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DYNAMICS - THEORY

   

Chapter 6 introduced moment of inertia to describe the angular motion of a rigid body in a plane. To analyze motion in 3-D, moments of inertia is still used but an additional term, product of inertia, must also be determined.

     
    Moment of Inertia


Coordinate System
 

Recall, the moment of inertia of the element dm about the x axis is defined as

     dIxx = rx2 dm     (noting rx2 = y2 + z2)

Integrating over the mass, gives,

     

Thus, all three directions are
 
            (1a)
            (1b)
            (1c)
 

     
    Product of Inertia

   

The product of inertia of the element dm is defined in relation to a set of two orthogonal planes as the product of the mass of the element and the perpendicular distance from the planes to the element.

For example, the product of inertia of an element with respect to the x-z and y-z planes is

     dIxy = xy dm   

Integrating over the mass, gives,

    

For all three directions are

 
                      (2a)
                      (2b)
                      (2c)
 
     
    Parallel Axis and Parallel Plane Theorems


Center of Gravity Dimensions
 

If the center of gravity of the body is located at xcg, ycg, zcg within the xyz system, the parallel axis equations are

 
Ixx = (Ix'x')cg + m(ycg2 + zcg2)          (3a)

Iyy = (Iy'y')cg + m(xcg2 + zcg2)          (3b)

Izz = (Iz'z')cg + m(xcg2 + ycg2)           (3c)
 

Using a similar analysis, the parallel plane equations can be derived that transfer the products of inertia from a set of orthogonal planes passing through the body's center of gravity to a parallel set of planes passing through some other point O,

 
Ixy = (Ix'y')cg + mxcgycg                 (4a)

Iyz = (Iy'z')cg + mycgzcg                (4b)

Ixz = (Ix'z')cg + mxcgzcg                (4c)
 
     
    Angular Momentum


Angular Acceleration Dimensions

 

Consider a rigid body moving in general motion relative to an inertial reference frame XYZ. The origin A of the xyz axes translates and rotates with the rigid body, which has an angular velocity ω relative to XYZ.

Now find the angular momentum of the body about point A relative to XYZ. The angular momentum of the ith particle, with mass mi, about point A is

     HA-i = ρi/A × mivi

where ρi/A represents the position vector from A to the mass element, mi. For simplicity, it will be noted as ρi. Recall, the velocity vi is given by

     vi = vA + ω × ρi

     
   

Substituting and integrating over the mass, gives

     

If A is located at the center of gravity, the first term in H is zero. This gives,

 
        (5)
 

If A is a fixed point O about which the mass rotates, vA = 0, then it is similar to the cg form,

 
           (6)
 

If A is an arbitrary point on the body, the total angular momentum reduces to

 
HA = ρcg/A × mvcg + Hcg                 (7)
 
     
   

Note that Eqs. 5, 6, and 7 all contain the general form

     

If the coordinate system is rectangular (xyz), then the three components are,

     

Expanding the cross products, combining terms, and making use of Eqs. 1, the angular momentum for xyz coordinate system is

 
Hx = Ixxωx + Ixyωy + Ixzωz                 (8a)

Hy = Iyxωx + Iyyωy + Iyzωz                 (8b)

Hz = Izxωx + Izyωy + Izzωz                 (8c)
 

These represent the scalar forms of Eqs. 5 or 6.

     
   
 
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