So we found I_x’ in terms of I_x , I_y and I_xy .  Assuming we can find the moments of inertia for one orientation, we can now find them for any other orientation.

 

This equation is typically used in a different form.  Knowing the trig identities

 

             cos² q = 1/2 ( 1 + cos 2q )

 

             sin² q = 1/2 ( 1 - cos 2q )         and

 

             2 sin q cos q = sin 2q

 

we can substitute to find

 

             I_x’ = [ ( I_x + I_y )/2 ] + [ ( I_x - I_y )/2 ] cos 2q - I_xy sin 2q

 

Following the same procedure, we also find

 

             I_y’ = [ ( I_x + I_y )/2 ] - [ ( I_x - I_y )/2 ] cos 2q + I_xy sin 2q

 

and        I_xy’ = [ ( I_x - I_y )/2 ] sin 2q + I_xy cos 2q

 

The importance of these equations is that we can determine the maximum or minimum values of the moments of inertia.  Consider finding the extreme values of I_x’ by taking a derivative and setting to zero.

 

             dI_x’/dq = 0 = - [ ( I_x - I_y )/2 ] 2 sin 2q - I_xy 2 cos 2q

 

             [ ( I_x - I_y )/2 ] sin 2q = I_xy cos 2q

 

             tan 2q = - 2 I_xy / ( I_x - I_y )

 

This locates the extreme values of I_x’ .  There are two solutions for 2q which are different by 180^o or p.  Your calculator always gives the value between +90^o and - 90^o or between + p/2 and - p/2.  Solving for q, our two roots differ by 90^o.  This means that we only need one of the roots, because the second root is the opposite axis.  One of the roots gives a maximum value while the other gives a minimum value.  Therefore, the maximum and minimum values of moment of inertia occur for the same coordinate system - one being I_x’ and the other I_y’. 

 

 

Lecture #16, page 2                 Back                        Lecture #16, page 3