÅÅÅÅÅ 4 Hm+ML Written as two pieces: h2 + L2 -L2 -h L y i j z j z mM j z 2 2 2 j z ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ j + -L h + L -h L z j z z 4 Hm + ML j j z -h L 2 L2 { k -h L 2 h3 L + 3 h2 L2 + 4 h L3 + L4 i j j M j ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ j j 0 2 j 12 H2 h L + L L j j 0 k 0 3 2 2 3 2h L+3h L +4hL +L 0 y z z z z 0 z z z 3 4 z 8hL +2L { 0 4