ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid-points of AB, AC and G i

Q: Why is the mass of the removed triangle m/4 and not m/2?

Because the lamina is a 2D object, its mass is proportional to its area. Since the side length of $ADE$ is half that of $ABC$, its area (and thus its mass) is $(1/2)^2 = 1/4$ of the total.

Q: What is the moment of inertia of an equilateral triangle about its centroid?

For a uniform equilateral triangle of mass $m$ and side $a$, the moment of inertia about an axis through its centroid and perpendicular to its plane is $ma^2/12$.

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

The moment of inertia of the removed part $ADE$ must be calculated about the axis passing through $G$ (the centroid of the large triangle). However, the standard formula gives the moment of inertia about its *own* centroid $G'$. The parallel axis theorem bridges this gap.

Q: Can this problem be solved using integration?

Yes, you could set up a double integral over the trapezoidal region $DBCE$. However, using the principle of superposition (subtracting the small triangle from the large one) is vastly more elegant and less prone to algebraic errors.

Question

ABC is a plane lamina of the shape of an equilateral triangle. D, E are mid-points of AB, AC and G is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is $I_0$. If part ADE is removed, the moment of inertia of the remaining part about the same axis is $\frac{NI_0}{16}$, where N is an integer. Value of N is ……… .

Sigma Percentile · Accepted Answer

### The Principle of Superposition

When dealing with complex geometries, especially those with "holes" or removed sections, the principle of superposition is our most powerful tool. Instead of trying to integrate over an awkward trapezoidal shape, we can treat the remaining lamina as a complete, solid equilateral triangle with a smaller equilateral triangle "subtracted" from it.

Mathematically, this means the net moment of inertia is simply:
$$I_{net} = I_{total} - I_{removed}$$

### Analyzing the Removed Triangle

Let the original equilateral triangle $ABC$ have a mass $m$ and side length $a$. The moment of inertia of a uniform equilateral triangle about an axis passing through its centroid and perpendicular to its plane is a standard result:
$$I_0 = \frac{ma^2}{12}$$

The removed part, $	riangle ADE$, is formed by connecting the midpoints of $AB$ and $AC$. This makes $	riangle ADE$ an equilateral triangle with a side length of $a/2$. Because the lamina is uniform, its mass is directly proportional to its area. Since the side length is halved, the area is scaled by a factor of $(1/2)^2 = 1/4$. Therefore, the mass of the removed triangle is:
$$m_1 = \frac{m}{4}$$

### The Parallel Axis Theorem in Action

Here is where many students make a critical error. You cannot simply subtract the moment of inertia of $	riangle ADE$ about its *own* centroid from $I_0$. Both moments of inertia must be calculated about the **same axis**—the axis passing through the main centroid $G$.

Let $G'$ be the centroid of the small triangle $ADE$. We need to find the distance $d$ between $G$ and $G'$. In an equilateral triangle, the distance from a vertex to the centroid is the side length divided by $\sqrt{3}$.

For the large triangle $ABC$:
$$AG = \frac{a}{\sqrt{3}}$$

For the small triangle $ADE$:
$$AG' = \frac{a/2}{\sqrt{3}} = \frac{a}{2\sqrt{3}}$$

The distance $d$ between the two centroids is:
$$d = AG - AG' = \frac{a}{\sqrt{3}} - \frac{a}{2\sqrt{3}} = \frac{a}{2\sqrt{3}}$$

Now, we apply the **Parallel Axis Theorem** ($I = I_{cm} + md^2$) to find the moment of inertia of the removed part about $G$:
$$I_1 = I_{G'} + m_1 d^2$$
$$I_1 = \frac{m_1 (a/2)^2}{12} + m_1 \left(\frac{a}{2\sqrt{3}}ight)^2$$

Substituting $m_1 = m/4$:
$$I_1 = \frac{(m/4)(a^2/4)}{12} + \frac{m}{4} \left(\frac{a^2}{12}ight)$$
$$I_1 = \frac{ma^2}{192} + \frac{ma^2}{48} = \frac{5ma^2}{192}$$

### The Final Assembly

With both moments of inertia calculated about the same axis $G$, we can finally apply superposition:
$$I_{net} = I_0 - I_1$$
$$I_{net} = \frac{ma^2}{12} - \frac{5ma^2}{192}$$

To subtract these, find a common denominator (which is 192):
$$I_{net} = \frac{16ma^2}{192} - \frac{5ma^2}{192} = \frac{11ma^2}{192}$$

The problem asks us to express this in the form $\frac{N I_0}{16}$. Let's factor out $I_0 = \frac{ma^2}{12}$:
$$I_{net} = \frac{11}{16} \left(\frac{ma^2}{12}ight) = \frac{11}{16} I_0$$

Comparing this to the given expression, we can clearly see that **$N = 11$**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Moment of Inertia of A B C$

$Analyzing the Removed Part A D E$

$Mass of the Removed Part$

$Locating the Centroid G^{'}$

$Distance Between Centroids$

$Parallel Axis Theorem$

$Substituting Values$

$Evaluating I_{1}$

$Net Moment of Inertia$

$Finding N$

The Sigma Insight: Moment of Inertia

The Principle of Superposition

Analyzing the Removed Triangle

The Parallel Axis Theorem in Action

The Final Assembly

Similar Questions

Consider a uniform square plate of side $a$ and mass $m$ . The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

Moment of inertia of a square plate of side $l$ about the axis passing through one of the corner and perpendicular to the plane of square plate is given by

One quarter section is cut from a uniform circular disc of radius $R$ . This section has a mass $M$ . It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

The linear mass density of a thin rod $A B$ of length $L$ varies from $A$ to $B$ as $λ (x) = λ_{0} (1 + \frac{x}{L})$ , where $x$ is the distance from $A$ . If $M$ is mass of the rod, then its moment of inertia about an axis passing through $A$ and perpendicular to the rod is

Enter Numerical Value:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Locating the Centroid G′

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

.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; } Substituting Values

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

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

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