← 2025 Paper 2
UPSC Maths 2025 Paper 2 Q5d — Solution 10 marks · Section B
Question
A bead of mass m m m x = a ( θ − sin θ ) x = a(\theta - \sin\theta) x = a ( θ − sin θ ) y = a ( 1 + cos θ ) y = a(1 + \cos\theta) y = a ( 1 + cos θ ) ( 0 ≤ θ ≤ 2 π ) (0 \leq \theta \leq 2\pi) ( 0 ≤ θ ≤ 2 π ) d 2 u d t 2 + g 4 a u = 0 \dfrac{d^2 u}{dt^2} + \dfrac{g}{4a}u = 0 d t 2 d 2 u + 4 a g u = 0 u = cos ( θ 2 ) u = \cos\left(\dfrac{\theta}{2}\right) u = cos ( 2 θ )
Technique
Use Lagrangian mechanics with the single generalised coordinate θ \theta θ L = T − V L = T - V L = T − V u = cos ( θ / 2 ) u = \cos(\theta/2) u = cos ( θ /2 )
Solution
Kinetic and potential energy
Differentiate the parametric coordinates:
x ˙ = a ( 1 − cos θ ) θ ˙ , y ˙ = − a sin θ θ ˙ . \dot x = a(1 - \cos\theta)\dot\theta,\qquad \dot y = -a\sin\theta\,\dot\theta. x ˙ = a ( 1 − cos θ ) θ ˙ , y ˙ = − a sin θ θ ˙ .
Speed squared:
x ˙ 2 + y ˙ 2 = a 2 θ ˙ 2 [ ( 1 − cos θ ) 2 + sin 2 θ ] . \dot x^2 + \dot y^2 = a^2\dot\theta^2\big[(1-\cos\theta)^2 + \sin^2\theta\big]. x ˙ 2 + y ˙ 2 = a 2 θ ˙ 2 [ ( 1 − cos θ ) 2 + sin 2 θ ] .
Expand the bracket:
( 1 − cos θ ) 2 + sin 2 θ = 1 − 2 cos θ + cos 2 θ + sin 2 θ = 2 − 2 cos θ = 2 ( 1 − cos θ ) . (1-\cos\theta)^2 + \sin^2\theta = 1 - 2\cos\theta + \cos^2\theta + \sin^2\theta = 2 - 2\cos\theta = 2(1-\cos\theta). ( 1 − cos θ ) 2 + sin 2 θ = 1 − 2 cos θ + cos 2 θ + sin 2 θ = 2 − 2 cos θ = 2 ( 1 − cos θ ) .
Using 1 − cos θ = 2 sin 2 ( θ / 2 ) 1 - \cos\theta = 2\sin^2(\theta/2) 1 − cos θ = 2 sin 2 ( θ /2 ) x ˙ 2 + y ˙ 2 = a 2 θ ˙ 2 ⋅ 4 sin 2 ( θ / 2 ) . \dot x^2 + \dot y^2 = a^2\dot\theta^2\cdot 4\sin^2(\theta/2). x ˙ 2 + y ˙ 2 = a 2 θ ˙ 2 ⋅ 4 sin 2 ( θ /2 ) .
Kinetic energy:
T = 1 2 m ( x ˙ 2 + y ˙ 2 ) = 2 m a 2 sin 2 ( θ 2 ) θ ˙ 2 . T = \tfrac12 m(\dot x^2 + \dot y^2) = 2ma^2\sin^2\!\left(\tfrac{\theta}{2}\right)\dot\theta^2. T = 2 1 m ( x ˙ 2 + y ˙ 2 ) = 2 m a 2 sin 2 ( 2 θ ) θ ˙ 2 .
Potential energy (taking V = m g y V = mgy V = m g y y y y V = m g a ( 1 + cos θ ) = m g a ⋅ 2 cos 2 ( θ 2 ) = 2 m g a cos 2 ( θ 2 ) . V = mg\,a(1 + \cos\theta) = mga\cdot 2\cos^2\!\left(\tfrac{\theta}{2}\right) = 2mga\cos^2\!\left(\tfrac{\theta}{2}\right). V = m g a ( 1 + cos θ ) = m g a ⋅ 2 cos 2 ( 2 θ ) = 2 m g a cos 2 ( 2 θ ) .
