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UPSC 2022 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

Find the moment of inertia of a right circular solid cone about one of its slant sides (generator) in terms of mass $M$, height $h$, and base radius $a$.

Technique

Use inertia tensor at apex (diagonal by symmetry); take general axis direction; $I_{\hat n}=\vec n^T I\vec n=n_x^2 I_x+n_z^2 I_z$.

Solution

Setup. Place cone with apex at origin, axis along positive $z$-axis, base at $z=h$, base radius $a$. The slant side from apex to a base point makes angle $\alpha$ with the axis, where $\tan\alpha=a/h$.

Strategy. Use the perpendicular- and parallel-axis theorems combined with known moments of inertia for the cone about its axis and through the apex perpendicular to the axis.

Step 1 — Standard moments of inertia of cone

Let $\rho$ = density, mass $M=\rho\cdot V=\rho\cdot\tfrac{1}{3}\pi a^2 h$, so $\rho=3M/(\pi a^2 h)$.

Moment of inertia about axis ($z$-axis), $I_z$: Standard result:

Moment of inertia about axis perpendicular to symmetry axis, passing through apex ($I_x$ or $I_y$):

(Standard derivation: integrate $\int(x^2+z^2)\,dm$ over the cone.)

Step 2 — Moment of inertia about slant side

The slant side is a line from apex $(0,0,0)$ to a base point, e.g., $(a,0,h)$. Direction: $(a,0,h)/\sqrt{a^2+h^2}$.

Use the formula for moment of inertia about a general axis through the origin:

where $I$ is the inertia tensor at origin and $\vec n$ is the axis direction.

By symmetry of the cone (rotational about $z$-axis), the inertia tensor at apex is diagonal:

Wait, let me re-derive $I_{x}$ for a cone with apex at origin. Use $I_x=\int(y^2+z^2)dm$.

For a cone, this evaluates (standard) to $\dfrac{3Mh^2}{5}+\dfrac{3Ma^2}{20}=\dfrac{3M(4h^2+a^2)}{20}=\dfrac{3M(a^2+4h^2)}{20}$.

So $I_x=I_y=\dfrac{3M(a^2+4h^2)}{20}$.

Step 3 — Apply to slant axis $\vec n=(a,0,h)/\sqrt{a^2+h^2}$

$n_x=a/\sqrt{a^2+h^2}$, $n_y=0$, $n_z=h/\sqrt{a^2+h^2}$.

$I_{\hat n}=n_x^2 I_x+n_z^2 I_z$ (since $n_y=0$ and inertia tensor diagonal).

$=\dfrac{a^2}{a^2+h^2}\cdot\dfrac{3M(a^2+4h^2)}{20}+\dfrac{h^2}{a^2+h^2}\cdot\dfrac{3Ma^2}{10}$.

$=\dfrac{3M}{20(a^2+h^2)}\!\left[a^2(a^2+4h^2)+2a^2 h^2\right]$.

$=\dfrac{3Ma^2}{20(a^2+h^2)}\!\left[a^2+4h^2+2h^2\right]$.

$=\dfrac{3Ma^2(a^2+6h^2)}{20(a^2+h^2)}$.

Answer