A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a d

Q: Why is the mass of the disc the same as the sphere?

Because the sphere is melted to form the disc, the total amount of matter remains conserved. No mass is lost or gained in the process.

Q: What is the moment of inertia of a disc about its central perpendicular axis?

For a uniform disc of mass $M$ and radius $r$, the moment of inertia about an axis passing through its center and perpendicular to its plane is $\frac{1}{2} M r^2$.

Q: Why do we use the parallel axis theorem here?

The required axis is tangential and perpendicular to the disc's plane. This axis is parallel to the central perpendicular axis passing through the center of mass, making the parallel axis theorem the perfect tool.

Q: Does the thickness $t$ of the disc affect its moment of inertia?

No, for an axis perpendicular to the plane of the disc, the moment of inertia depends only on its mass and radius, not its thickness. The mass distribution radially is what matters.

Q: What if the axis was in the plane of the disc?

Then the moment of inertia about the diameter would be $\frac{1}{4} M r^2$, and the tangential axis in the plane would have $I = \frac{1}{4} M r^2 + M r^2 = \frac{5}{4} M r^2$.

Question

A solid sphere of radius $R$ has moment of inertia $I$ about its geometrical axis. It is melted into a disc of radius $r$ and thickness $t$. If it's moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to

Options:

(A) $\frac{2}{\sqrt{15}} R$

(B) $\frac{2}{\sqrt{5}} R$

(C) $\frac{3}{\sqrt{15}} R$

(D) $\frac{\sqrt{3}}{\sqrt{15}} R$

Sigma Percentile · Accepted Answer

The problem presents a fascinating transformation: a solid sphere is melted and recast into a disc. We are tasked with finding the radius of this new disc, given a specific condition about their moments of inertia. Let's break down the physics step-by-step.

### The Setup: From Sphere to Disc

Imagine a solid sphere of mass $M$ and radius $R$. Its moment of inertia about its geometrical axis (an axis passing through its center) is a standard result in rotational mechanics:

$$I_{	ext{sphere}} = \frac{2}{5} M R^2$$

Now, this sphere is melted down and reshaped into a disc of radius $r$ and thickness $t$. The crucial physical principle here is the conservation of mass. Since no material is lost during the melting process, the mass of the newly formed disc is exactly the same as the original sphere, $M$.

### The Master Equation: Parallel Axis Theorem

The problem states that the moment of inertia of this disc about a specific axis is equal to the sphere's moment of inertia, $I$. This specific axis is described as a "tangential axis which is perpendicular to the plane of the disc."

To find the moment of inertia about this tangential axis, we must employ the **Parallel Axis Theorem**. This theorem states that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis passing through the center of mass, plus the product of the mass and the square of the perpendicular distance between the two axes.

For a disc, the moment of inertia about its central axis perpendicular to its plane is:

$$I_{	ext{CM}} = \frac{1}{2} M r^2$$

The tangential axis is parallel to this central axis and is located at the edge of the disc. Therefore, the distance between these two parallel axes is simply the radius of the disc, $r$. Applying the theorem:

$$I_{	ext{disc}} = I_{	ext{CM}} + M r^2$$

$$I_{	ext{disc}} = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2$$

### The Grand Equivalence

We are given that the moment of inertia of the sphere equals the moment of inertia of the disc about this tangential axis. We can now set up our master equation:

$$I_{	ext{sphere}} = I_{	ext{disc}}$$

Substituting the expressions we derived:

$$\frac{2}{5} M R^2 = \frac{3}{2} M r^2$$

Notice how the mass $M$ elegantly cancels out from both sides, leaving us with a purely geometric relationship between the radii. Now, we simply solve for $r^2$:

$$r^2 = \frac{2}{5} 	imes \frac{2}{3} R^2$$

$$r^2 = \frac{4}{15} R^2$$

Taking the square root of both sides yields the final radius of the disc:

**$$r = \frac{2}{\sqrt{15}} R$$**

This perfectly matches option (a). The beauty of this problem lies in how it seamlessly connects the concepts of mass conservation, standard moments of inertia, and the parallel axis theorem into a single, elegant calculation.

Select Answer:

Visualized Solution

$I_{sphere}$

Mass Conservation

Axis of the Disc

Parallel Axis Theorem

$I_{disc}$

Total $I_{disc}$

Equating Inertias

Solving for $r^{2}$

Final Answer

The Sigma Insight: Moment of Inertia

The Setup

The Master Equation

The Grand Equivalence

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