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UPSC 2021 Maths Optional Paper 1 Q7c — Step-by-Step Solution

15 marks · Section B

Constrained motion · Dynamics & Statics · asked 8× in 14 yrs · Read the full method →

Question

Heavy particle hangs by inextensible string of length $a$ from a fixed point; projected horizontally with velocity $\sqrt{2gh}$. If $5a/2>h>a$, prove circular motion ceases at height $\dfrac{1}{3}(a+2h)$ from point of projection. Also prove the greatest height above projection point is $\dfrac{(4a-h)(a+2h)^2}{27 a^2}$.

Technique

Energy conservation gives $v^2$ as function of height; tension condition $T=0$ gives angle where circular motion ceases; subsequent projectile motion gives additional height.

Solution

Setup. Particle at bottom of string (length $a$ from fixed point $O$). Projected horizontally with $v_0=\sqrt{2gh}$.

For circular motion, string must remain taut (tension $\ge 0$).

Let the particle rise through angle $\theta$ from vertical (initial position). Height risen: $a(1-\cos\theta)$.

Wait — re-read: “height … from the point of projection”. So measure heights from projection point (bottom of vertical string).

At angle $\theta$ from vertical:

• Velocity squared: $v^2=v_0^2-2g\cdot a(1-\cos\theta)=2gh-2ga(1-\cos\theta)=2g[h-a(1-\cos\theta)]$.

Step 1 — Condition for circular motion (string taut)

At angle $\theta$, centripetal force = $mv^2/a$. Provided by component of gravity toward $O$ plus tension:

If $\theta<\pi/2$ (below horizontal through $O$): gravity has component $g\cos\theta$ toward $O$.

If $\theta>\pi/2$ (above horizontal): gravity has component $g\cos\theta<0$ (away from $O$, i.e., outward).

Force balance toward $O$: $T+mg\cos\theta=mv^2/a$ for $\theta<\pi/2$.

Wait, careful with signs. Let me redo.

Take $\theta$ as angle from upward vertical (so at start, $\theta=\pi$ means at bottom). Easier: take $\theta$ from downward vertical (so $\theta=0$ at start, $\theta=\pi$ at top).

At angle $\theta$ from downward vertical: position $(a\sin\theta,a-a\cos\theta)$ taking projection point as origin and $y$-up.

Height above projection: $a(1-\cos\theta)$. ✓

Radial direction (away from $O$): from $O=(0,a)$ to particle position $(a\sin\theta,a-a\cos\theta)$, direction $(\sin\theta,-\cos\theta)$.

Gravity = $(0,-mg)$. Radial component: $(0,-mg)\cdot(\sin\theta,-\cos\theta)=mg\cos\theta$.

So radial component of gravity (outward) = $mg\cos\theta$.

For $\theta\in[0,\pi/2)$: $\cos\theta>0$, gravity outward (away from $O$).

Net force toward $O$ (centripetal): $T-mg\cos\theta$ (tension inward, gravity component outward).

For circular motion: $T-mg\cos\theta=mv^2/a$, so $T=m(g\cos\theta+v^2/a)$.

For $T\ge 0$: $v^2\ge-ag\cos\theta$.

For $\theta<\pi/2$, $\cos\theta>0$, so $-ag\cos\theta<0$ ≤ $v^2$ — automatic.

For $\theta>\pi/2$, $\cos\theta<0$, so $-ag\cos\theta>0$, and we need $v^2\ge-ag\cos\theta=ag|\cos\theta|$.

Condition for string to go slack: $v^2=-ag\cos\theta$ (i.e., $T=0$), with $\theta>\pi/2$.

Step 2 — Apply energy + tension condition

$v^2=2g[h-a(1-\cos\theta)]$.

Set $v^2=-ag\cos\theta$:

$2g[h-a+a\cos\theta]=-ag\cos\theta$,

$2h-2a+2a\cos\theta=-a\cos\theta$,

$2h-2a=-3a\cos\theta$,

$\cos\theta=\dfrac{2a-2h}{3a}=\dfrac{2(a-h)}{3a}$.

