Mass Moment of Inertia (Example continued)

Procedure (continued):

To facilitate integration:

y = a+h x = 2a

I^A_z,_K = ∫ ∫ (x/a) [ (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

s=h/a +1 r = 2

I^A_z,_K = a⁴ ∫ ∫ r [ (r ˗ 1)² + (s ˗ 1)² ] dr ds

s=1 r=1

Also let v = s ˗ 1 ; and v = h/a. and dv = ds

The result is:

v=h/a r=2

I^A_z,_K = a⁴ ∫ ∫ r [ (r˗1)² + v² ] dr dv

v=0 r=1

v=h/a r=2 v=h/a

I^A_z,_K = a⁴ ∫ [((r⁴/4) ˗ (2/3)r³+r²/2 + v²(r²/2) ]| dv = a⁴ ∫ [(1/4) + (v²)(3/2) ] dv

v=0 r=1 v=0

I^A_z,_K = a⁴ [ (1/4)(h/a) +(1/2)( h/a)³ ] = (ha/4) [ a² + 2h² ]

s=h/a+1 r=2

Note: m = ∫ ∫ (x/a) dx dy = a² ∫ ∫ r dr ds = (3/2)a² (h/a)

s=1 r=1

So m = (3/2) ha

and I^A_z,_K = (1/6)m [ a² +2 h² ] (result)

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