ME317 - Dynamics

Mass and Inertia Properties

1. Mass Center of a Rigid Body

Mathematically the mass the center is given by the following volume integral.

The above definition of mass center can be extended to a collection of components that make a rigid body. The expression for the mass center then is,

where,


   : Total mass of the system.
  : Mass of the ith body.
    : Number of bodies
: Position vector of the CG of the ith body with respect to a coordinate system fixed on the body

The mass center changes with either gain in mass or loss of mass.  Let the mass center of system  be defined as,

If some amount of mass  is lost from the system, then the new mass center will be given by,

Dividing the numerator and denominator of the above equation by ,

where, 

.

 

2. Mass Moment of Inertia Matrix

For every rigid body, the mass moment of inertia matrix is defined as a symmetric matrix.

Elements of the mass moment of inertia matrix are defined with the formulas below. 

 

For a rigid body that is modeled as a point mass, the mass moment of inertia matrix of the body about a reference frame centered at the point is zero.  Note that the mass moment of inertia matrix of a point mass about a reference frame not located at the mass location is not zero.  For a rigid body that is symmetric about a plane, the mass moment of inertia matrix about a reference frame that is centered at the mass center and aligned with the plane of symmetry will have two of the three off diagonal inertia matrix components equal to zero

 

 

3. Example Problems

3.1 –Problem 1

Find the CG for the following block configuration. The lower block is made of aluminum with density . The upper block is made of Iron with a density of .

Solution : The equation to be used is

 where

Let  be the masses of the aluminum block and the iron block respectively.

 and

 Let  be the centroidal radius vectors of the 2 blocks. These can be evaluated by inspection as,

 and

Expand the expression for the centroidal radial vector.

 

 

3.2 –Problem 2

A 3kg slender bar is attached to a 2 kg thin circular disk. Find the resulting moment of inertia of the composite body about z axis at the CG. The radius of the bar is 50 mm.

Solution : The CG of the composite body is to be found first with respect to the co-ordinate frame shown. This follows a procedure similar to the one in Problem 1. The value calculated is

The moment of inertia of the thin circular disk about an axis coming out of the plane of the paper can be looked up in any standard reference to be,

The moment of inertia of the slender bar can be looked up as,

Using the parallel axis theorem these inertias can be transferred to the CG of the composite body. Thus,

·        Disk


·        Bar

The total moment of Inertia at the CG of the composite body is the sum of the two inertias.