UPSC 2021 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

D'Alembert's Principle · Mechanics & Fluid Dynamics · Read the full method →

Question

Particle constrained to circle in vertical $xy$ -plane. Using D’Alembert’s principle, show equation of motion is $\ddot x y-\ddot y x-gx=0$ .

Technique

D’Alembert’s principle: $(\vec F_{\text{applied}}-m\ddot{\vec r})\cdot\hat\tau=0$ along the constraint tangent.

Solution

Setup. Particle of mass $m$ on a circle in $xy$ -plane. Gravity $\vec g=-g\hat\jmath$ (downward, so $\hat\jmath$ is up, $g>0$ standard).

Wait — depending on convention. Let me re-read. The result $\ddot x y-\ddot y x-gx=0$ . The $-gx$ term suggests gravity acts in the $-x$ direction or the angle parametrisation is such.

Let me parametrise the particle position as $(x,y)$ on a circle of radius $a$ centred at origin: $x=a\cos\theta$ , $y=a\sin\theta$ .

Step 1 — D’Alembert’s principle

For a particle constrained to a curve, the generalised force on the constraint direction must vanish — i.e., the projection of “applied force minus inertial force” on the tangent direction is zero.

Constraint: $x^2+y^2=a^2$ (circle). Tangent direction: perpendicular to radial $(x,y)$ , i.e., $(-y,x)$ (up to sign).

Applied force: gravity $(0,-mg)$ (assuming $y$ up).

Inertial force: $-(m\ddot x,m\ddot y)$ .

Net force in tangent direction: $(applied-m\ddot{\vec r})\cdot\hat\tau=0$ .

$\vec F_{\text{applied}}-m\ddot{\vec r}=(0-m\ddot x,-mg-m\ddot y)=-m(\ddot x,\ddot y+g)$ .

Projection onto tangent $(-y,x)$ (any positive scaling): $-m(\ddot x\cdot(-y)+(\ddot y+g)\cdot x)=-m(-y\ddot x+x\ddot y+gx)=0$ .

So $-y\ddot x+x\ddot y+gx=0$ , i.e., $x\ddot y-y\ddot x+gx=0$ .

Rearrange: $y\ddot x-x\ddot y-gx=0$ , or $\ddot x y-\ddot y x-gx=0$ . ✓ Matches the given form.

Answer

\boxed{\;\ddot x\,y-\ddot y\,x-gx=0.\;}