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Step-by-Step Solution
Step 1: Understand the Mass Distribution
The mass per unit area of the disc is given by
$ \sigma(r) = k\,r^2 $,
where $r$ is the distance from the center. The total mass $M$ of the disc can be expressed in terms of $k$ by integrating over the entire surface of the disc.
Step 2: Express the Total Mass of the Disc
The differential area element in polar coordinates is $ dA = 2\pi r \, dr $. Therefore, the mass element $ dm $ at a distance $r$ from the center is
$ dm = \sigma(r)\,dA = k\,r^2 \times 2\pi r\,dr = 2\pi k \,r^3\,dr $.
Integrating from $r = 0$ to $r = R$ gives the total mass $M$:
$$
M = \int_{0}^{R} 2\pi k \, r^3 \, dr = 2\pi k \int_{0}^{R} r^3 \, dr \,.
$$
Evaluating,
$$
M = 2\pi k \left[ \frac{r^4}{4} \right]_{0}^{R} = 2\pi k \, \frac{R^4}{4} = \frac{\pi k R^4}{2}.
$$
Step 3: Determine the Constant $k$ in Terms of $M$
From
$ M = \frac{\pi k R^4}{2}, $
we can solve for $k$:
$$
k = \frac{2M}{\pi R^4}.
$$
Step 4: Set Up the Moment of Inertia Integral
The moment of inertia of a continuous mass distribution about an axis perpendicular to the plane and passing through the center is given by
$$
I = \int r^2 \, dm \,.
$$
Using $ dm = 2\pi k \, r^3 \, dr $ from above, we get
$$
I = \int_{0}^{R} r^2 \bigl(2\pi k \, r^3 \, dr \bigr) = 2\pi k \int_{0}^{R} r^5 \, dr.
$$
Step 5: Evaluate the Moment of Inertia Integral
Substitute $ k = \frac{2M}{\pi R^4} $:
$$
I = 2\pi \left(\frac{2M}{\pi R^4}\right) \int_{0}^{R} r^5 \, dr
= \frac{4M}{R^4} \int_{0}^{R} r^5 \, dr
= \frac{4M}{R^4} \left[\frac{r^6}{6}\right]_{0}^{R}.
$$
Thus,
$$
I = \frac{4M}{R^4}\times \frac{R^6}{6} = \frac{4M}{R^4} \times \frac{R^6}{6} = \frac{4M R^2}{6} = \frac{2 M R^2}{3}.
$$
Step 6: State the Final Answer
The moment of inertia of the disc about the given axis is
$$
I = \frac{2 M R^2}{3}.
$$
