Moment of inertia question

Question

The boundary of a thin plate is an ellipse with semiaxes a and b.

Let L denote a line in the plane of the plate passing through the center of the ellipse and making an angle k with the axis of length 2a

If the density is constant and if the mass of the plate is m,

find out the moment of inertia about line L.

 

how can I find out the equation for perpendicular distance?

Answer 1

You don't need the perpendicular distance to L.

 

Tables of moments of inertia usually mention the circular disk plate about the axis of rotation, which is mr²/2

Placing it on the x-y plane, this is I_z and by the perpendicular axis theorem for flat plates, I_x+I_y = I_z so I_x = I_y = mr²/4

 

For an ellipse with axis of length 2a on the x-axis and 2b on the y-axis, we have

 

I_x = mb²/4, I_y = ma²/4 and I_z = m(a²+b²)/4

 

The products of inertia I_xy = ∫xy dm, etc. are all 0 by symmetry.

 

So the inertia tensor is I =

 

[I_x     0      0]
[0      I_y     0]
[0     0      I_z]

 

We need the inertia tensor I' in the rotated system by k  counterclockwise about the z-axis where the x-axis will be along line L. The matrix of the rotation transformation is S =

 

[cos k     -sin k     0]
[sin k      cos k     0]
[  0            0          1]

 

And S^-1 is a clockwise rotation.

 

[ cos k      sin k     0]
[-sin k     cos k      0]
[   0           0           1]

 

The transformed inertia tensor is the matrix product I' = S^-1IS =

 

[ I_x cos² k + I_y sin² k            (I_x-I_y)sin k cos k       0]
[   (I_x-I_y)sin k cos k            I_x sin² k + I_y cos² k     0]
[            0                                     0                       I_z]

 

So The moment of inertia about L is

 

I_L = I_x' = I_x cos² k + I_y sin² k = m(b²cos²k + a²sin²k)/4.