# What is the moment of inertia of a cylinder about a skew axis? [duplicate]

Suppose that there is a cylinder of uniform density with radius $r$ , height $h$ and mass $M$.

An axis $\ell$ is chosen through its geometric center making an angle $\theta$ with the horizontal.

What is the cylinder's moment of inertia about $\ell$?

Is it possible to do this by using the rotation of axes formulae?

• MMOI transforms according to the congruent transformation rules $$\mathbf{I}_{\rm new} = \mathrm{R} \mathbf{I}_{\rm old} \mathrm{R}^\top$$ Commented Jan 1, 2020 at 8:26

If the skew axis makes angles $$\alpha, \beta, \gamma$$ with the $$x, y, z$$ axes respectively then the moment of inertia about this axis for any rigid body is
$$I=I_{xx}\cos^2\alpha+I_{yy}\cos^2\beta+I_{zz}\cos^2\gamma-2I_{xy}\cos\alpha \cos\beta -2I_{yz}\cos\beta \cos\gamma -2I_{zx}\cos\gamma \cos\alpha$$

Here $$I_{xx}=\int \rho x^2 dV$$ etc are the moments of inertia about the $$x, y, z$$ axes and $$I_{xy}=I_{yx}=\int \rho xy dV$$ etc are products of inertia.

Note that $$\cos\alpha, \cos\beta, \cos\gamma$$ are direction cosines so that $$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$$. They are related to spherical polar angles by
$$\cos\alpha=\cos\theta \sin\phi, \space \cos\beta=\sin\theta \sin\phi, \space \cos\gamma=\cos\phi$$

The products of inertia $$I_{xy}$$ etc vanish for 3 particular mutually orthogonal axes through the centre of mass called the principal axes. These axes can be identified practically because no torque is required when the body is rotating about such an axis; the rotation is dynamically balanced. Geometrically they can often be identified as axes of symmetry.

Mathematically the principal axes and their moments of inertia can be found by diagonalising the inertia matrix/tensor formed by the elements $$I_{xx}, I_{xy}, I_{xz}$$ etc.

With the origin at the CM of the cylinder, and $$Ox$$ along its axis, then all 3 planes $$Oxy, Oyz, Ozx$$ are planes of symmetry, with the result that $$I_{xy}=I_{yz}=I_{zx}=0$$.

The cylinder is an elongated disk so $$I_{xx}=\frac12 MR^2$$. If the cylinder were a planar disk ($$H\ll R$$) then $$I_{yy}=I_{zz}=\frac12 I_{xx}=\frac14 MR^2$$, whereas if it were a rod ($$H \gg R$$) then $$I_{yy}=I_{zz}=\frac{1}{12} MH^2$$. Combining both forms we get $$I_{yy}=I_{zz}=\frac{1}{12} M(3R^2+H^2)$$.

If we were to take an axis through the rim of the cylinder (still passing though the centre of mass) then $$\tan\alpha=\frac{R}{H}$$. With a suitable choice of $$y, z$$ axes we get $$\beta=0$$ and $$\gamma=\frac{\pi}{2}-\alpha$$ so that $$\cos\gamma=\sin\alpha$$. It would make no difference if we chose the $$y, z$$ axes differently.

Reference : Chapter 2: Centroids and Moments of Inertia in Advanced Dynamics : Analytical and Numerical Calculations with MATLAB by Marghitu and Dupac, Springer Science.