Question 21.6: The 10-kg flywheel (or thin disk) shown in Fig. 21-14a rotat......

SOLUTION I

Free-Body Diagram. Fig. 21-14b. The origin of the x, y, z coordinate
system is located at the center of mass G of the flywheel. Here we will let these coordinates have an angular velocity of Ω = ω_{p} = {3k} rad/s. Although the wheel spins relative to these axes, the moments of inertia remain constant,* i.e.,

I_{x} = I_{z} =\frac{1}{4} (10   kg)(0.2   m)²  = 0.1   kg· m²

I_{y} = \frac{1}{2} (10   kg)(0.2   m)² = 0.2   kg· m²

Kinematics. From the coincident inertial X, Y, Z frame of reference, Fig. 21-14c, the flywheel has an angular velocity of ω = {6j + 3k} rad/s, so that

ω_{x} = 0      ω_{y} = 6   rad/s              ω_{z} = 3   rad/s

The time derivative of ω must be determined relative to the x, y, z
axes. In this case both ω_{p} and ω_{s} do not change their magnitude or direction, and so

\dot{ω}_{x} = 0       \dot{ω}_{y} = 0              \dot{ω}_{z} =0

Equations of Motion. Applying Eqs. 21-26 (Ω ≠ ω) yields

Σ M_{x} = I_{x}\dot{ω}_{x}    –   I_{y}Ω_{z}ω_{z}   +    I_{z})Ω_{y}ω_{z}  

– A_{z}(0.5)   +  B_{z}(0.5) = 0    –   (0.2)(3)(6)   +   0 = –  3.6

Σ M_{y} = I_{y}\dot{ω}_{y}    –   I_{z}Ω_{x}ω_{z}   +   I_{x})Ω_{z}ω_{x} 

0    =  0    –   0  +   0

Σ M_{z} = I_{z}\dot{ω}_{z}    –   I_{x}Ω_{y}ω_{x}   +   I_{y})Ω_{x}ω_{y}

A_{x}(0.5)   –    B_{x}(0.5)   =   0   –   0   +   0 

Applying Eqs. 21-19, we have

ΣF_{x}  = m(a_{G})_{x};                    A_{x}  + B_{x}   = 0   

ΣF_{y}  = m(a_{G})_{y};                    A_{y}  = – 10(0.5)(3)²

ΣF_{z}  = m(a_{G})_{z};                    A_{z}   +  B_{z}  –  10(9.81) = 0 

Solving these equations, we obtain

A_{x} = 0       A_{y} = –  45.0   N         A_{z} = 52.6   N      

B_{x} = 0                               B_{z} = 45.4 N

NOTE: If the precession ω_{p} had not occurred, the z component of force at A and B would be equal to 49.05 N. In this case, however, the difference in these components is caused by the gyroscopic moment created whenever a spinning body processes about another axis. We will study this effect in detail in the next section.

SOLUTION II

This example can also be solved using Euler’s equations of motion, Eqs. 21-25. In this case Ω = ω = {6j + 3k} rad/s, and the time derivative (\dot{ω})_{xyz} can be conveniently obtained with reference to the fixed X, Y, Z axes since \dot{ω} = (\dot{ω})_{ xyz}. This calculation can be performed by choosing x’ , y’, z’ axes to have an angular velocity of Ω’ = ω_{P} , Fig. 21-14c, so that

\dot{ω}= (\dot{ω})_{x'y'z'} + ω_{p} × ω = 0 + 3k ×  (6j + 3k) = {-18i}   rad/s² 

\dot{ω}_{x} =  –  18   rad/s          \dot{ω}_{y}  =   0         \dot{ω}_{z} = 0

The moment equations then become

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

– A_{z}(0.5)   +   B_{z}(0.5) = 0.1(- 18)   –   (0.2  –   0.1)(6)(3) = – 3.6

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

0  =   0   –   0

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

A_{x}(0.5)      – B_{x}(0.5) = 0   –   0

The solution then proceeds as before.

* This would not be true for the propeller in Example 21.5.