For $h>a$: $\cos\theta<0$, so $\theta>\pi/2$ ✓ (top half of circle).

Height where slack: $a(1-\cos\theta)=a\!\left(1-\dfrac{2(a-h)}{3a}\right)=a\cdot\dfrac{3a-2a+2h}{3a}=\dfrac{a+2h}{3}$. ✓

So circular motion ceases at height $\dfrac{a+2h}{3}$. $\blacksquare$

Step 3 — Greatest height (projectile motion after string slackens)

At the moment string slackens:

• Height: $H_1=\dfrac{a+2h}{3}$.
• Speed: $v^2=-ag\cos\theta=ag\cdot\dfrac{2(h-a)}{3a}=\dfrac{2g(h-a)}{3}$.
• Direction: tangent to circle, i.e., perpendicular to the radius at angle $\theta$. Tangent direction at angle $\theta$ from downward vertical: $(\cos\theta,\sin\theta)$ (perpendicular to $(\sin\theta,-\cos\theta)$, rotated 90° CCW).

Velocity components at slack: $v_x=v\cos\theta$ (where $\cos\theta=2(a-h)/(3a)<0$ since $h>a$, so $v_x<0$). $v_y=v\sin\theta$ ($\sin\theta>0$ in upper half, so $v_y>0$).

The particle now undergoes projectile motion.

Additional height gained as projectile: $\Delta H=v_y^2/(2g)=v^2\sin^2\theta/(2g)$.

$\sin^2\theta=1-\cos^2\theta=1-\dfrac{4(a-h)^2}{9a^2}=\dfrac{9a^2-4(a-h)^2}{9a^2}$.

Numerator: $9a^2-4(a-h)^2=9a^2-4(a^2-2ah+h^2)=9a^2-4a^2+8ah-4h^2=5a^2+8ah-4h^2$.

Hmm, let me try factoring: $5a^2+8ah-4h^2$. Discriminant ($h$): $64a^2+80a^2=144a^2$, so $h=(-8a\pm 12a)/(-8)$ — giving $h=-a/2$ or $h=5a/2$. So $5a^2+8ah-4h^2=-4(h-5a/2)(h+a/2)=(5a/2-h)(4(h+a/2))=2(5a-2h)(h+a/2)\cdot\ldots$ — let me just expand $(5a-2h)(a+2h)/(\text{something})$.

$(5a-2h)(a+2h)=5a^2+10ah-2ah-4h^2=5a^2+8ah-4h^2$ ✓.

So $\sin^2\theta=\dfrac{(5a-2h)(a+2h)}{9a^2}$.

For $h<5a/2$ (given), $5a-2h>0$, and $a+2h>0$, so $\sin^2\theta>0$ ✓.

$\Delta H=\dfrac{v^2\sin^2\theta}{2g}=\dfrac{(2g(h-a)/3)\cdot(5a-2h)(a+2h)/(9a^2)}{2g}=\dfrac{(h-a)(5a-2h)(a+2h)}{27 a^2}$.

Greatest height: $H_{\max}=H_1+\Delta H=\dfrac{a+2h}{3}+\dfrac{(h-a)(5a-2h)(a+2h)}{27 a^2}$.

$=\dfrac{(a+2h)}{27a^2}\!\left[9a^2+(h-a)(5a-2h)\right]$.

Expand $(h-a)(5a-2h)=5ah-2h^2-5a^2+2ah=7ah-2h^2-5a^2$.

$9a^2+7ah-2h^2-5a^2=4a^2+7ah-2h^2$.

Factor: $-2h^2+7ah+4a^2=(-1)(2h^2-7ah-4a^2)=(-1)(2h+a)(h-4a)=-(2h+a)(h-4a)=(2h+a)(4a-h)$.

So $H_{\max}=\dfrac{(a+2h)(2h+a)(4a-h)}{27 a^2}=\dfrac{(a+2h)^2(4a-h)}{27 a^2}$ ✓.

Answer