Lagrangian
L = T − V = 2 m a 2 sin 2 ( θ 2 ) θ ˙ 2 − 2 m g a cos 2 ( θ 2 ) . \boxed{\,L = T - V = 2ma^2\sin^2\!\left(\tfrac{\theta}{2}\right)\dot\theta^2 - 2mga\cos^2\!\left(\tfrac{\theta}{2}\right).} L = T − V = 2 m a 2 sin 2 ( 2 θ ) θ ˙ 2 − 2 m g a cos 2 ( 2 θ ) .
Equation of motion
Euler–Lagrange: d d t ∂ L ∂ θ ˙ − ∂ L ∂ θ = 0 \dfrac{d}{dt}\dfrac{\partial L}{\partial\dot\theta} - \dfrac{\partial L}{\partial\theta} = 0 d t d ∂ θ ˙ ∂ L − ∂ θ ∂ L = 0
∂ L ∂ θ ˙ = 4 m a 2 sin 2 ( θ 2 ) θ ˙ . \frac{\partial L}{\partial\dot\theta} = 4ma^2\sin^2\!\left(\tfrac{\theta}{2}\right)\dot\theta. ∂ θ ˙ ∂ L = 4 m a 2 sin 2 ( 2 θ ) θ ˙ . d d t ∂ L ∂ θ ˙ = 4 m a 2 sin 2 ( θ 2 ) θ ¨ + 4 m a 2 sin ( θ 2 ) cos ( θ 2 ) θ ˙ 2 . \frac{d}{dt}\frac{\partial L}{\partial\dot\theta} = 4ma^2\sin^2\!\left(\tfrac{\theta}{2}\right)\ddot\theta + 4ma^2\sin\!\left(\tfrac{\theta}{2}\right)\cos\!\left(\tfrac{\theta}{2}\right)\dot\theta^2. d t d ∂ θ ˙ ∂ L = 4 m a 2 sin 2 ( 2 θ ) θ ¨ + 4 m a 2 sin ( 2 θ ) cos ( 2 θ ) θ ˙ 2 .
For ∂ L / ∂ θ \partial L/\partial\theta ∂ L / ∂ θ d d θ sin 2 ( θ / 2 ) = sin ( θ / 2 ) cos ( θ / 2 ) \tfrac{d}{d\theta}\sin^2(\theta/2) = \sin(\theta/2)\cos(\theta/2) d θ d sin 2 ( θ /2 ) = sin ( θ /2 ) cos ( θ /2 ) d d θ cos 2 ( θ / 2 ) = − sin ( θ / 2 ) cos ( θ / 2 ) \tfrac{d}{d\theta}\cos^2(\theta/2) = -\sin(\theta/2)\cos(\theta/2) d θ d cos 2 ( θ /2 ) = − sin ( θ /2 ) cos ( θ /2 ) ∂ L ∂ θ = 2 m a 2 θ ˙ 2 sin ( θ 2 ) cos ( θ 2 ) + 2 m g a sin ( θ 2 ) cos ( θ 2 ) . \frac{\partial L}{\partial\theta} = 2ma^2\dot\theta^2\sin\!\left(\tfrac{\theta}{2}\right)\cos\!\left(\tfrac{\theta}{2}\right) + 2mga\sin\!\left(\tfrac{\theta}{2}\right)\cos\!\left(\tfrac{\theta}{2}\right). ∂ θ ∂ L = 2 m a 2 θ ˙ 2 sin ( 2 θ ) cos ( 2 θ ) + 2 m g a sin ( 2 θ ) cos ( 2 θ ) .
Subtract:
4 m a 2 sin 2 ( θ 2 ) θ ¨ + 2 m a 2 sin ( θ 2 ) cos ( θ 2 ) θ ˙ 2 − 2 m g a sin ( θ 2 ) cos ( θ 2 ) = 0. 4ma^2\sin^2\!\left(\tfrac{\theta}{2}\right)\ddot\theta + 2ma^2\sin\!\left(\tfrac{\theta}{2}\right)\cos\!\left(\tfrac{\theta}{2}\right)\dot\theta^2 - 2mga\sin\!\left(\tfrac{\theta}{2}\right)\cos\!\left(\tfrac{\theta}{2}\right) = 0. 4 m a 2 sin 2 ( 2 θ ) θ ¨ + 2 m a 2 sin ( 2 θ ) cos ( 2 θ ) θ ˙ 2 − 2 m g a sin ( 2 θ ) cos ( 2 θ ) = 0.
