Let's say I have a solid cylinder of uniform mass density, radius $R$ and height $h$. I know that the moment of inertia of this cylinder rotating about the axis parallel to the height and passing through the center of mass is $\frac{MR^2}{2}$. How would the Moment of Inertia (about the same axis) change if I were to cut this cylinder in half? (The cut goes along the length of the cylinder)


1 Answer 1


If you do the calculations, you get that the Moment of inertia of a cilinder it's

$$ I=\rho\int_{z_1}^{z_2}\int_0^{2\pi}\int_0^R r^3drd\theta dz $$ With

$$ \rho=\frac{M}{V}=\frac{M}{h\pi R^2} $$

Half of the cilinder means that $\theta$ goes from $0$ to $\pi$. Also you conserve the density.

Since there is no angular dependence, it's just the half of the Moment of inertia of the initial cilinder.

  • 2
    $\begingroup$ So it still remains $\frac{MR^2}{2}$. Just that this value is half because the mass is half. $\endgroup$ Oct 14, 2016 at 2:39
  • $\begingroup$ If you deform the cilinder to a half cilinder, yes. The density will double compensating the angular decrease. It's like the moment of inertia of a point, that it's conservated if you deform the point to be a ring of radius equal to the distance from the rotation axis. $\endgroup$
    – Grego_gc
    Oct 14, 2016 at 3:25

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