A uniform thin bar of mass 6 kg and length 2.4 m is bent to make an equilateral hexagon. The moment

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

The axis of rotation passes through the centre of the hexagon, not through the centre of mass of the individual rods. The parallel axis theorem allows us to shift the axis from the rod's centre of mass to the hexagon's centre.

Q: How do we know the distance from the centre to the rod?

A regular hexagon can be perfectly divided into 6 equilateral triangles. The perpendicular distance from the centre to a side is simply the height (altitude) of one of these equilateral triangles, which is $\frac{\sqrt{3}}{2}$ times the side length.

Q: Can we use the perpendicular axis theorem instead?

No, the perpendicular axis theorem relates moments of inertia about two perpendicular axes in the plane to a third axis perpendicular to the plane. Here, we are dealing with an axis already perpendicular to the plane, shifted away from the rod's centre, making the parallel axis theorem the correct choice.

Q: Why do we just multiply the moment of inertia of one side by 6?

Moment of inertia is a scalar additive quantity. Since all 6 sides are identical in mass and length, and are located at the exact same perpendicular distance from the central axis, their individual moments of inertia are identical. Thus, we can simply sum them up or multiply by 6.

Q: What if the axis of rotation was in the plane of the hexagon?

If the axis was in the plane of the hexagon (e.g., passing through two opposite vertices), the perpendicular distances from the axis to the various sides would be different. You would have to calculate the moment of inertia for each side individually based on its specific distance and orientation relative to that new axis.

Question

A uniform thin bar of mass 6 kg and length 2.4 m is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is ……… $	imes 10^{-1} 	ext{ kg -m}^2$.

Sigma Percentile · Accepted Answer

### The Transformation: From Bar to Hexagon

Imagine a straight, uniform bar of length $2.4 	ext{ m}$ and mass $6 	ext{ kg}$. We are tasked with bending this bar into a perfect equilateral hexagon and finding its moment of inertia about an axis passing through its geometric centre, perpendicular to its plane.

To tackle this, we must first break the problem down into its fundamental building blocks. Since the hexagon is composed of 6 identical sides, we can analyze just one side and then use symmetry to find the total moment of inertia.

Let's determine the properties of a single side:
- **Mass of one side ($m'$):** $\frac{6 	ext{ kg}}{6} = 1 	ext{ kg}$
- **Length of one side ($a$):** $\frac{2.4 	ext{ m}}{6} = 0.4 	ext{ m}$

### The Geometry of the Hexagon

Consider one side of the hexagon, say $AB$. To find its moment of inertia about the central axis $O$, we need to know its perpendicular distance from $O$.

A regular hexagon can be divided into 6 equilateral triangles. The perpendicular distance $d$ from the centre $O$ to the side $AB$ is simply the altitude (or apothem) of one of these equilateral triangles of side $a$.

Using basic geometry, this distance is:
$$d = \frac{\sqrt{3}}{2}a$$

### Applying the Parallel Axis Theorem

The moment of inertia of a uniform rod of mass $m'$ and length $a$ about an axis passing through its own centre of mass is $I_{CM} = \frac{m'a^2}{12}$.

However, our axis of rotation passes through $O$, which is at a distance $d$ from the rod's centre of mass. This is where the **Parallel Axis Theorem** comes to the rescue! The theorem states that the moment of inertia about a parallel axis is:
$$I_{OP} = I_{CM} + m'd^2$$

Let's substitute our known values into this master equation:
$$I_{OP} = \frac{m'a^2}{12} + m'\left(\frac{\sqrt{3}a}{2}ight)^2$$
$$I_{OP} = \frac{m'a^2}{12} + \frac{3m'a^2}{4}$$

To add these fractions, we find a common denominator (12):
$$I_{OP} = \frac{m'a^2}{12} + \frac{9m'a^2}{12} = \frac{10m'a^2}{12} = \frac{5}{6}m'a^2$$

Now, we plug in the numerical values for $m'$ and $a$:
$$I_{OP} = \frac{5}{6} 	imes (1) 	imes (0.4)^2 = \frac{5}{6} 	imes 0.16 	ext{ kg-m}^2$$

### The Final Calculation

Since the hexagon is perfectly symmetric and composed of 6 identical sides, the total moment of inertia $I_{net}$ is simply 6 times the moment of inertia of a single side:
$$I_{net} = 6 	imes I_{OP}$$
$$I_{net} = 6 	imes \left(\frac{5}{6} 	imes 0.16ight)$$
$$I_{net} = 5 	imes 0.16 = 0.80 	ext{ kg-m}^2$$

The question asks for the answer in the format $N 	imes 10^{-1} 	ext{ kg-m}^2$. We can rewrite our result as:
$$0.80 = 8 	imes 10^{-1} 	ext{ kg-m}^2$$

Therefore, the integer value we are looking for is **8**.

Animated Solution for Physics - Rotational Motion: A uniform thin bar of mass 6 kg and length 2.4 m is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is ……… $\times 1 0^{- 1} kg -m^{2}$ .

Enter Numerical Value:

Visualized Solution

$Visualizing the Hexagon$

$Properties of One Side$

$Distance from Centre$

$Parallel Axis Theorem$

$Moment of Inertia of One Side$

$Total Moment of Inertia$

$Final Answer$

The Sigma Insight: Moment of Inertia

The Transformation

The Geometry of the Hexagon

Applying the Parallel Axis Theorem

The Final Calculation

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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 Hexagon

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

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

.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; } Moment of Inertia of One Side

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

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