Question #a1395

1 Answer
Oct 26, 2016

Answer:

see below

Explanation:

For point mass #m# at generalised position #vec r # with respect to origin, its angular momentum is:
#vec L = vec r times m vec v = m vecr times ( vec omega times vec r)#

From the bac-cab rule

#(vec L)/m = vec omega (r^2) - vec r (vec r * vec omega) #

In cartesian
#(vec L)/m = ((omega_x),(omega_y),(omega_z)) (x^2+y^2+z^2) - ((x),(y),(z)) (((x),(y),(z)) * ((omega_x),(omega_y),(omega_z)) ) #

#(vec L)/m = ((omega_x),(omega_y),(omega_z)) (x^2+y^2+z^2) - ( (x),(y),(z)) (x omega_x + y omega_y + z omega_z) #

If we pick out the x-component of angular momentum, we have

#L_x/m = omega_x (x^2+y^2+z^2) - x (x omega_x + y omega_y + z omega_z) #

# = omega_x (y^2+z^2) - x y omega_y - x z omega_z #

#L_x = color(red)( omega_x) m(y^2+z^2) - color(green)(omega_y) mx y - color(blue)( omega_z) m x z #

#= I_(x x) omega_x + I_(xy) omega_y + I_(xz) omega_z#

We can pattern match to find a 3x3 matrix:

# vec L = m ( ( y^2+z^2, -xy, -xz), ( -yx, x^2 + z^2, -yz), (-zx, -zy, x^2+y^2 )) ((omega_x), (omega_y), (omega_z))#

# = ( ( I_{x x}, I_{xy}, I_{xz}), ( I_{yx}, I_{yy}, I_{yz}), ( I_{zx}, I_{zy}, I_{zz} )) ((omega_x), (omega_y), (omega_z))#

So that

#I = m ( ( y^2+z^2, -xy, -xz), ( -yx, x^2 + z^2, -yz), (-zx, -zy, x^2+y^2 ))#

In this generalised form, inertia is a 3x3 tensor (ie of rank 2).

If we consider simple circular rotation of the mass in a plane about the origin, we can set the co-ordinate system so that rotation is about the z-axis at z = 0, ie with #omega_x = omega_y = 0# and #omega_z = omega#.

Angular momentum then is

# L = m ( ( y^2, -xy, 0), ( -yx, x^2 , 0), (0, 0, x^2+y^2 )) ((0),(0),(omega_z)) #

#= m (x^2 +y^2) omega_z = m r^2 omega_z = I omega#

In this simplified form, #I = mr^2# which is a scalar (but still also a tensor of rank zero).