Question

Mass per unit area of a circular disc of radius$a$ depends on the distance$r$ from its center as$\sigma \left(r\right)=A+Br$. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is

• A. $2{\mathrm{\pi a}}^4\left(\frac{\mathrm{A}}{4}+\frac{\mathrm{aB}}{5}\right)$
• B. $2{\mathrm{\pi a}}^4\left(\frac{\mathrm{aA}}{4}+\frac{B}{5}\right)$
• C. ${\mathrm{\pi a}}^4\left(\frac{\mathrm{A}}{4}+\frac{\mathrm{aB}}{5}\right)$
• D. $2{\mathrm{\pi a}}^4\left(\frac{\mathrm{A}}{4}+\frac{B}{5}\right)$

Answer 1

The correct option is A

$2{\mathrm{\pi a}}^4\left(\frac{\mathrm{A}}{4}+\frac{\mathrm{aB}}{5}\right)$

Step 1. Given data:

The radius of the circular disc$=a$

The distance from the center$=r$

Mass per unit area of the disc$\sigma \left(r\right)=A+Br$

Step 2. Formula used:

The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is

$\begin{array}{rcl}dI& =& dm{r}^2\\ I& =& \int dm{r}^2\end{array}$ ($dm=$Total mass distribution)

Step 3. Calculating the moment of inertia

The total mass distribution$m=$ mass per unit area$\times$ total area of the circular disc

$\begin{array}{rcl}m& =& \sigma \left(r\right).\pi r^2\\ dm& =& \sigma \left(r\right).2\pi rdr\\ I& =& {\int }_0^a\left(A+Br\right)2\pi r.r^{2}dr\\ & =& 2\pi {{\left[\left(\frac{A{r}^4}{4}\right)+\left(\frac{B{r}^5}{5}\right)\right]}^a}_0\\ & =& \left[\left(\frac{A{a}^4}{4}\right)+\left(\frac{B{a}^5}{5}\right)\right]2\pi \\ & =& 2\pi a^4\left[\left(\frac{A}{4}\right)+\left(\frac{Ba}{5}\right)\right]\end{array}$

Thus, the moment of inertia at the center is $2{\mathrm{\pi a}}^4\left(\frac{\mathrm{A}}{4}+\frac{\mathrm{aB}}{5}\right)$

Hence, option A is the correct answer.