Question 21.5: The airplane shown in Fig. 21-13a is in the process of makin......

Free-Body Diagram. Fig. 21-13b. The reactions of the connecting
shaft on the propeller are indicated by the resultants F_{R} and M_{R} . (The propeller’s weight is assumed to be negligible.) The x, y, z axes will be taken fixed to the propeller, since these axes always represent the principal axes of inertia for the propeller. Thus, 0 = ω. The moments of inertia I_{x} and I_{y} are equal (I_{x} = I_{y} = 1) and I_{z}   = 0.

Kinematics. The angular velocity of the propeller observed from the X, Y, Z axes, coincident with the x, y, z axes, Fig. 21-13c, is ω = ω_{s}   +  ω_{p}   =  ω_{s}i   + ω_{p}k,   so that the x, y, z components of w are

ω_{x} =ω_{s}              ω_{y} = 0                    ω_{z} =ω_{p}

Since Ω = ω, then \dot{ω} = (\dot{ω})_{xyz} To find ω, which is the time derivative with respect to the fixed X, Y, Z axes, we can use Eq. 20-6 since W changes direction relative to X, Y, Z. The time rate of change of each of these components \dot{ω} = \dot{ω}_{s} + \dot{ω}_{p}   relative to the X, Y, Z axes can be obtained by introducing a third coordinate system x’, y’, Z’, which has an angular velocity Ω’ = ω_{p}  and is coincident with the X, Y, Z axes at the instant shown. Thus

  \dot{ω} = (\dot{ω})_{x'y'z'} +    ω_{p} × ω

  = (\dot{ω}_{s} )_{x'y'z'} + (\dot{ω}_{p})_{x'y'z'} +ω_{p} × (ω_{s} + ω_{p})

  = 0 + 0 + ω_{p} × ω _{s }  +  ω_{p} ×  ω_{p} 

  = 0 + 0 + ω_{p}k ×  ω_{s }i + 0 = ω_{p}ω _{s }j

Since the X, Y, Z axes are coincident with the x, y, z axes at the instant
shown, the components of ω along x, y, z are therefore

  \dot{ω}_{x} = 0            \dot{ω}_{y} = ω_{p}ω _{s }                  \dot{ω}_{z} = 0

These same results can also be determined by direct calculation of (\dot{ω})_{xyz} ; however, this will involve a bit more work. To do this, it will be necessary to view the propeller (or the x, y, z axes) in some general position such as shown in Fig. 21-13d. Here the plane has turned through an angle Φ (phi) and the propeller has turned through an angle ψ (psi) relative to the plane. Notice that ω_{p}   is always directed along the fixed Z axis and ω_{s}   follows the x axis. Thus the general components of ω are

  ω_{x} = ω_{s}              ω_{y} = ω_{p}   sin   ψ                  ω_{z} = ω_{p}  cos   ψ 

Since ω, and ω_{p} are constant, the time derivatives of these components become

  \dot{ω}_{x} = 0            \dot{ω}_{y} = ω_{p}cos   ψ   \dot{ψ}                 \dot{ω}_{z} = –  ω_{p}   sin    ψ   \dot{ψ}

But Φ =  ψ = 0° and \dot{ψ}    = ω_{s} at the instant considered. Thus,

ω_{x} = ω_{s}              ω_{y} = 0                  ω_{z} = ω_{p}

  \dot{ω}_{x} = 0            \dot{ω}_{y} =ω_{p}ω _{s }                  \dot{ω}_{z} = 0

which are the same results as those obtained previously.

Equations of Motion. Using Eqs. 21-25, we have

Σ M_{x} = I_{x}\dot{ω}_{x}    –   (I_{y}   –    I_{z})ω_{y}ω_{z}= I(0) – (I – 0)(0)ω_{p}

M_{x}   = 0

ΣM_{y} = I_{y}\dot{ω}_{y}  –   (I_{z}   –    I_{x})ω_{z}ω_{x} I(ω_{p}ω_{s}) –   (0   –   I)ω_{p}ω_{s}

  M_{y} =  2 Iω_{p}ω_{s}     

ΣM_{z} = I_{z}\dot{ω}_{z}    –    (I_{x}    –     I_{y})ω_{x}ω_{y}= 0(0)   – (I   –    I)ω_{s}(0) 

M_{z} = 0