Three solid sphere each of mass m and diameter d are stuck together such that the lines connecting t

Q: Why do we use $d/2$ for the radius of the sphere?

The problem states the diameter of each sphere is $d$. The formula for the moment of inertia of a solid sphere uses its radius, which is half the diameter, so $r = d/2$.

Q: Why is the distance to the centroid $d/\sqrt{3}$?

In an equilateral triangle of side $d$, the altitude is $d\sqrt{3}/2$. The centroid divides the altitude in a 2:1 ratio, so the distance from a vertex to the centroid is $(2/3) \times (d\sqrt{3}/2) = d/\sqrt{3}$.

Q: Can I find $I_A$ by calculating the MI of each sphere about A individually?

Yes, you can! One sphere is at distance 0, and the other two are at distance $d$. $I_A = I_{cm} + (I_{cm} + md^2) + (I_{cm} + md^2) = 3I_{cm} + 2md^2$. Substituting $I_{cm} = \frac{1}{10}md^2$, you get $\frac{3}{10}md^2 + 2md^2 = \frac{23}{10}md^2$. It gives the exact same result!

Q: Why did we multiply by 3 to get $I_0$?

Moment of inertia is a scalar quantity (more precisely, a tensor, but it adds like a scalar for a fixed axis). Since all three spheres are identical and equidistant from the centroid, their individual moments of inertia about the centroidal axis are equal. We just sum them up.

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

If the axis was in the plane, you would need to use the Perpendicular Axis Theorem or calculate the perpendicular distances from each sphere's center to that specific in-plane axis. The distances would be different, changing the result.

Question

Three solid sphere each of mass $m$ and diameter $d$ are stuck together such that the lines connecting the centres form an equilateral triangle of side of length $d$. The ratio $I_0/I_A$ of moment of inertia $I_0$ of the system about an axis passing the centroid and about centre of any of the spheres $I_A$ and perpendicular to the plane of the triangle is

Options:

(A) $\frac{15}{13}$

(B) $\frac{13}{15}$

(C) $\frac{13}{23}$

(D) $\frac{23}{13}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine three identical solid spheres, each of mass $m$ and diameter $d$, arranged such that their centers form an equilateral triangle of side $d$. This means the spheres are perfectly touching each other. We are tasked with finding the ratio of the moment of inertia of this entire system about two different axes:
1. An axis passing through the centroid $O$ of the triangle and perpendicular to its plane ($I_0$).
2. An axis passing through the center of any one of the spheres, say $A$, and perpendicular to the plane ($I_A$).

### The Master Equation: Parallel Axis Theorem
To navigate this problem, our primary tool will be the **Parallel Axis Theorem**. It states that the moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis passing through its center of mass, plus the product of its mass and the square of the perpendicular distance between the two axes.
$$I = I_{cm} + md^2$$

### Calculating Moment of Inertia about the Centroid ($I_0$)
First, let's determine the distance $h$ from the center of any sphere (a vertex of the triangle) to the centroid $O$. For an equilateral triangle of side $d$, the distance from a vertex to the centroid is:
$$h = \frac{d}{\sqrt{3}}$$

Next, we find the moment of inertia of a single solid sphere about its own central axis. Since the radius is $r = d/2$, we have:
$$I_{cm} = \frac{2}{5}mr^2 = \frac{2}{5}m\left(\frac{d}{2}ight)^2 = \frac{1}{10}md^2$$

Now, we apply the Parallel Axis Theorem to shift this axis to the centroid $O$:
$$I'_{sphere} = I_{cm} + mh^2 = \frac{1}{10}md^2 + m\left(\frac{d}{\sqrt{3}}ight)^2$$
$$I'_{sphere} = \frac{1}{10}md^2 + \frac{1}{3}md^2 = \frac{13}{30}md^2$$

Since the system consists of three identical spheres symmetrically placed around the centroid, the total moment of inertia $I_0$ is simply three times the moment of inertia of one sphere:
$$I_0 = 3 	imes I'_{sphere} = 3 	imes \frac{13}{30}md^2 = \frac{13}{10}md^2$$

### Calculating Moment of Inertia about a Vertex ($I_A$)
To find $I_A$, we could calculate the moment of inertia of each sphere about $A$ and sum them up. However, there is a much more elegant way! We can treat the entire three-sphere system as a single rigid body.

The center of mass of this entire system is at the centroid $O$. The total mass of the system is $M_{sys} = 3m$. We already know the moment of inertia of the system about its center of mass, which is $I_0$.

We can apply the Parallel Axis Theorem to the **entire system** to shift the axis from $O$ to $A$. The distance between these axes is again $h = d/\sqrt{3}$.
$$I_A = I_0 + M_{sys}h^2$$
$$I_A = \frac{13}{10}md^2 + (3m)\left(\frac{d}{\sqrt{3}}ight)^2$$
$$I_A = \frac{13}{10}md^2 + md^2 = \frac{23}{10}md^2$$

### Final Calculation
Finally, we take the ratio of $I_0$ to $I_A$:
$$\frac{I_0}{I_A} = \frac{\frac{13}{10}md^2}{\frac{23}{10}md^2} = \frac{13}{23}$$

This elegant application of the Parallel Axis Theorem on the entire system saves us from tedious individual calculations and leads us straight to the correct answer!

Select Answer:

Visualized Solution

$System Geometry$

$Parallel Axis Theorem$

$Distance to Centroid$

$MI of One Sphere about O$

$Total MI about O$

$System Parallel Axis Theorem$

$Total MI about A$

$Ratio I_{0} / I_{A}$

$Concept Variation$

The Sigma Insight: Moment of Inertia

Analyzing the Setup Imagine three identical solid spheres, each of mass $m$ and diameter $d$ , arranged such that their centers form an equilateral triangle of side $d$

The Master Equation

Calculating Moment of Inertia about the Centroid ( $I_{0}$ ) First, let's determine the distance $h$ from the center of any sphere (a vertex of the triangle) to the centroid $O$

Calculating Moment of Inertia about a Vertex ( $I_{A}$ ) To find $I_{A}$ , we could calculate the moment of inertia of each sphere about $A$ and sum them up

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; } System Geometry

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MI of One Sphere about O

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Parallel Axis Theorem

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio I0​/IA​

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