An equilateral triangle ABC is cut from a thin solid sheet of wood. (see figure) D, E and F are the

Q: Why can we just subtract the moment of inertia of the smaller triangle?

Moment of inertia is a scalar additive quantity (like mass). The principle of superposition states that the moment of inertia of a composite body is the sum of the moments of inertia of its parts, provided they are all calculated about the exact same axis.

Q: How do we know the centroid of $\Delta DEF$ is the same as $\Delta ABC$?

$\Delta DEF$ is the medial triangle of $\Delta ABC$. By geometric symmetry, the centroid of a triangle and its medial triangle always perfectly coincide.

Q: Why is the mass of the smaller triangle $M/4$ and not $M/2$?

The sheet is two-dimensional and uniform, meaning mass is directly proportional to area. Since the side length is halved, the area scales by the square of the scaling factor: $(1/2)^2 = 1/4$.

Q: Do we need to know the exact formula for the moment of inertia of a triangle?

Not at all! Because we are taking a ratio, the geometric constant $k$ in the formula $I = kMl^2$ cancels out completely. This is the beauty of scaling laws.

Q: What if the removed triangle was not centered at $G$?

If the removed triangle had a different centroid, we would first calculate its moment of inertia about its own centroid, and then use the Parallel Axis Theorem to find its moment of inertia about $G$ before subtracting.

Question

An equilateral triangle $ABC$ is cut from a thin solid sheet of wood. (see figure) $D, E$ and $F$ are the mid points of its sides as shown and $G$ is the centre of the triangle. The moment of inertia of the triangle about an axis passing through $G$ and perpendicular to the plane of the triangle is $I_0$. If the smaller triangle $DEF$ is removed from $ABC$, the moment of inertia of the remaining figure about the same axis is $I$. Then

Options:

(A) $I' = \frac{3}{4}I_0$

(B) $I' = \frac{15}{16}I_0$

(C) $I' = \frac{I_0}{4}$

(D) $I' = \frac{9}{16}I_0$

Sigma Percentile · Accepted Answer

### The Geometry of the Problem

Imagine you are holding a perfectly uniform, solid wooden triangle. The beauty of physics lies in its symmetries and scaling laws. When we are asked to find the moment of inertia of a complex shape—like a triangle with a piece missing—we don't need to perform a terrifying double integral over the triangular domain. Instead, we can use the power of dimensional analysis and the principle of superposition.

Let the mass of the original large triangle $\Delta ABC$ be $M$ and its side length be $l$. For any uniform two-dimensional shape, the moment of inertia about its centroid is always proportional to its mass and the square of its linear dimension.

Therefore, we can write the initial moment of inertia as:

$$I_0 = k M l^2$$

Here, $k$ is a dimensionless geometric constant that depends purely on the shape (an equilateral triangle in this case). The exact value of $k$ doesn't matter, as we will soon see it magically cancel out!

### The Power of Dimensional Scaling

Now, let's analyze the smaller triangle $\Delta DEF$ that is being removed. This triangle is formed by joining the midpoints of the original triangle.

By basic geometry, the side length of this medial triangle is exactly half of the original triangle:

$$l_{DEF} = \frac{l}{2}$$

Because the wooden sheet is uniform and two-dimensional, its mass is directly proportional to its area. When you scale the linear dimensions of a 2D object by a factor of $1/2$, its area scales by the square of that factor, which is $(1/2)^2 = 1/4$.

Thus, the mass of the removed triangle is:

$$M_{DEF} = \frac{M}{4}$$

### Analyzing the Removed Triangle

Here is where the symmetry of the problem makes our life incredibly easy. The centroid of the medial triangle $\Delta DEF$ perfectly coincides with the centroid $G$ of the original triangle $\Delta ABC$.

Because they share the exact same axis of rotation, we can use the exact same formula to find the moment of inertia of the removed piece. We just plug in its scaled mass and scaled side length:

$$I_{DEF} = k \left(\frac{M}{4}\right) \left(\frac{l}{2}\right)^2$$

Let's expand the squared term:

$$I_{DEF} = k \left(\frac{M}{4}\right) \left(\frac{l^2}{4}\right)$$

$$I_{DEF} = \frac{k M l^2}{16}$$

Notice that the term $k M l^2$ is exactly our original moment of inertia, $I_0$. Substituting this back in, we get a beautifully simple relation:

$$I_{DEF} = \frac{I_0}{16}$$

### The Principle of Superposition

Finally, we arrive at the last step. The moment of inertia is a scalar additive quantity, much like mass. The principle of superposition states that the moment of inertia of a composite body is the sum of the moments of inertia of its individual parts, provided they are all calculated about the **exact same axis**.

Conversely, if we remove a piece from a body, we can simply subtract its moment of inertia.

$$I_{remaining} = I_{original} - I_{removed}$$

Substituting the values we found:

$$I' = I_0 - \frac{I_0}{16}$$

$$I' = \frac{15}{16}I_0$$

### Final Conclusion

And there we have it! By leveraging scaling laws and the principle of superposition, we bypassed complex calculus entirely. The moment of inertia of the remaining figure is **$\frac{15}{16}I_0$**. This elegant technique is a powerful tool in your physics arsenal, especially for competitive exams where time is of the essence.

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Visualized Solution

$Analyzing the Setup$

$Properties of the Removed Triangle$

$Moment of Inertia of Δ D E F$

$Calculating I_{D E F}$

$Principle of Superposition$

$The Power of Scaling$

The Sigma Insight: Moment of Inertia

The Geometry of the Problem

The Power of Dimensional Scaling

Analyzing the Removed Triangle

The Principle of Superposition

Final Conclusion

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Properties of the Removed Triangle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moment of Inertia of ΔDEF

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

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

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