A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass o

Q: Why is the density constant?

The problem states the parts are 'converted' into new shapes, implying the material itself isn't changed or compressed, so its density remains the same.

Q: Why is the moment of inertia of the disc $\frac{1}{2}mr^2$?

This is the standard formula for the moment of inertia of a uniform circular disc about its central perpendicular axis.

Q: How did we get $r = R/2$?

Since density is constant, mass is proportional to volume ($r^3$). The mass is $1/8$th of the original, so the volume is $1/8$th, meaning the radius must be $(1/8)^{1/3} = 1/2$ of the original radius.

Q: What if the disc was rotated about its diameter?

The problem says 'about its axis', which conventionally means the central axis perpendicular to its plane. If it were about a diameter, it would be $\frac{1}{4}mr^2$.

Q: Can we assume the thickness of the disc is negligible?

Yes, for a 'uniform disc', the standard moment of inertia formula $\frac{1}{2}mr^2$ applies regardless of its uniform thickness, as long as it's a right circular cylinder/disc.

Question

A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The first part has a mass of $7M/8$ and is converted into a uniform disc of radius $2R$. The second part is converted into a uniform solid sphere. Let $I_1$ be the moment of inertia of the disc about its axis and $I_2$ be the moment of inertia of the new sphere about its axis. The ratio $I_1/I_2$ is given by

Options:

(A) 285

(B) 185

(C) 65

(D) 140

Sigma Percentile · Accepted Answer

The problem of reshaping a solid sphere into two distinct objects is a beautiful exercise in understanding the conservation of mass, the constancy of density, and the geometric dependence of the moment of inertia. Let's embark on this journey step by step.

### Analyzing the Setup

We begin with a uniform solid sphere of mass $M$ and radius $R$. This sphere is divided into two unequal parts. The first part, which takes the lion's share of the mass, has a mass of $m_1 = \frac{7M}{8}$. This part is then flattened and reshaped into a uniform circular disc with a new radius $R_1 = 2R$.

The remaining mass is used to form a second object. By simple subtraction, the mass of this second part is $m_2 = M - \frac{7M}{8} = \frac{M}{8}$. This smaller chunk of material is molded into a new, smaller uniform solid sphere of radius $r$.

Our ultimate goal is to find the ratio of their moments of inertia, $I_1 / I_2$, about their respective central axes.

### The Moment of Inertia of the Disc

Let's focus on the first part—the disc. The moment of inertia of a uniform circular disc about its central axis (perpendicular to its plane) is given by the standard formula:
$$I = \frac{1}{2} m r^2$$

Substituting the specific mass and radius for our disc, we get:
$$I_1 = \frac{1}{2} m_1 R_1^2$$
$$I_1 = \frac{1}{2} \left(\frac{7M}{8}ight) (2R)^2$$

Now, we carefully expand the squared term and simplify the expression:
$$I_1 = \frac{1}{2} \cdot \frac{7M}{8} \cdot 4R^2$$
$$I_1 = \frac{7}{16} M \cdot 4R^2 = \frac{7}{4} MR^2$$

This gives us the moment of inertia for the first object.

### The Secret of the Small Sphere

Now, we turn our attention to the second part—the small solid sphere. We know its mass is $m_2 = \frac{M}{8}$, but to find its moment of inertia, we desperately need its radius $r$.

Here lies the crucial physical insight: **the material remains the same**. When you reshape a piece of clay or metal, its density $ho$ does not change. Since density is mass divided by volume, and density is constant, the volume of the object must be directly proportional to its mass.

For a sphere, the volume is proportional to the cube of its radius ($V \propto r^3$). Therefore, we can write:
$$\frac{r^3}{R^3} = \frac{m_2}{M}$$

Substituting the mass of the small sphere:
$$\frac{r^3}{R^3} = \frac{M/8}{M} = \frac{1}{8}$$

Taking the cube root of both sides reveals the radius of the new sphere:
$$r = \frac{R}{2}$$

### Calculating the Small Sphere's Moment of Inertia

With the radius $r$ in hand, we can now calculate the moment of inertia of the small solid sphere. The formula for a solid sphere about its central axis is:
$$I = \frac{2}{5} m r^2$$

Substituting our values for $m_2$ and $r$:
$$I_2 = \frac{2}{5} m_2 r^2 = \frac{2}{5} \left(\frac{M}{8}ight) \left(\frac{R}{2}ight)^2$$

Expanding the squared term and multiplying the fractions:
$$I_2 = \frac{2}{5} \cdot \frac{M}{8} \cdot \frac{R^2}{4}$$
$$I_2 = \frac{2}{160} MR^2 = \frac{1}{80} MR^2$$

### The Final Calculation

We have successfully found both moments of inertia. The final step is to compute their ratio, $I_1 / I_2$:
$$\frac{I_1}{I_2} = \frac{\frac{7}{4} MR^2}{\frac{1}{80} MR^2}$$

The $MR^2$ terms cancel out beautifully, leaving us with a simple fraction division:
$$\frac{I_1}{I_2} = \frac{7}{4} 	imes 80$$
$$\frac{I_1}{I_2} = 7 	imes 20 = 140$$

**The final ratio is 140.** This problem elegantly combines the principles of mass conservation, constant density, and rotational inertia into a single, cohesive narrative.

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Visualized Solution

$M, R \to m_{1}, R_{1} and m_{2}, r$

$I_{1} of Disc$

$I_{1} Substitution$

$I_{1} Calculation$

$Radius of Small Sphere r$

$Calculating r$

$I_{2} Substitution$

$I_{2} Calculation$

$Ratio I_{1} / I_{2}$

The Sigma Insight: Moment of Inertia

Analyzing the Setup

The Moment of Inertia of the Disc

The Secret of the Small Sphere

Calculating the Small Sphere's Moment of Inertia

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M,R→m1​,R1​ and m2​,r

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ of Disc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius of Small Sphere r

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio I1​/I2​