JEE Advanced 2025 Paper 1 Q01 Physics Rotational Mechanics Combined Rotation Hard

JEE Advanced 2025 Paper 1 · Q01 · Combined Rotation

The center of a disk of radius $r$ and mass $m$ is attached to a spring of spring constant $k$, inside a ring of radius $R > r$. The other end of the spring is attached to the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can roll along the inside periphery of the ring without slipping. The spring can only be stretched or compressed along the periphery of the ring (Hooke's law). In equilibrium, the disk is at the bottom of the ring. For small displacements of the disk, the time period of oscillation of its center of mass is $T = 2\pi/\omega$. Find $\omega$ ($g$ = acceleration due to gravity).

  1. A. $\sqrt{\tfrac{2}{3}\left(\tfrac{g}{R-r} + \tfrac{k}{m}\right)}$
  2. B. $\sqrt{\tfrac{2g}{3(R-r)} + \tfrac{k}{m}}$
  3. C. $\sqrt{\tfrac{1}{6}\left(\tfrac{g}{R-r} + \tfrac{k}{m}\right)}$
  4. D. $\sqrt{\tfrac{1}{4}\left(\tfrac{g}{R-r} + \tfrac{k}{m}\right)}$
Reveal answer + step-by-step solution

Correct answer:A

Solution

Let $\theta$ be the small angular displacement of the disk's CM measured from the bottom of the ring's center. The CM of the disk moves on a circle of radius $(R-r)$. KE has translational and rotational parts; for rolling without slipping on the inside of the ring with disk moment of inertia $I = mr^2/2$, total KE $= \tfrac{1}{2}\cdot\tfrac{3m}{2}(R-r)^2(\dot\theta)^2 = \tfrac{3}{4} m (R-r)^2 \dot\theta^2$. PE: gravitational $\approx \tfrac{1}{2} m g (R-r)\theta^2$ (small-angle pendulum-like term). Spring PE: arc-length stretch along the ring is approximately $R\theta$ (spring along the periphery; for small motion the relative arc between disk attachment and fixed end is proportional to $(R-r)\theta$ with effective stretch term giving $\tfrac{1}{2} k (R-r)^2 \theta^2$ in the small-oscillation expansion). Lagrangian gives equation of motion $\tfrac{3}{2} m (R-r)^2 \ddot\theta = -\bigl[m g (R-r) + k (R-r)^2\bigr]\theta$, so $\omega^2 = \tfrac{2}{3}\!\left[\tfrac{g}{R-r} + \tfrac{k}{m}\right]$. Hence $\omega = \sqrt{\tfrac{2}{3}\left(\tfrac{g}{R-r} + \tfrac{k}{m}\right)}$. Answer: (A).

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