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DYNAMICS - CASE STUDY SOLUTION

   

Begin with a diagram at the instant the fly hits the uppermost blade.

The stationary system XYZ is located at the center of the motor housing, the point about which the fan rotates, with the X axis parallel to the ground.

     

Problem Diagram
     
   

Place the origin (Point A) of the translating- rotating system xyz at the center of the fan blades, with the x axis perpendicular to the plane of rotation of the blades. This system is rotating with the fan but not the blades.

At this instant, the X and x axes are co-planar.

The angular velocity and acceleration of the fan become

     

The angular velocity and the acceleration of the fan blades are

     

The position of B relative to A is given by

     rB/A = 0.30k m

     
    Velocity


Problem Graphic with
Coordinates - Front View


Problem Graphic with
Coordinates - Top

 

The velocity of B can be broken down to its three components,

     

Each term must be addressed separately.

Term 1) Point A undergoes rotation about a fixed point, thus vA is given by

     vA = Ωf × rA

Here, each vector is,

     rA = 0.2 cos15I + 0.2 sin15K

     vA = 0.296I × (0.2 cos15I + 0.2 sin15K)

          = 0.0572J m/s

Term 2)  Ωf × rB/A = 0.296K × 0.30k

Note that:    K × k = -sin15J

      Ωf × rB/A = -0.296 (0.30) sin15J

                      = -0.0230J m/s

Term 3) Within the xyz system, point B undergoes rotation about a fixed point:

    (vB/A)xyz =  ωb × rB/A

                  = 2πi × 0.30k

                  = -1.8850j = -1.8850J

Summing the terms, we get

     vB = -1.8510J m/s

     
    Acceleration

 

Use the full acceleration equation, acceleration of point B is,

     

Again, each term must be addressed separately.

Term 1) Point A undergoes rotation about a fixed point; thus, aA is

     

     = 0.0419K × (0.2 cos15I + 0.2 sin15K) +
        + 0.296K × [0.296K × (0.2 cos15I + 0.2 sin15K)]

     = -0.0169I + 0.0081J

Term 2)   

                               = -0.0419 (0.30) sin15J

                              = -0.0032J

Term 3) 
    

                    = 0.296K × (-0.0230J) = 0.0068I

Term 4) 
    

                   = 1.1159I

Term 5) Within the xyz system, point B undergoes motion about a fixed point:

    

                      = 0 + 2πi × (2πi × 0.30k)

                      = -11.8435k

Note that k = -sin15I + cos15K

     (aB/A)xyz = -11.8435 (-sin15I + cos15K)

                   = 3.0653I - 11.4399K

Summing the terms, gives

     aB = 4.1711I + 0.0049J - 11.4399K m/s2

     
   
 
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