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DYNAMICS - CASE STUDY SOLUTION

    Solution of a)

   

Begin with a diagram, modeling the hull as a circular cylinder and the con as a rectangular prism.

Use Eqs. 1 to determine the moments of inertia of the con about its center of gravity:

     

(Note, dm = ρ c dy dz)

Using a similar derivation, we get

     

These expressions for the moment of inertia could also be found from Appendix B, or a table of inertial properties.

Equations 2 give the products of inertia of the con about its center of gravity:

     

(Note, dm = ρ a dx dy)

Using a similar derivation, we get

     Iy'z' = Ix'z' = 0

     


Problem Simplification- Skew View


Problem Simplification - Side View

 

The expressions for the products of inertia could have been determined simply by noting that the product of inertia about a plane of symmetry is always zero.

Use the parallel axis theorem (Eqs. 3) to transfer the moment of inertia of the con about its center of gravity to the center of gravity of the sub:

     Ixx = Ix'x' + mc (dy2 + dz2)
          = 1/12 mc(a2 + b2) + mc (dy2 + dz2)

     Iyy = 1/12 mc(a2 + c2) + mc (dx2 + dz2)

     Izz = 1/12 mc(b2 + c2) + mc (dx2 + dy2)

Use the parallel plane theorem (Eqs. 4) to transfer the product of inertia of the con about its center of gravity to the center of gravity of the sub:

     Ixy = Ix'y' + mrdxdy

     Iyz = mcdydz

     Ixz = 0

     
    Solution of b)

 

Use Eq. 8 to express the angular momentum of the con about its center of gravity (Eq. 5) in scalar form:

     Hcg = Hxi + Hyj + Hzk

Here, we have

     Hx = -Ixxωx

     Hy = Iyyωy

     Hz = -Iyzωy

Express the position and velocity of the con in rectangular components, and determine the moment of the linear momentum of the con about the center of gravity of the sub:

     ρcon = dyj + dzk

     vcon = vxj + vyk

     

Use Eq. 7 to determine the angular momentum of the con about the center of gravity of the sub:

     HA = ρcon × mcvcon + Hcg  

          = (-Ixxωx - dzvy)i + (-Iyyωy - dzvx)j
                     + (-Iyzωy - dyvx)k

     
   
 
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