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UPSC 2021 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Equilibrium of a system of particles · Dynamics & Statics · asked 8× in 13 yrs · Read the full method →

Question

Two rods $LM$ and $MN$ joined rigidly at $M$ with $(LM)^2+(MN)^2=(LN)^2$ are hanged freely from $L$. Uniform density $\omega$/length. Determine the angle $LM$ makes with the vertical.

Technique

Compound CG must lie below pivot for equilibrium; coordinate the two rods with $LM$ at angle $\theta$ from vertical and $MN$ perpendicular; solve $\text{CG}_x=0$ for $\tan\theta$.

Solution

Setup. The condition $(LM)^2+(MN)^2=(LN)^2$ means triangle $LMN$ has a right angle at $M$ (Pythagoras). The rods are rigidly joined at $M$.

Let $LM=a$, $MN=b$, $LN=c$ with $a^2+b^2=c^2$.

The compound rod hangs from $L$. In equilibrium, the centre of gravity of the system lies directly below $L$.

Step 1 — Centre of gravity

Weight of $LM$: $\omega a$ at midpoint of $LM$, which is at distance $a/2$ along $LM$ from $L$.

Weight of $MN$: $\omega b$ at midpoint of $MN$, which is at $M$ plus $b/2$ along $MN$. Since $LM\perp MN$, this midpoint is at perpendicular distance $b/2$ from $LM$.

In coordinates with $L$ at origin, $LM$ along some direction at angle $\theta$ with vertical:

Take $LM$ to be along direction $(\sin\theta,-\cos\theta)$ (down-and-tilt). So midpoint of $LM$ is at $(a/2)(\sin\theta,-\cos\theta)$.

$M$ is at $a(\sin\theta,-\cos\theta)$.

$MN$ perpendicular to $LM$: direction $(\cos\theta,\sin\theta)$ or $(-\cos\theta,-\sin\theta)$. Take the one such that the CG is on the “lower” side (gravity pulls the joint outward).

Midpoint of $MN$: $M + (b/2)\cdot(\text{direction of }MN)$.

Step 2 — Equilibrium condition

CG must lie below $L$ (i.e., on the vertical through $L$, which is $(0,-y)$ axis for some $y>0$).

CG = $\dfrac{\omega a\cdot G_1+\omega b\cdot G_2}{\omega(a+b)}$ where $G_1=(a/2)(\sin\theta,-\cos\theta)$ and $G_2=a(\sin\theta,-\cos\theta)+(b/2)(\cos\theta,\sin\theta)$ (taking $MN$ in direction $(\cos\theta,\sin\theta)$).

Hmm — let me reconsider. Both $LM$ and $MN$ are physical rods; the angle $\theta$ between $LM$ and the vertical is what we want to find.

Let me set up more carefully: $L$ at origin, vertical = downward $y$-axis (so down is $+y$). $LM$ makes angle $\theta$ with downward vertical, so $M=(a\sin\theta, a\cos\theta)$ (with $a\cos\theta$ being depth below $L$).

$MN$ perpendicular to $LM$, but the direction of $MN$ from $M$ is one of two perpendicular choices. Let $MN$ from $M$ in direction making angle $\theta+\pi/2$ with vertical (i.e., perpendicular to $LM$, rotated clockwise). Then $N=M+b(\sin(\theta+\pi/2),\cos(\theta+\pi/2))=M+b(\cos\theta,-\sin\theta)$.

So $N=(a\sin\theta+b\cos\theta,\;a\cos\theta-b\sin\theta)$.

Midpoint of $MN$: $M+\dfrac{b}{2}(\cos\theta,-\sin\theta)=(a\sin\theta+\dfrac{b\cos\theta}{2},\;a\cos\theta-\dfrac{b\sin\theta}{2})$.

Midpoint of $LM$: $(\dfrac{a\sin\theta}{2},\dfrac{a\cos\theta}{2})$.

Step 3 — CG of system

$\text{CG}_x=\dfrac{(\omega a)(a\sin\theta/2)+(\omega b)(a\sin\theta+b\cos\theta/2)}{\omega(a+b)}$.

$=\dfrac{a^2\sin\theta/2+ab\sin\theta+b^2\cos\theta/2}{a+b}$.

For CG below $L$ (on vertical), $\text{CG}_x=0$: $\dfrac{a^2\sin\theta}{2}+ab\sin\theta+\dfrac{b^2\cos\theta}{2}=0$,

$\sin\theta(\dfrac{a^2}{2}+ab)+\dfrac{b^2}{2}\cos\theta=0$,

$\sin\theta\cdot\dfrac{a(a+2b)}{2}=-\dfrac{b^2}{2}\cos\theta$,

$\tan\theta=-\dfrac{b^2}{a(a+2b)}$.

The negative sign indicates the rod hangs to the other side; we want magnitude:

$|\tan\theta|=\dfrac{b^2}{a(a+2b)}$.

Or, with appropriate orientation (I had the $MN$ direction in the “right” rotation; flipping to clockwise direction gives the other sign):

Answer