Question 21.1: Determine the moment of inertia of the bent rod shown in Fig......

Before applying Eq. 21-5, it is first necessary to determine the moments and products of inertia of the rod with respect to the x, y, z axes. This is done using the formula for the moment of inertia of a slender rod, I = \frac{1}{12} ml², and the parallel-axis and parallel-plane theorems, Eqs. 21-3 and 21-4. Dividing the rod into three parts and locating the mass center of each segment, Fig. 21-5b, we have

I_{xx} = (I_{x'x'})_{G} + m(y²_{G} + z²_{G})
I_{yy} = (I_{y'y'})_{G} + m(x²_{G} + z²_{G})           (21.3)
I_{zz} = (I_{z'z'})_{G}  + m(x²_{G}   + y²_{G})

I_{xy =} (I_{x'y'})_{G} + mx_{G}y_{G} 
I_{yz} = (I_{y'z'})_{G} + my_{G}z_{G}          (21.4)
I_{zx} = (I_{z'x'})_{G} + mz_{G}x_{G}

I_{Oa} = I_{xx}u²_{x}    +   I_{yy}u²_{y}    +   I_{zz}u²_{z}    –   2I_{xy}u_{x}u_{y}   –    2I_{yz}u_{y}u_{z}   –    2I_{zx}u_{z}u_{x}       (21.5)

I_{xx} = [\frac{1}{12}(2) (0.2)² + 2(0.1)²] + [0 + 2(0.2)²]

+ [\frac{1}{12}(4)(0.4)² + 4( (0.2)² + (0.2)²)] = 0.480   kg. m²

I_{yy} = [\frac{1}{12}(2) (0.2)² + 2(0.1)²] + [\frac{1}{12}(2)(0.2)² + 2((-0.1)² + (0.2)²)]

+ [0 + 4( (-0.2)² + (0.2)²)] = 0.453 kg .m²

I_{zz} = [0 + 0] + [\frac{1}{12}(2) (0.2)² + 2(-0.1)²] + [\frac{1}{12}(4)(0.4)² +  4( (-0.2)² + (0.2)²)] = 0.400  kg. m²

I_{xy} = [0 + 0] + [0 + 0] + [0 + 4(-0.2) (0.2)] = –  0.160  kg. m²

I_{yz} = [0 + 0] + [0 + 0] + [0 + 4(0.2) (0.2)] = 0.160   kg. m²

I_{zx} = [0 + 0] + [0 + 2(0.2)( -0.1)] + [0 + 4(0.2)(-0.2)] = -0.200   kg. m²

The Aa axis is defined by the unit vector

u_{Aa}  =  \frac{r_{D}}{r_{D}} = \frac{-0.2i + 0.4j + 0.2k }{\sqrt{(-0.2)²   +   (0.4)²  +  (0.2)²} } = -0.408i + 0.816j + 0.408k

Thus,

U_{x} = -0.408     U_{y} = 0.816             U_{z} = 0.408

Substituting these results into Eq. 21-5 yields

I_{Aa} = I_{xx}u²_{x}    +   I_{yy}u²_{y}   +   I_{zz}u²_{z}   – 2I_{xy}u_{x}u_{y}    –   2I_{yz}u_{y}u_{z}  –   2I_{zx}u_{z}u_{x}

= 0.408 ( -0.408 )² + (0.453)(0.816)² + 0.400( 0.408 )²

– 2( – 0.160)( – 0.408)(0.816)   –  2(0.160) (0.816)(0.408)

– 2( -0.200)(0.408)( – 0.408)

= 0.169 kg .m²