Question

A square laminar sheet with side a and mass M, has mass per unit area given by \(\sigma(x)=\sigma_0[1-\frac{x}{a}]\), (see figure). Moment of inertia of the sheet about y-axis is

• A. \(\frac{Ma^2}{2}\)
• B. \(\frac{Ma^2}{4}\)
• C. \(\frac{Ma^2}{6}\)
• D. \(\frac{Ma^2}{12}\)

Answer 1

The Correct Option is C

To find the moment of inertia of the square laminar sheet about the y-axis, we need to integrate over the entire area of the sheet considering the given mass per unit area \(\sigma(x)\).

• The mass per unit area is given as \(\sigma(x) = \sigma_0\left[1-\frac{x}{a}\right]\).
• The small strip at a distance \(x\) from the origin along the x-axis has a width \(dx\) and height \(a\), making its area \(a \cdot dx\).
• The mass of the strip, \(dm\), is given by multiplying the area of the strip by the mass per unit area: \(dm = \sigma(x) \cdot a \cdot dx = \sigma_0\left[1-\frac{x}{a}\right]a \cdot dx\).

The moment of inertia of this strip about the y-axis is:

• \(dI = x^2 \cdot dm = x^2 \cdot \sigma_0\left[1-\frac{x}{a}\right]a \cdot dx\)

To find the total moment of inertia, integrate over the width of the sheet from \(x = 0\) to \(x = a\):

• \(I = \int_{0}^{a} x^2 \cdot \sigma_0\left[1-\frac{x}{a}\right]a \cdot dx\)

Now, let's calculate the integral:

• \(I = a\sigma_0\int_{0}^{a} \left[x^2 - \frac{x^3}{a}\right]dx = a\sigma_0 \left[\frac{x^3}{3} - \frac{x^4}{4a}\right]_{0}^{a}\)
• After evaluating the definite integral:
  • \(= a\sigma_0\left[\frac{a^3}{3} - \frac{a^4}{4a}\right]\)
  • \(= a\sigma_0\left[\frac{a^3}{3} - \frac{a^3}{4}\right]\)
  • \(= a\sigma_0\left[\frac{4a^3 - 3a^3}{12}\right]\)
  • \(= a\sigma_0\left[\frac{a^3}{12}\right]\)
  • \(= \frac{a^4\sigma_0}{12}\)

According to the problem, the total mass \(M\) of the sheet is:

• \(M = \int_{0}^{a} \sigma(x) \cdot a \cdot dx = a\sigma_0 \left[\frac{x}{1} - \frac{x^2}{2a}\right]_{0}^{a} = a\sigma_0\left [ a - \frac{a}{2}\right]\)
• \(M = \frac{a^2\sigma_0}{2}\)

Solving for \(\sigma_0\), we get \(\sigma_0 = \frac{2M}{a^2}\).

Substitute back to get \(I\):

• \(I = \frac{a^4 \cdot \sigma_0}{12} = \frac{a^4 \cdot \frac{2M}{a^2}}{12} = \frac{2Ma^2}{12} = \frac{Ma^2}{6}\)

Therefore, the moment of inertia of the sheet about the y-axis is \(\frac{Ma^2}{6}\)