The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a r

Q: Why is the spring extension exactly $(R-r)\theta$?

The problem states that the spring is stretched along the periphery of the ring. The center of the disk moves along a circular arc of radius $(R-r)$. When the disk's center sweeps an angle $\theta$, the arc length it covers is exactly $(R-r)\theta$, which corresponds to the extension of the spring.

Q: Why is the kinetic energy $\frac{3}{4}mv^2$ instead of just $\frac{1}{2}mv^2$?

Because the disk is rolling, not just sliding. It has translational kinetic energy $\frac{1}{2}mv^2$ and rotational kinetic energy $\frac{1}{2}I\omega^2$. For a uniform disk, $I = \frac{1}{2}mr^2$, and rolling without slipping means $\omega = \frac{v}{r}$. Adding these together gives $\frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.

Q: Why do we differentiate the total energy equation?

In a conservative system, the total mechanical energy $E$ is constant over time. Therefore, its rate of change with respect to time, $\frac{dE}{dt}$, must be zero. Differentiating the energy equation naturally produces the equation of motion (the differential equation for SHM) without having to manually resolve complex torque and force vectors.

Q: What does 'rolling without slipping' mean in this specific geometry?

It means the point on the disk that is instantaneously in contact with the outer ring has zero velocity. This kinematic constraint forces a strict relationship between how fast the disk's center moves along its arc and how fast the disk spins around its own center ($v = \omega_{\text{disk}} r$).

Q: Why do we use the small angle approximation $1-\cos\theta \approx \frac{\theta^2}{2}$?

Simple Harmonic Motion requires a linear restoring force, which corresponds to a purely quadratic potential energy profile ($U \propto \theta^2$). The exact gravitational potential energy involves a cosine term, which makes the motion a non-linear pendulum. For very small displacements, the Taylor expansion allows us to approximate the curve as a perfect parabola, yielding standard SHM.

Question

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 $T = \frac{2\pi}{\omega}$. The correct expression for $\omega$ is ($g$ is the acceleration due to gravity):

Options:

(A) $\sqrt{\frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}\right)}$

(B) $\sqrt{\frac{2g}{3(R-r)} + \frac{k}{m}}$

(C) $\sqrt{\frac{1}{6}\left(\frac{g}{R-r} + \frac{k}{m}\right)}$

(D) $\sqrt{\frac{1}{4}\left(\frac{g}{R-r} + \frac{k}{m}\right)}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a small disk of mass $m$ and radius $r$ rolling back and forth inside a larger stationary ring of radius $R$. The center of the disk is tethered to the bottom of the ring by a spring of constant $k$. When the disk is displaced by a small angle $	heta$ from the vertical, its center of mass traces out a circular arc of radius $(R-r)$.

Because the disk rolls without slipping and we are ignoring any non-conservative forces like air resistance or rolling friction, the total mechanical energy of the system remains perfectly conserved. This makes the **Energy Method** the most elegant tool to find the time period of the resulting Simple Harmonic Motion (SHM).

### The Master Equation: Total Energy
The total mechanical energy $E$ is the sum of the kinetic energy $K$ and the potential energy $U$.

Let's break down the kinetic energy first. The disk is undergoing combined translation and rotation. The velocity of its center of mass is $v = (R-r)\dot{	heta}$. Because it rolls without slipping, its angular velocity about its own center of mass is $\omega_{	ext{disk}} = \frac{v}{r}$.

The total kinetic energy is:
$$ K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega_{	ext{disk}}^2 $$

Substituting the moment of inertia of a uniform disk, $I = \frac{1}{2}mr^2$, we get:
$$ K = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2ight)\left(\frac{v}{r}ight)^2 = \frac{3}{4}mv^2 $$

Expressing this in terms of $\dot{	heta}$:
$$ K = \frac{3}{4}m(R-r)^2\dot{	heta}^2 $$

Now, let's look at the potential energy. It has two components: the elastic potential energy stored in the spring and the gravitational potential energy due to the slight rise of the disk's center of mass.

The spring stretches along the arc, so its extension is $x = (R-r)	heta$. The elastic potential energy is:
$$ U_s = \frac{1}{2}kx^2 = \frac{1}{2}k(R-r)^2	heta^2 $$

The height gained by the center of mass is $h = (R-r)(1-\cos	heta)$. For small angular displacements, we can use the Taylor series expansion $1-\cos	heta \approx \frac{	heta^2}{2}$. The gravitational potential energy becomes:
$$ U_g = mgh \approx \frac{1}{2}mg(R-r)	heta^2 $$

Combining these, the total potential energy is:
$$ U = \frac{1}{2}\left[k(R-r)^2 + mg(R-r)ight]	heta^2 $$

### Final Calculation
Since the total energy $E = K + U$ is constant, its time derivative must be zero:
$$ \frac{dE}{dt} = \frac{d}{dt} \left( \frac{3}{4}m(R-r)^2\dot{	heta}^2 + \frac{1}{2}\left[k(R-r)^2 + mg(R-r)ight]	heta^2 ight) = 0 $$

Differentiating with respect to time $t$ yields:
$$ \frac{3}{2}m(R-r)^2\dot{	heta}\ddot{	heta} + \left[k(R-r)^2 + mg(R-r)ight]	heta\dot{	heta} = 0 $$

Assuming the disk is in motion ($\dot{	heta} 
eq 0$), we can divide out the common $\dot{	heta}$ term and rearrange to form the standard SHM differential equation $\ddot{	heta} = -\omega^2	heta$:
$$ \ddot{	heta} + \frac{k(R-r)^2 + mg(R-r)}{\frac{3}{2}m(R-r)^2}	heta = 0 $$

From this, we can directly extract the square of the angular frequency $\omega$:
$$ \omega^2 = \frac{k(R-r) + mg}{\frac{3}{2}m(R-r)} = \frac{2}{3}\left(\frac{k}{m} + \frac{g}{R-r}ight) $$

Taking the square root gives us the final expression for $\omega$:
$$ \omega = \sqrt{\frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}ight)} $$

This perfectly matches Option (A).

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Energy Method for SHM$

$Kinetic Energy Setup$

$Evaluating Kinetic Energy$

$Potential Energy Setup$

$Small Angle Approximation$

$Total Energy Equation$

$Differentiating the Energy$

$Final Angular Frequency$

$The Way Forward$

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup Imagine a small disk of mass $m$ and radius $r$ rolling back and forth inside a larger stationary ring of radius $R$

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Method for SHM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Energy Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential Energy Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Small Angle Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Energy Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiating the Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Angular Frequency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward