Mass Moment of Inertia  (Example continued)

 

 

Procedure (continued):

                        

 

 

 

6.

 

 

 

 

To facilitate integration:

 

                                        y = a+h    x = a + b

                          IAz,K   =     ∫              ∫      [ (x ˗ a)2  +  (y ˗ a)2 ] dx dy

                                        y=a          x=a

 

Let      r = x/a  and  s = y/a .   So  dx = a dr,  dy = a ds

 

 

 

 

 

7.

 

 

                                            s=h/a +1      r = b/a +1

                          IAz,K   = a4    ∫                  ∫      [ (r ˗ 1)2  +  (s ˗ 1)2 ] dr ds

                                            s=1              r=1

 

Also let     u = r ˗1 and  v = s˗1 ;   So   u = b/a  and  v = h/a.

 

and         du = dr,  dv  =  ds

 

 

 

8.

 

The result is:

                                            v=h/a          u=b/a

                          IAz,K   = a4    ∫                  ∫      [ u2  +  v2 ] du dv

                                            v=0             u=0

 

 

 

 

 

 

9.

 

                          v=h/a                  u=b/a            v=h/a

       IAz,K   = a4   ∫ [(1/3)  u3 + v2u ]|  dv   =   a4    ∫   [(1/3) (b3/a3)  + (v2)(b/a) ] dv

                         v=0                     u=0                v=0

 

 

                          IAz,K   =  (1/3) [ b3h + h3b ]

 

Note:  m  = ρ bh  =  bh

 

So                         IAz,K   =  (1/3)m [ b2 + h2 ]    (result)

 

 

Click here for another example.

 

 

 

 

                            



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