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Step-by-Step Solution
Step 1: Understand the Problem
We have a thin circular plate of radius $R$ and total mass $M$, with a density that varies according to
$$\rho(r) = \rho_{0} \, r \,,$$
where $r$ is the distance from the center of the plate and $\rho_{0}$ is a constant. We want to find the moment of inertia of this plate about an axis perpendicular to the plate and passing through its edge. The result can be written in the form
$$I = a \, M R^{2}\,.$$
We need to determine the coefficient $a$.
Step 2: Express the Total Mass $M$
First, we calculate the total mass $M$ by integrating the mass density over the entire area of the circular plate. Let us consider a small annular ring of radius $r$ and thickness $dr$.
• The mass of this small ring is density $\times$ area of the ring.
• The density at radius $r$ is $\rho_{0} \, r$.
• The circumference of the ring is $2\pi r$, and its width is $dr$, so the area of the ring is $2\pi r \, dr$.
Therefore, the mass of the ring element is:
$$dM = \rho_{0} \, r \,\times 2\pi r \,dr = 2\pi \rho_{0} \, r^{2}\, dr\,.$$
Integrating from $r=0$ to $r=R$ gives the total mass:
$$
M = \int_{0}^{R} 2\pi \rho_{0} \, r^{2}\, dr
= 2\pi \rho_{0} \int_{0}^{R} r^{2}\, dr
= 2\pi \rho_{0} \left[\frac{r^{3}}{3}\right]_{0}^{R}
= \frac{2\pi \rho_{0} R^{3}}{3}\,.
$$
Step 3: Moment of Inertia About the Center ($I_C$)
Next, we find the moment of inertia of the plate about an axis perpendicular to the plate and passing through its center (let us call it $I_{C}$). For the same annular ring element:
• The mass element is $dM = 2\pi \rho_{0} r^{2} \,dr$.
• Each ring is at distance $r$ from the center, so the moment of inertia of the ring about the center is $r^{2} \times dM = r^{2}\,(2\pi \rho_{0} r^{2} \,dr) = 2\pi \rho_{0} r^{4} \,dr$.
Hence,
$$
I_{C} = \int_{0}^{R} 2\pi \rho_{0} \, r^{4}\,dr
= 2\pi \rho_{0} \int_{0}^{R} r^{4}\, dr
= 2\pi \rho_{0} \left[\frac{r^{5}}{5}\right]_{0}^{R}
= \frac{2\pi \rho_{0} R^{5}}{5}\,.
$$
(Note: In many references, the integral is sometimes written with additional steps, but the core process remains the same.)
Step 4: Apply Parallel Axis Theorem for the Axis Through the Edge
To find the moment of inertia about an axis perpendicular to the plate and passing through its edge, we use the parallel axis theorem. The theorem states:
$$
I = I_{C} + M R^{2}\,.
$$
Substituting $I_{C} = \frac{2\pi \rho_{0} R^{5}}{5}$ and
$M = \frac{2\pi \rho_{0} R^{3}}{3}$, we get
$$
I
= \frac{2\pi \rho_{0} R^{5}}{5}
+ \left(\frac{2\pi \rho_{0} R^{3}}{3}\right) R^{2}\,.
$$
Simplify:
$$
\frac{2\pi \rho_{0} R^{5}}{5}
+ \frac{2\pi \rho_{0} R^{5}}{3}
= 2\pi \rho_{0} R^{5}
\left(\frac{1}{5} + \frac{1}{3}\right)
= 2\pi \rho_{0} R^{5}
\left(\frac{3}{15} + \frac{5}{15}\right)
= 2\pi \rho_{0} R^{5}
\left(\frac{8}{15}\right)
= \frac{16\pi \rho_{0} R^{5}}{15}\,.
$$
Step 5: Express $I$ in the Form $a M R^{2}$
We have
$$
I = \frac{16\pi \rho_{0} R^{5}}{15}\,.
$$
Also, from Step 2,
$$
M = \frac{2\pi \rho_{0} R^{3}}{3}\,.
$$
Therefore,
$$
I
= \frac{16\pi \rho_{0} R^{5}}{15}
= \left(\frac{16\pi \rho_{0} R^{5}}{15}\right)
\cdot \frac{1}{1}\,.
$$
Factor out $M = \frac{2\pi \rho_{0} R^{3}}{3}$:
$$
M R^{2}
= \left(\frac{2\pi \rho_{0} R^{3}}{3}\right) R^{2}
= \frac{2\pi \rho_{0} R^{5}}{3}\,.
$$
Compare $\frac{16\pi \rho_{0} R^{5}}{15}$ with $\frac{2\pi \rho_{0} R^{5}}{3}$. Note that
$$
\frac{16\pi \rho_{0} R^{5}}{15}
= \frac{8}{5} \times \left(\frac{2\pi \rho_{0} R^{5}}{3}\right)
= \frac{8}{5} \, M R^{2}\,.
$$
Hence,
$$
a = \frac{8}{5}\,.
$$
Final Answer
The value of the coefficient $a$ in the expression
$I = a\,M\,R^{2}$
is
$$
\boxed{\frac{8}{5}}\,.
$$
