A hollow cylinder has a greater torsional rigidity than a solid one of same mass, length, and material:

Let us consider a solid cylinder of radius \( r \), length \( l \). The modulus of rigidity of the material of the cylinder is \( \eta \). So the twisting couple per unit twist of this solid cylinder is given by,

\( \displaystyle{\tau=\frac{\pi\eta{r^4}}{2l}}\tag{1} \)
[Read In Detail]

Now, let us consider another hollow cylinder of the same length \( l \), inner radius \( r_1 \) and outer radius \( r_2 \), The modulus of rigidity of the material of the hollow cylinder is \( \eta \). So the twisting couple per unit twist, i.e., torsional rigidity of this hollow cylinder is given by,

\( \displaystyle{\tau'=\frac{\pi\eta}{2l}({r_2}^4-{r_1}^4)}\tag{2} \)
[Read In Detail]

Since these two cylinders are same in length and mass, and of the same material of density ( \( \rho \)), then we can write,
\( \pi({r_2}^2-{r_1}^2)l\rho=\pi{r^2}l\rho \)
\( or,\ {r_2}^2-{r_1}^2=r^2\tag{3} \)

From equation (1) and (2) we get,

\( \displaystyle{\frac{\tau'}{\tau}=\frac{{r_2}^4-{r_1}^4}{r^4}\\=\frac{({r_2}^2+{r_1}^2)({r_2}^2-{r_1}^2)}{r^4}\\=\frac{({r_2}^2+{r_1}^2)r^2}{r^4}\\=\frac{({r_2}^2+{r_1}^2)}{r^2}\\=\frac{{r_2}^2+{r_1}^2}{{r_2}^2-{r_1}^2}} \)

Therefore, \( \displaystyle{\tau'>\tau} \).

So the torsional rigidity of a hollow cylinder is greater than that of a solid cylinder of same length, mass, and material. This is the reason, why a hollow cylinder is stronger than a solid cylinder.