JEE Challenger
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PhyMedAdv 2025 • Paper 1

Oscillation of a Rolling Disk Attached to Spring Inside a Ring

The center of a disk of radius rr and mass mm is attached to a spring of spring constant kk, inside a ring of radius R>rR > r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T=2πωT = \frac{2\pi}{\omega}. The correct expression for ω\omega is (gg is the acceleration due to gravity):

Question Diagram 1

Options

A

23(gR−r+km)\sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}

Correct
B

2g3(R−r)+km\sqrt{\frac{2g}{3(R-r)}+\frac{k}{m}}

C

16(gR−r+km)\sqrt{\frac{1}{6}\left(\frac{g}{R-r}+\frac{k}{m}\right)}

D

14(gR−r+km)\sqrt{\frac{1}{4}\left(\frac{g}{R-r}+\frac{k}{m}\right)}

Topics & Concepts

OscillationsSHMRolling Motion

Step-by-Step Solution

To find the angular frequency ω\omega of small oscillations of the disk, we can use the energy method.

1. Kinematics of Pure Rolling:

Let θ\theta be the angular displacement of the center of mass of the disk with respect to the vertical passing through the center of the ring. The radius of the circular path traversed by the center of mass of the disk is (R−r)(R - r).

The speed of the center of mass of the disk is given by: v=(R−r)θ˙v = (R - r)\dot{\theta}

For pure rolling without slipping on the inside surface of the ring: v=ωdiskr  ⟹  ωdisk=vr=(R−r)θ˙rv = \omega_{\text{disk}} r \implies \omega_{\text{disk}} = \frac{v}{r} = \frac{(R - r)\dot{\theta}}{r}

2. Kinetic Energy of the Disk:

The moment of inertia of the disk about its center of mass is: Icm=12mr2I_{\text{cm}} = \frac{1}{2}mr^2

The total kinetic energy KK is the sum of translational and rotational kinetic energies: K=12mv2+12Icmωdisk2K = \frac{1}{2}mv^2 + \frac{1}{2}I_{\text{cm}}\omega_{\text{disk}}^2 K=12m[(R−r)θ˙]2+12(12mr2)((R−r)θ˙r)2K = \frac{1}{2}m\left[(R-r)\dot{\theta}\right]^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{(R-r)\dot{\theta}}{r}\right)^2 K=12m(R−r)2θ˙2+14m(R−r)2θ˙2=34m(R−r)2θ˙2K = \frac{1}{2}m(R-r)^2\dot{\theta}^2 + \frac{1}{4}m(R-r)^2\dot{\theta}^2 = \frac{3}{4}m(R-r)^2\dot{\theta}^2

3. Potential Energy of the System:

For a small angular displacement θ\theta, taking the equilibrium position as the reference for gravitational and elastic potential energy:

  • The rise in height of the center of mass is h=(R−r)(1−cos⁡θ)≈12(R−r)θ2h = (R-r)(1 - \cos\theta) \approx \frac{1}{2}(R-r)\theta^2.
  • The extension/compression of the spring along the path is x=(R−r)θx = (R-r)\theta.

Thus, the total potential energy UU of the system is: U=mgh+12kx2≈12mg(R−r)θ2+12k(R−r)2θ2U = mgh + \frac{1}{2}kx^2 \approx \frac{1}{2}mg(R-r)\theta^2 + \frac{1}{2}k(R-r)^2\theta^2 U=12(R−r)[mg+k(R−r)]θ2U = \frac{1}{2}(R-r)\left[mg + k(R-r)\right]\theta^2

4. Equation of Motion:

Since the system is conservative, the total mechanical energy E=K+UE = K + U is constant: E=34m(R−r)2θ˙2+12(R−r)[mg+k(R−r)]θ2=constantE = \frac{3}{4}m(R-r)^2\dot{\theta}^2 + \frac{1}{2}(R-r)\left[mg + k(R-r)\right]\theta^2 = \text{constant}

Differentiating with respect to time tt: dEdt=32m(R−r)2θ˙θ¨+(R−r)[mg+k(R−r)]θθ˙=0\frac{dE}{dt} = \frac{3}{2}m(R-r)^2\dot{\theta}\ddot{\theta} + (R-r)\left[mg + k(R-r)\right]\theta\dot{\theta} = 0

Dividing through by (R−r)2θ˙(R-r)^2\dot{\theta} (for θ˙≠0\dot{\theta} \neq 0): 32mθ¨+[mgR−r+k]θ=0\frac{3}{2}m\ddot{\theta} + \left[\frac{mg}{R-r} + k\right]\theta = 0 θ¨+23(gR−r+km)θ=0\ddot{\theta} + \frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}\right)\theta = 0

Comparing this with the standard simple harmonic motion equation θ¨+ω2θ=0\ddot{\theta} + \omega^2\theta = 0, we obtain: ω=23(gR−r+km)\omega = \sqrt{\frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}\right)}

Therefore, the correct option is (A).

Oscillation of a Rolling Disk Attached to Spring Inside a Ring | Physics PYQ Solution - JEE Challenger