Divide by 2 m a sin ( θ / 2 ) 2ma\sin(\theta/2) 2 ma sin ( θ /2 ) sin ( θ / 2 ) ≠ 0 \sin(\theta/2)\neq 0 sin ( θ /2 ) = 0 2a\sin\!\left(\tfrac{\theta}{2}\right)\ddot\theta + a\cos\!\left(\tfrac{\theta}{2}\right)\dot\theta^2 - g\cos\!\left(\tfrac{\theta}{2}\right) = 0. \tag{$\ast$}
Reduction to SHM
Let u = cos ( θ / 2 ) u = \cos(\theta/2) u = cos ( θ /2 ) u ˙ = − 1 2 sin ( θ 2 ) θ ˙ , \dot u = -\tfrac12\sin\!\left(\tfrac{\theta}{2}\right)\dot\theta, u ˙ = − 2 1 sin ( 2 θ ) θ ˙ , u ¨ = − 1 2 sin ( θ 2 ) θ ¨ − 1 4 cos ( θ 2 ) θ ˙ 2 . \ddot u = -\tfrac12\sin\!\left(\tfrac{\theta}{2}\right)\ddot\theta - \tfrac14\cos\!\left(\tfrac{\theta}{2}\right)\dot\theta^2. u ¨ = − 2 1 sin ( 2 θ ) θ ¨ − 4 1 cos ( 2 θ ) θ ˙ 2 .
Multiply ( ∗ ) (\ast) ( ∗ ) − 1 4 -\tfrac14 − 4 1 − 1 2 a sin ( θ 2 ) θ ¨ − 1 4 a cos ( θ 2 ) θ ˙ 2 + 1 4 g cos ( θ 2 ) = 0. -\tfrac12 a\sin\!\left(\tfrac{\theta}{2}\right)\ddot\theta - \tfrac14 a\cos\!\left(\tfrac{\theta}{2}\right)\dot\theta^2 + \tfrac14 g\cos\!\left(\tfrac{\theta}{2}\right) = 0. − 2 1 a sin ( 2 θ ) θ ¨ − 4 1 a cos ( 2 θ ) θ ˙ 2 + 4 1 g cos ( 2 θ ) = 0.
The first two terms are exactly a u ¨ a\,\ddot u a u ¨ cos ( θ / 2 ) = u \cos(\theta/2) = u cos ( θ /2 ) = u a u ¨ + g 4 u = 0. a\,\ddot u + \tfrac{g}{4}u = 0. a u ¨ + 4 g u = 0.
Divide by a a a d 2 u d t 2 + g 4 a u = 0. \boxed{\,\frac{d^2 u}{dt^2} + \frac{g}{4a}u = 0.} d t 2 d 2 u + 4 a g u = 0.
This is SHM with angular frequency ω = g / 4 a = 1 2 g / a \omega = \sqrt{g/4a} = \tfrac12\sqrt{g/a} ω = g /4 a = 2 1 g / a 4 π a / g 4\pi\sqrt{a/g} 4 π a / g
Answer
Lagrangian: L = 2 m a 2 sin 2 ( θ / 2 ) θ ˙ 2 − 2 m g a cos 2 ( θ / 2 ) L = 2ma^2\sin^2(\theta/2)\,\dot\theta^2 - 2mga\cos^2(\theta/2) L = 2 m a 2 sin 2 ( θ /2 ) θ ˙ 2 − 2 m g a cos 2 ( θ /2 )
The equation of motion reduces to d 2 u d t 2 + g 4 a u = 0 \dfrac{d^2u}{dt^2} + \dfrac{g}{4a}u = 0 d t 2 d 2 u + 4 a g u = 0 u = cos ( θ / 2 ) u = \cos(\theta/2) u = cos ( θ /2 )