A wire of lenght l and mass m is bent in the form of a rectangle ABCD with (AB/BC) = 2.

The moment of inertia of this wire frame about the side BCC is

To determine the moment of inertia of the wire frame bent in the shape of a rectangle ABCD about the side BC, we first need to understand the dimensions and properties of the rectangle based on the given ratio of the sides. Let's break this down step by step.

Understanding the Dimensions

Given that the ratio of the lengths of sides AB to BC is 2, we can denote the length of side AB as 2x and the length of side BC as x. Therefore, the rectangle can be defined as follows:

• AB = 2x
• BC = x
• CD = 2x (opposite side to AB)
• DA = x (opposite side to BC)

The total length of the wire, which is the perimeter of the rectangle, can be calculated as:

Perimeter = AB + BC + CD + DA = 2x + x + 2x + x = 6x

Relating Mass and Length

We know the mass of the wire is m and its total length is l. Therefore, we can establish a relationship between the mass per unit length (linear density) and the total length:

Linear density (λ) = m / l

From the perimeter, we can express the total length in terms of x:

l = 6x

Calculating the Moment of Inertia

The moment of inertia (I) of a wire frame about a given axis can be calculated using the formula:

I = Σ(m_i * r_i²)

where m_i is the mass of each segment and r_i is the distance from the axis of rotation. In this case, we will consider each side of the rectangle and its contribution to the moment of inertia about the axis BC.

Contributions from Each Side

1. **Side AB (2x)**: This side is at a distance of x from the axis BC. The mass of this side can be calculated as:

Mass of AB = (Length of AB / Total Length) * m = (2x / 6x) * m = (1/3)m

The moment of inertia contribution from AB is:

I_AB = (1/3)m * (x)² = (1/3)m * x²

2. **Side CD (2x)**: This side is also at a distance of x from the axis BC, and its mass is the same as that of AB:

I_CD = (1/3)m * (x)² = (1/3)m * x²

3. **Side BC (x)**: This side is on the axis of rotation, so its contribution to the moment of inertia is zero:

I_BC = 0

4. **Side DA (x)**: This side is at a distance of 2x from the axis BC. Its mass can be calculated as:

Mass of DA = (Length of DA / Total Length) * m = (x / 6x) * m = (1/6)m

The moment of inertia contribution from DA is:

I_DA = (1/6)m * (2x)² = (1/6)m * 4x² = (2/3)m * x²

Summing Up the Contributions

Now, we can sum the contributions from all sides to find the total moment of inertia about the axis BC:

I_total = I_AB + I_CD + I_BC + I_DA

I_total = (1/3)m * x² + (1/3)m * x² + 0 + (2/3)m * x²

I_total = (1/3)m * x² + (1/3)m * x² + (2/3)m * x² = (4/3)m * x²

Final Expression

Substituting back for x in terms of l, we have:

x = l / 6

Thus, the moment of inertia can be expressed as:

I_total = (4/3)m * (l/6)² = (4/3)m * (l² / 36) = (1/27)m * l²

In summary, the moment of inertia of the wire frame about the side BC is:

I = (1/27)m * l²