Irodov 1.242: The Shell a Sphere Grows – Amit & Deepali Sinha Foundation

Problem Irodov 1.242

Starting from the standard result for a solid uniform sphere, whose moment of inertia about a diameter is $I = (2/5)MR^2$ , find the moment of inertia of a thin uniform spherical shell of mass $m$ and radius $R$ about an axis through its center (a diameter).

The thin uniform shell of mass m and radius R in section, with the axis along a diameter drawn dashed. — The thin uniform shell of mass $m$ and radius $R$ in section, with the axis along a diameter drawn dashed.

Given

$m$
$R$
uniform sphere: $I = (2/5)MR^2$

Find

I

of a thin spherical layer about a diameter

Solution

Step 1Turn the layer into a change of radius

We are told to start from the solid sphere, so the layer must be produced from solid spheres. A uniform sphere of radius $R$ and one of radius $R+dR$ made of the same material differ by precisely one thin uniform layer of radius $R$ : the larger body is the smaller body plus that shell, and the interior of the smaller body is untouched by the addition.

The uniform sphere in section. Growing its radius from R to R+dR at fixed density adds exactly the thin layer of mass dm, shaded here, and adds nothing to the interior. — The uniform sphere in section. Growing its radius from $R$ to $R+dR$ at fixed density adds exactly the thin layer of mass $dm$ , shaded here, and adds nothing to the interior.

Step 2Both quantities as functions of the radius alone

Fix the density $\rho$ and write the mass and the moment of inertia of a solid sphere of radius $R$ as functions of $R$ . The mass is the density times the volume,

M(R) = \rho\,\frac{4}{3}\pi R^{3},

and substituting this into the quoted formula $I = \frac{2}{5}MR^{2}$ gives

I(R) = \frac{2}{5}\left(\rho\,\frac{4}{3}\pi R^{3}\right)R^{2} = \frac{8}{15}\pi\rho R^{5}.

Step 3Differentiate to peel off one layer

Increase the radius by $dR$ . The mass added and the moment of inertia added are the differentials of the two expressions above,

\begin{aligned} dm &= \frac{dM}{dR}\,dR = 4\pi\rho R^{2}\,dR,\\[2pt] dI &= \frac{dI}{dR}\,dR = \frac{8}{3}\pi\rho R^{4}\,dR. \end{aligned}

Both refer to the same shell, so dividing one by the other eliminates the density and the thickness at once, which is the whole point of the maneuver:

\frac{dI}{dm} = \frac{\frac{8}{3}\pi\rho R^{4}\,dR}{4\pi\rho R^{2}\,dR} = \frac{2}{3}R^{2}

(1)

Step 4Restore the layer’s own mass

The ratio in (1) is the moment of inertia carried by unit mass of a shell of radius $R$ , and it does not depend on how thin the shell is. A layer whose total mass is $m$ therefore has

I = \frac{2}{3}mR^{2}

(2)

Check

Check it against the shell’s geometry directly. Every particle of the layer lies at distance $R$ from the center, so $\langle x^{2}\rangle + \langle y^{2}\rangle + \langle z^{2}\rangle = R^{2}$ averaged over the layer, and by symmetry the three averages are equal, each $R^{2}/3$ . About the $z$ axis the moment of inertia is $m\langle x^{2}+y^{2}\rangle = \frac{2}{3}mR^{2}$ , which reproduces the answer without any differentiation at all.

Remark

The shell’s $\frac{2}{3}mR^{2}$ exceeds the solid sphere’s $\frac{2}{5}mR^{2}$ for the same mass and the same radius, as it must: hollowing the body out moves every particle to the largest distance available.