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  y2m  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 x2m  x  2 m  y      rF xy0  FyiFxkz z   xy i j  4m    y y 0 0 4m  y2m  x  2 m  y 0 0 Fz

xzFy 2 m  x  2 m y x  y xFz 4 m y yzFx 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)

(-1,-1)

Lx2 m 2 m 0  x   L 2 mm 2 0  Lx2 m x  2 m  y y  y   L2 m  2 m  0 0 0 4m  0 y x y Lx2 m x  2 m  y

Ly2 m x  2 m  y    L Lix  Lj y  Lm(2xy  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  L4 mimjy  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 32myy 2 64 m  y

0  L and  are parallel L  x x Angular momentum and Angular velocity have same slopes Ly y   L 4 myy ,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      Lmm2 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?

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?

 21 1  2 1 1    L Macos   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?

 21 1  2 1 1    L Macos   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)

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

(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 

zI 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. Calculate the Moment of Inertia matrix of a cube when the origin is shifted to the center of the cube. b. Intuitively find the principal axis of rotation

L L

L X

Y Moment of Inertia Tensor

 y2 zdm 2   y 2  z 2   dxdydz

L/ 2 L / 2 y2 z 2 dxdydz   L    y 2  z 2 dydz L/ 2  L / 2

LL/ 2 / 2 L / 2 L3 L y2 z 2  dydz   L  (  z 2 L ) dz LL/ 2 / 2  L / 2 12

1 y2 z 2  dm  ML 2  6 Moment of Inertia Tensor  xy dm   xy dxdydz

 xy dm  0

1/ 6 0 0  2   I   ML 0 1/ 6 0   1 2 0 0 1/ 6  L ML    6 (2) b. Intuitively find the principal axis of rotation

If the orogin is chosen at the centre and co-ordinate axis as shown in the figure, Z the MI tensor will be diagonal and then the coordinate is symmetrically selected. L L Thus this will be one of the proncipal axis.

L X