Equation (10) gives
$∑j=1nf(xj)≈1Δx∫0Lf(x) dx$
where
$f(xj)=mjρALϕi(xj)ϕk(xj)$
Considering that all disks have the same mass mD, then
$f(xj)=mDρALϕi(xj)ϕk(xj)$
Hence, from Eq. (10)
$∑j=1NDmDρALϕi(xj)ϕk(xj)≈1Δx∫0LmDρALϕi(x)ϕk(x) dx=m̂ΔxLδik=m̂(ND+1)δik=μδik$
(11)

where $m̂=mD/ρAL$ and $μ=m̂(ND+1)$, which we can call the specific disk mass.

Similarly, if we consider that all disks have the same inertia JD, then
$∑j=1NDJDρALϕi(xj)d2ϕk(xj)dx2≈1Δx∫0LJDρALϕi(x)d2ϕk(x)dx2 dx$
(12)
Therefore, considering Eqs. (13) and (14)
$∑j=1NDJDρALϕi(xj)ϕk(xj)≈1Δx∫0L−JDρAL(kπL)2ϕi(x)ϕk(x) dx=−Ĵ(ND+1)(kπL)2δik=λ(kπL)2δik$
(15)

where $Ĵ=JD/ρAL$ and $λ=Ĵ(ND+1)$, which we can call the specific disk inertia.

The results shown in Figs. 15–17 were obtained using the present formulation of μ and λ.