JEE Advanced 2025 — Paper 1 — PHY-I solution

JEE Advanced2025Paper 1PHY-I

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 $R > r$ 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 $T = \frac{2 π}{ω}$ . The correct expression for $ω$ is ( $g$ is the acceleration due to gravity):

A
$\frac{2}{3} (\frac{g}{R - r} + \frac{k}{m})$
B
$\frac{2 g}{3 ( R - r )} + \frac{k}{m}$
C
$\frac{1}{6} (\frac{g}{R - r} + \frac{k}{m})$
D
$\frac{1}{4} (\frac{g}{R - r} + \frac{k}{m})$

Solution

Let $θ$ be the small angular displacement of the disk's center from the lowest point, measured about the ring center. The arc length displacement of the contact point is $x = (R - r) θ$ , and pure rolling gives the disk's angular velocity about its own center as $ω_{d} = v / r$ , where $v = (R - r) \dot{θ}$ .

Using energy conservation for small oscillations:

$E = \frac{1}{2} k (R - r)^{2} θ^{2} + m g (R - r) (1 - cos θ) + \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{1}{2} m r^{2}) ω_{d}^{2}$

For small $θ$ , use $1 - cos θ \approx θ^{2} /2$ , and the rolling constraint gives the rotational KE as $\frac{1}{4} m v^{2}$ . So the total KE is $\frac{3}{4} m v^{2} = \frac{3}{4} m (R - r)^{2} \dot{θ}^{2}$ .

Differentiating $E$ with respect to time and dividing through by $\dot{θ}$ :

$\frac{3}{2} m (R - r)^{2} \ddot{θ} + [k (R - r)^{2} + m g (R - r)] θ = 0$

$\ddot{θ} + \frac{2}{3} (\frac{k}{m} + \frac{g}{R - r}) θ = 0$

$∴ ω = \frac{2}{3} (\frac{g}{R - r} + \frac{k}{m})$ , which matches option (A).

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