Linear Momentum, Angular Momentum and Kinetic Energy of Rigid Bodies (Part - 2) Civil Engineering (CE) Notes | EduRev

6.4.1 Deriving the linear momentum formula

By definition   We ca re-write this as follows:

(we used the definition of the COM to get the last result)

6.4.2 Deriving the angular momentum formula

Start with the definition: 

Note that r_i =r_G + d_i and recall the relative velocity formula v_i −v_G = ω × (r_i− r_G) = ω ×d_i .  This means we can re-write the angular momentum as

Note that

Finally, recall the dreaded triple cross product formula

a × b × c= (a ⋅ c)b − (a ⋅b)c

This means that

d_i × ω× d_i = (d_i ⋅ d_i )ω − d_i (d_i ⋅ ω)

We can expand this out

This shows that

Finally collecting terms gives the required answer

6.4.3 Deriving the kinetic energy formula

We can use v_i −v_G = ω × (r_i− r_G) = ω ×d_i

Recall that 

and expand the dot product of two cross products using the formula

(a × b) ⋅ (c × d) = (a ⋅c)(b ⋅ d) − (b ⋅ c)(a ⋅d)

This shows that 

(ω x d_i) • (ω x d_i) = (ω • ω)(d_i • d_i) - (w • d_i)²

= ω ⋅I_G ω

6.4.5 Calculating the center of mass and inertia of a general rigid body

It is not hard to extend the results for a system of N particles to a general rigid body. We simply regard the body to be made up of an infinite number of vanishingly small particles, and take the limit of the sums as the particle volume goes to zero. The sums all turn into integrals.

3D problems:  For a body with mass density ρ (mass per unit volume) we have that

• The total mass is 
• The position of the center of mass is 
• The mass moment of inertia about the center of mass is

 where d = r − r_G

For 2D problems: We know the COM must lie in the i,j plane and we don’t need to calculate the whole matrix.  

For a body with mass per unit area µ we can therefore use the formulas

• The total mass is 
• The position of the center of mass is  
• The mass moment of inertia about the center of mass is 

where d = r − r_G

Example 1: To show how to use these, let’s calculate the total mass, center of mass, and mass moment of inertia of a rectangular prism with faces perpendicular to the i, j,k axes:

First the total mass (sort of trivial)

Now the COM

And finally the mass moment of inertia

Example 2: As a second example, let’s calculate the mass moment of inertia of a cylinder with mass density ρ , length L and radius a.  (We know the COM is at the center and we know the total mass so we won’t bother calculating those). We have to do the integral with polar coordinates

Now note that d _x =r cosθ d_y = r sinθ dz = z .  We can have Mupad do the dirty work:

[reset() :

[ds := r*cos(q): dy := r*sin(q): dz := z:

Example 3: Let’s finish up with a 2D example. Find the mass, center of mass, and out of plane mass moment of inertia of the triangle shown in the figure.

The total mass is  

The position of the COM is 

The 2D mass moment of inertia is

This is all a big pain, and you may be contemplating a life of crime instead of an engineering career.   Fortunately, it is very rare to have to do these sorts of integrals in practice, because all the integrals for common shapes have already been done. You can google most of them. The tables below give a short list of all the objects we will encounter in this course.

Table of mass moment of inertia tensors for selected 3D objects

Prism M = ρ abc
Solid Cylinder M = πρ a2L
Solid Cone M =π/h ρα2h
Solid Sphere M = 4/3 πρα3
Solid Ellipsoid M =4/3 πραbc
Hollow Cylinder M = πρ (b2 − a2)L

Table of mass moment of inertia about perpendicular axis for selected 2D objects

Square		IGzz = M/12 (α2 + b2)
Disk		IGzz = M/2R2
Thin ring		IGzz = -MR2
Hollow disk		IGzz = M/2 (α2 + b2)
Slender rod		IGzz =  M/12 L2
Triangular Plate		M/18 (α2 + b2)