UPSC 2020 Maths Optional Paper 2 Q5e — Step-by-Step Solution

10 marks · Section B

Moment of inertia · Mechanics & Fluid Dynamics · asked 7× in 13 yrs · Read the full method →

Question

Prove that the moment of inertia of a triangular lamina $ABC$ about any axis through $A$ in its plane is $\frac{M}{6}(\beta^2+\beta\gamma+\gamma^2)$ , where $M$ is the mass of the lamina and $\beta,\gamma$ are respectively the lengths of perpendiculars from $B$ and $C$ on the axis.

Technique

Affine (area) coordinates $u,v$ ; perpendicular distance is linear, $\delta=u\beta+v\gamma$ ; integrate $\delta^2$ over the standard triangle using the moments $\tfrac1{12},\tfrac1{24}$ .

Solution

Let the axis through $A$ be the reference line; for any point $P$ of the lamina let $\delta(P)$ denote its perpendicular (signed) distance from the axis. The moment of inertia about the axis is

I=\iint \delta^2\,dm=\sigma\iint\delta^2\,dA,

where $\sigma=M/\text{Area}$ is the surface density.

Step 1 — Linear coordinate across the triangle

Set up affine (area) coordinates. Any point $P$ in triangle $ABC$ can be written

\vec{AP}=u\,\vec{AB}+v\,\vec{AC},\qquad u\ge0,\ v\ge0,\ u+v\le1.

Because perpendicular distance from a line through $A$ is a linear function of position vanishing at $A$ , the distance of $P$ from the axis is

\delta(P)=u\,\beta+v\,\gamma,

since the perpendicular distance of $A$ is $0$ , of $B$ (where $u=1,v=0$ ) is $\beta$ , and of $C$ (where $u=0,v=1$ ) is $\gamma$ . (Take $\beta,\gamma$ with appropriate signs if $B,C$ are on opposite sides; the formula below uses their squares and product so the stated result holds for perpendicular lengths in the standard configuration.)

Step 2 — Area element

The map $(u,v)\mapsto P$ has constant Jacobian: $dA=\big|\vec{AB}\times\vec{AC}\big|\,du\,dv=2\Delta\,du\,dv$ , where $\Delta$ is the area of the triangle. The mass element is

dm=\sigma\,dA=\frac{M}{\Delta}\,\Delta\,\cdots

more precisely $dm=\sigma\,2\Delta\,du\,dv$ and $\sigma=M/\Delta$ , so $dm=2M\,du\,dv$ . (Check of total mass: $\displaystyle\iint_{u+v\le1}2M\,du\,dv=2M\cdot\tfrac12=M$ . ✓)

Step 3 — Integrate

I=\iint_{\substack{u,v\ge0\\u+v\le1}}(u\beta+v\gamma)^2\,(2M)\,du\,dv =2M\iint(\beta^2u^2+2\beta\gamma\,uv+\gamma^2v^2)\,du\,dv.

Over the standard triangle $\{u,v\ge0,\ u+v\le1\}$ :

\iint u^2\,du\,dv=\frac{1}{12},\qquad \iint v^2\,du\,dv=\frac{1}{12},\qquad \iint uv\,du\,dv=\frac{1}{24}.

Hence

I=2M\left(\beta^2\cdot\frac1{12}+2\beta\gamma\cdot\frac1{24}+\gamma^2\cdot\frac1{12}\right) =2M\cdot\frac{1}{12}\big(\beta^2+\beta\gamma+\gamma^2\big).

Answer

\boxed{\;I=\frac{M}{6}\big(\beta^2+\beta\gamma+\gamma^2\big).\;}