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Moment of Inertia Tensor and Principal Axes RECAP

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Moment of Inertia Tensor and Principal Axes RECAP Chapter 4 RIGID BODY IN MOTION Lecture 19 & 20: Moment of Inertia tensor and principal axes RECAP dr Lx III xx xy xz x v r L III dt y yx yy yz y Lz III zx zy zz z 2 2 I mxy I mxz L r mv Ixx myz jj j xy jjj xz jjj LIx xxx I xyy I xzz L I y2 z 2 dm xydm zxdm I xydm z2 x 2 dm yzdm zxdm yzdm x2 y 2 dm y Recap (1,1) Angular momentum and Torque x o Lx2 m 2 m 0 x L 2 mm 2 0 y y (-1,-1) Lz0 0 4 m z x2m 2 m 0 x 2m 2 m x x y 2m 2 m 0 y2m x 2 m y y y z 4m z z0 0 4m z MI matrix is co-ordinate specific! Then, what about physical parameters? Will it remain same? Find the net angular acceleration in this case also compare it, when the whole system is rotated in x-z plane y y Easy Analysis! (1,1) Angular acceleration Fz in –z direction (- 2,0) ( 2,0) x x o o 2m 2 m 0 (-1,-1) I 2 mm 2 0 F in –z direction z 0 0 4m 0 0 0 i j k I 0 4 m 0 x2m x 2 m y rF xy0 FyiFxkz z xy i j 4m y y 0 0 4m y2m x 2 m y 0 0 Fz xzFy 2 m x 2 m y x y xFz 4 m y yzFx 2 m x 2 m y F y F y x 2 z z y 4m x 4m xFz Fz Fz y y=1 2 2m 4m 2 2m Find whether angular momentum and angular velocity are parallel in the below figure Problem 3 y y x-y plane (1,1) (- 2,0) ( 2,0) x-z plane o x o (-1,-1) 0 0 0 I 0 4 m 0 0 0 4m y Ly, y x-y plane (-1,0) (1,0) x-z plane LI 4 m y yyy y o x Ly and y are parallel to each other y x-y plane (-1,0) (1,0) o x z LIz zzz 4 m z Lz, z Lz and z are parallel to each other y (1,1) o (-1,-1) Lx2 m 2 m 0 x L 2 mm 2 0 Lx2 m x 2 m y y y L2 m 2 m 0 0 0 4m 0 y x y Lx2 m x 2 m y Ly2 m x 2 m y L Lix Lj y Lm(2xy 2 mi )(2 m xy 2 mj ) xi y j What is the angle between L and ? L. cos , also we know = - L x y When x= - y L4 mimjy 4 y yi y j L 4 m ( 4mimj 4 ).( i j ) 8 m 2 cos y yyy y 1 2 2 2 2 2 32myy 2 64 m y 0 L and are parallel L x x Angular momentum and Angular velocity have same slopes Ly y L 4 myy ,4 m ,0 and yy, ,0 In the matrix form we may write L I Lx2 m 2 m 0 x L 2 mm 2 0 y y 0 0 0 4m 0 x y Lx2 m 2 m 0 y L 2 mm 2 0 y y 0 0 0 4m 0 Lx2 m 2 m 0 y L 2 mm 2 0 y y 0 0 0 4m 0 4m Lx2 m 2 m 0 y y Lmm2 2 0 4 m y yy 0 0 0 4m 0 0 Lx y L 4 m y y 0 0 LI 4 m Although MI matrix is different in different co- ordinate system, physical parameters remain the same (Should be!). If L and are parallel, one can write LI Problem 4 Rotation of a square plate Consider rotation of a square plate of side a and mass M about an axis in the plane of the plate and making an angle with the x-axis. (a)What is the angular momentum L about the origin? (b)For what angle L and becomes parallel? (c) For square plate when the moment of inertia matrix becomes diagonal? M Also surface mass density is defined as . A M is the mass of the plate and A is the area Rotation of a square plate (a)What is the angular momentum L about the origin? y2 z 2 dm xydm zxdm L I 2 2 M I xydm z x dm yzdm M A ; a2 zxdm yzdm x2 y 2 dm M xy a a I( y2 z 2 ) dm y 2 z 2 dxdy xx 0 0 1M 1 a4 Ma 2 3a2 3 a a a4 1 I( z2 x 2 ) dm x 2 dxdy Ma 2 yy 0 0 3 3 a a 2 a4 2 Ma 2 I( x2 y 2 ) dm x 2 y 2 dxdy zz 0 0 3 3 1 Ma2 4 Rotation of a square plate (a)What is the angular moment L about the origin? L I Is L and parallel? 21 1 2 1 1 L Macos sin , Ma cos sin ,0 and cos , sin ,0 3 4 4 3 L x x Ly y L and are not parallel Rotation of a square plate (b) For what angle L and becomes parallel? 