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UPSC 2017 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

Show that the moment of inertia of an elliptic area of mass MM and semi-axis aa and bb about a semi-diameter of length rr is 14Ma2b2r2\dfrac{1}{4}M\dfrac{a^2b^2}{r^2}. Further, prove that the moment of inertia about a tangent is 5M4p2\dfrac{5M}{4}p^2, where pp is the perpendicular distance from the centre of the ellipse to the tangent.

Technique

Iα=∬(perp dist)2 dmI_\alpha=\iint(\text{perp dist})^2\,dm with Ixy=0I_{xy}=0; convert a2sin⁡2α+b2cos⁡2αa^2\sin^2\alpha+b^2\cos^2\alpha to a2b2/r2a^2b^2/r^2 via the ellipse pedal relation; for the tangent use the parallel central diameter plus the parallel-axes theorem.

Solution

Setup. Uniform elliptic lamina x2a2+y2b2≤1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\le1, total mass MM, centre at the origin, axes along Ox,OyOx,Oy. Standard moments of inertia about the principal axes:

IOx=Mb24,IOy=Ma24,I_{Ox}=\frac{Mb^2}{4},\qquad I_{Oy}=\frac{Ma^2}{4},

and the product of inertia about the axes vanishes by symmetry, Ixy=∬xy dm=0I_{xy}=\iint xy\,dm=0.

Step 1 — Moment of inertia about a central diameter

Take a diameter through OO in the direction making angle α\alpha with OxOx. The perpendicular distance of a point (x,y)(x,y) from this line is ∣xsin⁡α−ycos⁡α∣|x\sin\alpha-y\cos\alpha|, so

Iα=∬(xsin⁡α−ycos⁡α)2 dm=sin⁡2α∬x2 dm−2sin⁡αcos⁡α∬xy dm+cos⁡2α∬y2 dm.I_\alpha=\iint(x\sin\alpha-y\cos\alpha)^2\,dm=\sin^2\alpha\iint x^2\,dm-2\sin\alpha\cos\alpha\iint xy\,dm+\cos^2\alpha\iint y^2\,dm.

Now ∬x2 dm=IOy=Ma24\iint x^2\,dm=I_{Oy}=\dfrac{Ma^2}{4}, ∬y2 dm=IOx=Mb24\iint y^2\,dm=I_{Ox}=\dfrac{Mb^2}{4}, and ∬xy dm=0\iint xy\,dm=0. Hence

Iα=Ma24sin⁡2α+Mb24cos⁡2α=M4(a2sin⁡2α+b2cos⁡2α).I_\alpha=\frac{Ma^2}{4}\sin^2\alpha+\frac{Mb^2}{4}\cos^2\alpha=\frac{M}{4}\big(a^2\sin^2\alpha+b^2\cos^2\alpha\big).

Step 2 — Express through the semi-diameter length rr

The semi-diameter in direction α\alpha ends on the ellipse at (rcos⁡α,rsin⁡α)(r\cos\alpha,r\sin\alpha), so

r2cos⁡2αa2+r2sin⁡2αb2=1 ⟹ 1r2=cos⁡2αa2+sin⁡2αb2=b2cos⁡2α+a2sin⁡2αa2b2.\frac{r^2\cos^2\alpha}{a^2}+\frac{r^2\sin^2\alpha}{b^2}=1\ \Longrightarrow\ \frac1{r^2}=\frac{\cos^2\alpha}{a^2}+\frac{\sin^2\alpha}{b^2}=\frac{b^2\cos^2\alpha+a^2\sin^2\alpha}{a^2b^2}.

Therefore a2sin⁡2α+b2cos⁡2α=a2b2r2a^2\sin^2\alpha+b^2\cos^2\alpha=\dfrac{a^2b^2}{r^2}, and substituting into Step 1:

  Idiameter=M4⋅a2b2r2=14 M a2b2r2.  \boxed{\;I_{\text{diameter}}=\frac{M}{4}\cdot\frac{a^2b^2}{r^2}=\frac14\,M\,\frac{a^2b^2}{r^2}.\;}

Step 3 — Moment of inertia about a tangent

Let the tangent touch the ellipse, with pp the perpendicular distance from the centre OO to the tangent. Parametrise the tangent at (acos⁡t,bsin⁡t)(a\cos t,b\sin t) as xcos⁡ta+ysin⁡tb=1\dfrac{x\cos t}{a}+\dfrac{y\sin t}{b}=1; the distance from OO is

p=1cos⁡2ta2+sin⁡2tb2 ⟹ p2=a2b2a2sin⁡2t+b2cos⁡2t.p=\frac{1}{\sqrt{\dfrac{\cos^2 t}{a^2}+\dfrac{\sin^2 t}{b^2}}}\ \Longrightarrow\ p^2=\frac{a^2b^2}{a^2\sin^2 t+b^2\cos^2 t}.

Consider the central diameter parallel to the tangent. Its direction is the tangent direction (−asin⁡t, bcos⁡t)(-a\sin t,\,b\cos t), for which the semi-diameter length r′r' satisfies r′2=a2sin⁡2t+b2cos⁡2tr'^2=a^2\sin^2 t+b^2\cos^2 t. By Step 2 the moment of inertia about this parallel diameter is

Idiam=M4a2b2r′2=M4a2b2a2sin⁡2t+b2cos⁡2t=M4p2.I_{\text{diam}}=\frac{M}{4}\frac{a^2b^2}{r'^2}=\frac{M}{4}\frac{a^2b^2}{a^2\sin^2 t+b^2\cos^2 t}=\frac{M}{4}p^2.

By the parallel-axes theorem, the tangent is parallel to this central diameter at distance pp, so

Itangent=Idiam+Mp2=M4p2+Mp2=5M4p2.I_{\text{tangent}}=I_{\text{diam}}+Mp^2=\frac{M}{4}p^2+Mp^2=\frac{5M}{4}p^2.

Answer

  Itangent=5M4p2.  \boxed{\;I_{\text{tangent}}=\frac{5M}{4}p^2.\;}
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