The surface density (mass/area) of a circular disc of radius a depends on the distance from the centre as `rho(r)=A+Br.` Find its moment of inertia about the line perpendicular to the plane of the disc through its centre.

Options

A. `2pia^(4)((A)/(4)+(aB)/(5))`
B. `2pia^(4)((aA)/(4)+(B)/(5))`
C. `2pia^(4)((A)/(4)+(B)/(5))`
D. `2pia^(4)((A)/(4)+(aB)/(5))`

Answer

To find the moment of inertia of a circular disc with a surface density that varies with distance from the center, we can follow these steps:

### Step 1: Define the Surface Density
The surface density of the disc is given by:
\[ \rho(r) = A + Br \]
where \( A \) and \( B \) are constants, and \( r \) is the distance from the center of the disc.

### Step 2: Consider a Differential Ring
Consider a thin ring of radius \( r \) and thickness \( dr \) in the disc. The area \( dA \) of this ring is given by:
\[ dA = 2\pi r \, dr \]

### Step 3: Calculate the Mass of the Differential Ring
The mass \( dm \) of the differential ring can be calculated using the surface density:
\[ dm = \rho(r) \cdot dA = (A + Br) \cdot (2\pi r \, dr) = 2\pi r (A + Br) \, dr \]

### Step 4: Moment of Inertia of the Differential Ring
The moment of inertia \( dI \) of the differential ring about the axis perpendicular to the plane of the disc through its center is given by:
\[ dI = r^2 \, dm = r^2 \cdot (2\pi r (A + Br) \, dr) = 2\pi r^3 (A + Br) \, dr \]

### Step 5: Integrate to Find Total Moment of Inertia
To find the total moment of inertia \( I \) of the disc, we integrate \( dI \) from \( r = 0 \) to \( r = a \) (the radius of the disc):
\[ I = \int_0^a dI = \int_0^a 2\pi r^3 (A + Br) \, dr \]

### Step 6: Expand the Integral
Expanding the integral:
\[ I = 2\pi \int_0^a (A r^3 + Br^4) \, dr \]

### Step 7: Calculate the Integral
Now, we can calculate the integrals separately:
\[ \int_0^a A r^3 \, dr = A \left[\frac{r^4}{4}\right]_0^a = \frac{A a^4}{4} \]
\[ \int_0^a Br^4 \, dr = B \left[\frac{r^5}{5}\right]_0^a = \frac{B a^5}{5} \]

### Step 8: Combine the Results
Substituting these results back into the expression for \( I \):
\[ I = 2\pi \left( \frac{A a^4}{4} + \frac{B a^5}{5} \right) \]

### Step 9: Simplify the Expression
Thus, the moment of inertia \( I \) becomes:
\[ I = \frac{2\pi a^4}{4} A + \frac{2\pi a^5}{5} B \]
Factoring out common terms:
\[ I = \frac{2\pi a^4}{20} (5A + 4B a) \]

### Final Result
The moment of inertia of the disc about the axis perpendicular to its plane through the center is:
\[ I = \frac{2\pi a^4}{20} (5A + 4B a) \]