A ring is hung on a nail. It can oscillate without slipping or sliding
(i) in its plane with a time

Q: Why is the distance $d$ equal to $R$ in both cases?

The distance $d$ in the physical pendulum formula is the straight-line distance from the pivot point (the nail) to the center of mass (the center of the ring). Since the nail is on the circumference, this distance is exactly the radius $R$, regardless of the axis of rotation.

Q: Why is the axis of rotation a tangent in the second case?

When the ring swings perpendicular to its plane, it acts like a door swinging on a hinge. For the nail to act as the pivot without the ring slipping, the ring must rotate around the horizontal line that passes through the nail and lies in the plane of the ring.

Q: How do we know the moment of inertia of a ring about its diameter?

You can derive it using the Perpendicular Axis Theorem ($I_z = I_x + I_y$). For a uniform ring, the moments of inertia about any two perpendicular diameters are equal ($I_x = I_y$). Since $I_z = mR^2$, we get $2I_x = mR^2$, which means $I_x = \frac{1}{2}mR^2$.

Q: Why can't we just use the simple pendulum formula $T = 2\pi \sqrt{l/g}$?

A simple pendulum assumes all the mass is concentrated at a single point at the end of a massless string. A ring has its mass distributed over a large area. Therefore, we must account for its rotational inertia using the physical pendulum formula.

Q: Does the mass or radius of the ring affect the final ratio?

No! Both time periods are proportional to $\sqrt{mR^2}$. When you take the ratio $T_1 / T_2$, the mass $m$ and radius $R$ completely cancel out. The ratio depends purely on the geometry of the mass distribution relative to the axes.

Question

A ring is hung on a nail. It can oscillate without slipping or sliding
(i) in its plane with a time period $T_1$
(ii) back and forth in a direction perpendicular to its plane, with a period $T_2$.
The ratio $T_1 / T_2$ will be

Options:

(A) $\frac{2}{\sqrt{3}}$

(B) $\frac{2}{3}$

(C) $\frac{3}{\sqrt{2}}$

(D) $\frac{\sqrt{2}}{3}$

Sigma Percentile · Accepted Answer

The problem of the swinging ring is a classic exploration of the **physical pendulum**. Unlike a simple pendulum where all the mass is concentrated at a single point, a physical pendulum has its mass distributed over a volume.

When dealing with a physical pendulum, the time period is governed by the master equation:
$$T = 2\pi \sqrt{\frac{I}{mgd}}$$
Here, $I$ is the moment of inertia of the object about the axis of rotation, $m$ is its mass, $g$ is the acceleration due to gravity, and $d$ is the distance from the pivot point to the center of mass.

In our scenario, the ring is hung on a nail. This means the pivot point is on the circumference of the ring. The center of mass of a uniform ring is exactly at its geometric center. Therefore, the distance $d$ from the pivot to the center of mass is simply the radius of the ring, $R$. This holds true regardless of how the ring swings!

### Case (i): The In-Plane Swing

Imagine the ring swinging left and right, staying perfectly flat against the wall. For this motion to occur, the axis of rotation must be a line sticking straight out of the wall, passing through the nail.

To find the time period $T_1$, we need the moment of inertia $I_1$ about this axis. We start with the moment of inertia of a ring about its center, perpendicular to its plane, which is $I_{cm} = mR^2$.

Using the **Parallel Axis Theorem** ($I = I_{cm} + md^2$), we shift this axis from the center to the nail (a distance $R$ away):
$$I_1 = mR^2 + mR^2 = 2mR^2$$

### Case (ii): The Out-of-Plane Swing

Now, imagine pulling the bottom of the ring away from the wall and letting it swing back and forth, perpendicular to the wall. The axis of rotation is now a horizontal line lying flat against the wall, tangent to the top of the ring at the nail.

We need the moment of inertia $I_2$ about this new axis. The moment of inertia of a ring about its diameter is $I_{diameter} = \frac{1}{2}mR^2$.

Again, we apply the **Parallel Axis Theorem** to shift this axis from the center to the tangent at the nail:
$$I_2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$$

### The Grand Finale: Comparing the Swings

We now have everything we need to find the ratio of the time periods. Since $2\pi$, $m$, $g$, and $d$ are identical in both cases, the time period is directly proportional to the square root of the moment of inertia:
$$T \propto \sqrt{I}$$

Taking the ratio:
$$\frac{T_1}{T_2} = \sqrt{\frac{I_1}{I_2}}$$

Substituting our calculated moments of inertia:
$$\frac{T_1}{T_2} = \sqrt{\frac{2mR^2}{\frac{3}{2}mR^2}}$$

The $mR^2$ terms beautifully cancel out, leaving us with a purely geometric ratio:
$$\frac{T_1}{T_2} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$$

This elegant result shows that the in-plane swing is slightly slower than the out-of-plane swing, purely due to the way the mass is distributed relative to the axis of rotation!

Animated Solution for Physics - Oscillations: A ring is hung on a nail. It can oscillate without slipping or sliding (i) in its plane with a time period $T_{1}$ (ii) back and forth in a direction perpendicular to its plane, with a period $T_{2}$ . The ratio $T_{1} / T_{2}$ will be

Select Answer:

Visualized Solution

Physical Pendulum Formula

Case (i): In-Plane Oscillation

Calculating $I_{1}$

Case (ii): Out-of-Plane Oscillation

Calculating $I_{2}$

Ratio $\frac{T _{1}}{T _{2}}$

The Sigma Insight: Simple Harmonic Motion (SHM)

Case (i)

Case (ii)

The Grand Finale

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