21 1 2 1 1 L Macos sin , Ma cos sin ,0 and cos , sin ,0 3 4 4 3 L x x Ly y Rotation of a square plate (b) For what angle L and becomes parallel? 2 Lx L I Ma2 L 2 y Ma 12 2 0 0 12 RotationZ of a square plate (Implications) X Ixx I xy I xz I Iyx I yy I yz Izx I zy I zz L I L and are parallel! Z Concept behind the problemZ X X PHYSICAL NATURE OF ROTATION IS SAME AS IN 45 DEGREE ROTATION Use of symmetry will ensure diagonal Moment of Inertia tensor I xx 0 0 L I x xxx I 0 I 0 yy x I xxx 0 0 I zz Principal Axis I I I xx xy xz Cumbersome! I Iyx I yy I yz Principal axes are the orthogonal axes for Izx I zy I zz Which [I] is diagonal I xx 0 0 I 0 I 0 yy Z 0 0 I zz L I X Lx I xxx y yyy Lz I zzz Y Note! Whenever ω is parallel to L, choosing the corresponding direction of rotation as co-ordinate axis will be the principal axes. Let’s recollect the problems we did……… Whenever ω is parallel to L, choosing the corresponding direction of rotation as co-ordinate axis will be the principal axes y (1,1) x o 2m 2 m 0 (-1,-1) I 2 mm 2 0 0 0 4m Whenever ω is parallel to L, choosing the corresponding direction of rotation as co-ordinate axis will be the principal axes y x-y plane (- 2,0) ( 2,0) x-z plane 0 0 0 o x I 0 4 m 0 0 0 4m Whenever ω is parallel to L, choosing the corresponding direction of rotation as co-ordinate axis will be the principal axes Z Z X X I xx 0 0 I 0 I 0 yy I 0 0 I zz Find the diagonal elements after performing the integration! Is Moment of Inertia a Scalar? Vector? Or …. ???? Tensor Tensor Tensor represents a physical entity which may be characterized by magnitude and multiple directions simultaneously. The rank of a tensor is defined by the number of directionality required to describe a component of it. Tensor The rank of a tensor is defined by the number of directionality required to describe a component of it. Temperature, Mass, Potential ‘0’ SCALAR Temperature T Force, Velocity, Torque VECTOR ‘1’ F Fexxˆ Fe yy ˆ Fe zz ˆ Fx F F y Fz Tensor Ixx I xy I xz Moment of Inertia I Iyx I yy I yz ‘2’ Izx I zy I zz zI zxx I zyy I zzz Torque applied in a given direction need not guarantee angular acceleration also in same direction, rather it can have components in 2 other direction’s simultaneously as well. This is due to coupling between two directions ( example Z-X, Z-Y ……etc). This coupling is induced via Moment of Inertia Tensor The rank of a tensor is defined by the number of directionality required to describe a component of it. Rank 2 Tensor represents a physical entity which may be characterized by magnitude and bi-directionality. Rank of a tensor 0 Scalar 1 Vector 2 Tensor of rank 2 (Dyadic) Higher orders Higher rank tensors Tutorial 22-01-2021 (1) Find the (a) Moment of Inertia matrix and (b) angular momentum of a cube with Side length L and mass M, with co-ordinate axis parallel to the edges of the cube and the origin at a corner. Z L L L X Y Moment of Inertia Tensor (a) y2 zdm 2 xydm zxdm I xydm z2 x 2 dm yzdm zxdm yzdm x2 y 2 dm Z L L L X Y Moment of Inertia Tensor y2 zdm 2 y 2 z 2 dxdydz L L y2 z 2 dxdydz L y 2 z 2 dydz 0 0 LL L L3 L y2 z 2 dydz L ( z 2 L ) dz 0 0 0 3 2 2 y2 zdm 2 L 5 ML 2 3 3 Moment of Inertia Tensor xy dm xy dxdydz ML2 xy dm 4 2 / 3 1/ 4 1/ 4 2 I ML 1/ 4 2 / 3 1/ 4 1/ 4 1/ 4 2 / 3 (b) Angular momentum vector 2 / 3 1/ 4 1/ 4 2 I ML 1/ 4 2 / 3 1/ 4 1/ 4 1/ 4 2 / 3 2 2 1 1 Lx ML xyz 3 4 4 2 1 2 1 Ly ML xyz 4 3 4 2 1 1 2 Lz ML xyz 4 4 3 (2) a.
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