UPSC Maths 2025 Paper 2 Q6c — Solution
15 marks · Section B
Question
Calculate the moment of inertia of a uniform solid cylinder of mass , radius and length with respect to a set of axes passing through the centre of the cylinder, where -axis is the axis of the cylinder and is the constant density at any point of the cylinder. Also find for which the moment of inertia about - or -axis will be minimum for a given mass of the cylinder.
Technique
Integrate in cylindrical coordinates for the inertia tensor of the solid cylinder, then minimise the transverse moment under the constraint of fixed mass (hence fixed volume, since density is constant) using single-variable calculus.
Solution
Place the centre at the origin, axis along , so the cylinder occupies , , (use for the cylindrical radius). Density is constant; the volume element is .
Moment of inertia about the -axis (the cylinder axis)
Perpendicular distance from the -axis is :
= \rho\cdot 2\pi\cdot L\cdot\frac{R^4}{4} = \frac{\pi\rho L R^4}{2}.$$ Substituting $\rho = M/(\pi R^2 L)$: $$\boxed{I_{zz} = \tfrac12 M R^2.}$$ ### Moment of inertia about the $x$-axis (= $y$-axis by symmetry) Distance$^2$ from the $x$-axis is $y^2 + z^2$, with $y = s\sin\phi$: $$I_{xx} = \int\rho\,(y^2 + z^2)\,dV = \rho\int_0^R\!\!\int_0^{2\pi}\!\!\int_{-L/2}^{L/2}\big(s^2\sin^2\phi + z^2\big)s\,dz\,d\phi\,ds.$$ Split: - $\displaystyle\int s^2\sin^2\phi\,dV = \rho\Big(\int_0^R s^3\,ds\Big)\Big(\int_0^{2\pi}\sin^2\phi\,d\phi\Big)\Big(\int_{-L/2}^{L/2}dz\Big) = \rho\cdot\frac{R^4}{4}\cdot\pi\cdot L = \frac{\pi\rho L R^4}{4}.$ - $\displaystyle\int z^2\,dV = \rho\Big(\int_0^R s\,ds\Big)\Big(\int_0^{2\pi}d\phi\Big)\Big(\int_{-L/2}^{L/2}z^2\,dz\Big) = \rho\cdot\frac{R^2}{2}\cdot 2\pi\cdot\frac{L^3}{12} = \frac{\pi\rho R^2 L^3}{12}.$ Add and substitute $\pi\rho = M/(R^2 L)$: $$I_{xx} = \frac{\pi\rho L R^4}{4} + \frac{\pi\rho R^2 L^3}{12} = \frac{M R^2}{4} + \frac{M L^2}{12} = \frac{M}{12}\big(3R^2 + L^2\big).$$ $$\boxed{I_{xx} = I_{yy} = \frac{M}{12}\big(3R^2 + L^2\big),\qquad I_{zz} = \frac12 M R^2.}$$ (Off-diagonal products of inertia vanish by symmetry, so these are the principal moments.) ### Minimising $I_{xx}$ for given mass "For a given mass" with constant density means the **volume is fixed**: $V = \pi R^2 L = \text{const}$. So $R^2 = \dfrac{V}{\pi L}$. Substitute into $I_{xx}$: $$I_{xx}(L) = \frac{M}{12}\left(\frac{3V}{\pi L} + L^2\right).$$ Differentiate w.r.t. $L$ and set to zero: $$\frac{dI_{xx}}{dL} = \frac{M}{12}\left(-\frac{3V}{\pi L^2} + 2L\right) = 0 \implies 2L = \frac{3V}{\pi L^2} \implies L^3 = \frac{3V}{2\pi}.$$ The second derivative $\dfrac{M}{12}\big(\tfrac{6V}{\pi L^3} + 2\big) > 0$, so this is a minimum. Now $V = \pi R^2 L$, so $\dfrac{V}{\pi} = R^2 L$, giving $L^3 = \dfrac{3R^2 L}{2}$, i.e. $$L^2 = \frac{3}{2}R^2 \implies \frac{L}{R} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2} \approx 1.2247.$$ $$\boxed{\dfrac{L}{R} = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt6}{2}.}$$ ## Answer $I_{zz} = \tfrac12 MR^2$; $\;I_{xx} = I_{yy} = \tfrac{M}{12}(3R^2 + L^2)$. The transverse moment is minimised (for fixed mass/volume) when $\dfrac{L}{R} = \sqrt{3/2} = \dfrac{\sqrt6}{2}$.