Rod with Donut

You are given a rod having a linear mass density $\lambda(x) =\frac{2M}{2L}[1+(x/L)]$. Then you place a donut (as a cylinder) at a distance $d=0.75L$ from the lighter end, having a mass $m=0.25M$. Please find the center of the mass and the inertia moment with respect to the lighter end of the rod.

For the rod without donut:

The center of the mass coordinate is defined as

$M x_{cm}=\sum_i m_i x_i$  that is $x_{cm}\int_0^L \lambda(x)d x=\int_0^L x\lambda(x)d x$ because $d m=\lambda(x)d x$

$x_{cm}=\frac{\int_0^L x[1+ (x/L)]d x}{\int_0^L[1+(x/L))]}d x=\frac{(L^2/2)+(L^2/3)}{L+(L/2)}=\frac{5/6}{3/2}L=(10/18)L=(5/9)L$

from the lighter end  $(x=0)$

$M=\int_0^L\lambda(x)d x=(2M/2L)\int_0^L([1+(x/L)]d x=(2M/2L)[L+(L/2)]=(3/2)M$

Therefore the correct linear density should have been


The moment of inertia is defined as

$I=\int_0^L x^2*d m=\int_0^L x^2\lambda(x) dx=(2M/3L)\int_0^L x^2[1+(x/L)] dx=$


With the donut the CM is located at

$(M+0.25M)X_{cm}=M*(5/9)*L +0.25M*0.75L$  that is  $X_{cm}=0.5944L$

The new moment of inertia


If you want the results for $\lambda(x)=(2M/2L)[1+(x/L)]$ the CM is the same, the moment of inertia of the rod is $I=(7/12)ML^2$ and the moment of inertia with donut is $I_{tot}=0.7708ML^2$