Question

The density of a circular disc is given as $\sigma =p_{0}X$ where ‘X’ is the distance from the center. Its moment of inertia about an axis perpendicular to an axis perpendicular to its plane and passing through its edge is:

• A. $\frac{15}{16}{P}_0\pi R^5$
• B. $\frac{16}{15}{P}_0\pi R^5$
• C. $\frac{6}{5}{P}_0\pi R^5$
• D. $\frac{5}{6}{P}_0\pi R^5$

Answer 1

The correct option is B

$\frac{16}{15}{P}_0\pi R^5$

Step 1:Given data:

The density of a circular disc is $\sigma ={\rho }_{o}x$

where ‘$x$’ is the distance from the center

Step 2: Formula Used:

According to the parallel axis theorem

$I_o=I_c+M{d}^2$

Where $I_o$ moment of inertia about a point O

$I_c$ moment of inertia about a centroid

Mass density is defined as-

$density=\frac{mass}{volume}$

Step 3: Calculation:

Consider an element $dx$at a distance of $x$ from center.

The mass of that element using the mass density formula,

$dm=\sigma 2\mathrm{\pi }xdx$

it is given that $\sigma ={\rho }_{o}x$ now substituting the value

$\begin{array}{rcl}dm& =& \sigma 2\mathrm{\pi }xdx\\ & =& \left({\rho }_{o}x\right)2\pi xdx\end{array}$

Now use the parallel axis theorem

$dI=dm{x}^2+dm{R}^2$

Substituting the value of $dm$

$\begin{array}{c}dI={\int }_0^R{\rho }_{0}2\pi x^{4}dx+{\int }_0^R{\rho }_{0}2\pi R^2{x}^{2}dx\\ =\frac{{\rho }_{0}2\pi R^5}{5}+\frac{{\rho }_{0}2\pi R^5}{3}\\ =\frac{{\rho }_{0}6\pi R^5+{\rho }_{0}10\pi R^5}{15}\\ =\frac{16{\rho }_0\pi R^5}{15}\end{array}$

Therefore moment of inertia about an axis perpendicular to an axis perpendicular to its plane and passing through its edge is $\frac{16{\rho }_0\pi R^5}{15}$.

Hence, the correct option is B.