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UPSC 2025 Maths Optional Paper 2 Q6c — Step-by-Step Solution 15 marks · Section B
Moment of inertia · Mechanics & Fluid Dynamics · asked 7× in 13 yrs · Read the full method →
Question
Calculate the moment of inertia of a uniform solid cylinder of mass M M M R R R L L L z z z ρ \rho ρ L R \dfrac{L}{R} R L x x x y y y
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 z z z 0 ≤ s ≤ R 0\le s\le R 0 ≤ s ≤ R 0 ≤ ϕ ≤ 2 π 0\le\phi\le 2\pi 0 ≤ ϕ ≤ 2 π − L / 2 ≤ z ≤ L / 2 -L/2\le z\le L/2 − L /2 ≤ z ≤ L /2 s s s ρ = M π R 2 L \rho = \dfrac{M}{\pi R^2 L} ρ = π R 2 L M d V = s d s d ϕ d z dV = s\,ds\,d\phi\,dz d V = s d s d ϕ d z
Moment of inertia about the z z z
Perpendicular distance from the z z z s s s
I z z = ∫ ρ s 2 d V = ρ ∫ 0 R ∫ 0 2 π ∫ − L / 2 L / 2 s 2 ⋅ s d z d ϕ d s = ρ ⋅ 2 π ⋅ L ⋅ R 4 4 = π ρ L R 4 2 . I_{zz} = \int \rho\, s^2\,dV = \rho\int_0^{R}\!\!\int_0^{2\pi}\!\!\int_{-L/2}^{L/2} s^2\cdot s\,dz\,d\phi\,ds
= \rho\cdot 2\pi\cdot L\cdot\frac{R^4}{4} = \frac{\pi\rho L R^4}{2}. I z z = ∫ ρ s 2 d V = ρ ∫ 0 R ∫ 0 2 π ∫ − L /2 L /2 s 2 ⋅ s d z d ϕ d s = ρ ⋅ 2 π ⋅ L ⋅ 4 R 4 = 2 π ρ L R 4 . Substituting ρ = M / ( π R 2 L ) \rho = M/(\pi R^2 L) ρ = M / ( π R 2 L )
I z z = 1 2 M R 2 . \boxed{I_{zz} = \tfrac12 M R^2.} I z z = 2 1 M R 2 . Moment of inertia about the x x x y y y
Distance2 ^2 2 x x x y 2 + z 2 y^2 + z^2 y 2 + z 2 y = s sin ϕ y = s\sin\phi y = s sin ϕ
I x x = ∫ ρ ( y 2 + z 2 ) d V = ρ ∫ 0 R ∫ 0 2 π ∫ − L / 2 L / 2 ( s 2 sin 2 ϕ + z 2 ) s d z d ϕ d s . 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. I xx = ∫ ρ ( y 2 + z 2 ) d V = ρ ∫ 0 R ∫ 0 2 π ∫ − L /2 L /2 ( s 2 sin 2 ϕ + z 2 ) s d z d ϕ d s . Split:
∫ s 2 sin 2 ϕ d V = ρ ( ∫ 0 R s 3 d s ) ( ∫ 0 2 π sin 2 ϕ d ϕ ) ( ∫ − L / 2 L / 2 d z ) = ρ ⋅ R 4 4 ⋅ π ⋅ L = π ρ L R 4 4 . \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}. ∫ s 2 sin 2 ϕ d V = ρ ( ∫ 0 R s 3 d s ) ( ∫ 0 2 π sin 2 ϕ d ϕ ) ( ∫ − L /2 L /2 d z ) = ρ ⋅ 4 R 4 ⋅ π ⋅ L = 4 π ρ L R 4 . ∫ z 2 d V = ρ ( ∫ 0 R s d s ) ( ∫ 0 2 π d ϕ ) ( ∫ − L / 2 L / 2 z 2 d z ) = ρ ⋅ R 2 2 ⋅ 2 π ⋅ L 3 12 = π ρ R 2 L 3 12 . \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}. ∫ z 2 d V = ρ ( ∫ 0 R s d s ) ( ∫ 0 2 π d ϕ ) ( ∫ − L /2 L /2 z 2 d z ) = ρ ⋅ 2 R 2 ⋅ 2 π ⋅ 12 L 3 = 12 π ρ R 2 L 3 .
Add and substitute π ρ = M / ( R 2 L ) \pi\rho = M/(R^2 L) π ρ = M / ( R 2 L )
I x x = π ρ L R 4 4 + π ρ R 2 L 3 12 = M R 2 4 + M L 2 12 = M 12 ( 3 R 2 + L 2 ) . 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). I xx = 4 π ρ L R 4 + 12 π ρ R 2 L 3 = 4 M R 2 + 12 M L 2 = 12 M ( 3 R 2 + L 2 ) . I x x = I y y = M 12 ( 3 R 2 + L 2 ) , I z z = 1 2 M R 2 . \boxed{I_{xx} = I_{yy} = \frac{M}{12}\big(3R^2 + L^2\big),\qquad I_{zz} = \frac12 M R^2.} I xx = I y y = 12 M ( 3 R 2 + L 2 ) , I z z = 2 1 M R 2 . (Off-diagonal products of inertia vanish by symmetry, so these are the principal moments.)
Minimising I x x I_{xx} I xx
“For a given mass” with constant density means the volume is fixed : V = π R 2 L = const V = \pi R^2 L = \text{const} V = π R 2 L = const R 2 = V π L R^2 = \dfrac{V}{\pi L} R 2 = π L V I x x I_{xx} I xx
I x x ( L ) = M 12 ( 3 V π L + L 2 ) . I_{xx}(L) = \frac{M}{12}\left(\frac{3V}{\pi L} + L^2\right). I xx ( L ) = 12 M ( π L 3 V + L 2 ) . Differentiate w.r.t. L L L
d I x x d L = M 12 ( − 3 V π L 2 + 2 L ) = 0 ⟹ 2 L = 3 V π L 2 ⟹ L 3 = 3 V 2 π . \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}. d L d I xx = 12 M ( − π L 2 3 V + 2 L ) = 0 ⟹ 2 L = π L 2 3 V ⟹ L 3 = 2 π 3 V . The second derivative M 12 ( 6 V π L 3 + 2 ) > 0 \dfrac{M}{12}\big(\tfrac{6V}{\pi L^3} + 2\big) > 0 12 M ( π L 3 6 V + 2 ) > 0
Now V = π R 2 L V = \pi R^2 L V = π R 2 L V π = R 2 L \dfrac{V}{\pi} = R^2 L π V = R 2 L L 3 = 3 R 2 L 2 L^3 = \dfrac{3R^2 L}{2} L 3 = 2 3 R 2 L
L 2 = 3 2 R 2 ⟹ L R = 3 2 = 6 2 ≈ 1.2247. L^2 = \frac{3}{2}R^2 \implies \frac{L}{R} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2} \approx 1.2247. L 2 = 2 3 R 2 ⟹ R L = 2 3 = 2 6 ≈ 1.2247. L R = 3 2 = 6 2 . \boxed{\dfrac{L}{R} = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt6}{2}.} R L = 2 3 = 2 6 . Answer
I z z = 1 2 M R 2 I_{zz} = \tfrac12 MR^2 I z z = 2 1 M R 2 I x x = I y y = M 12 ( 3 R 2 + L 2 ) \;I_{xx} = I_{yy} = \tfrac{M}{12}(3R^2 + L^2) I xx = I y y = 12 M ( 3 R 2 + L 2 ) L R = 3 / 2 = 6 2 \dfrac{L}{R} = \sqrt{3/2} = \dfrac{\sqrt6}{2} R L = 3/2 = 2 6