Question

A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc vanes as $\left(\frac{{\sigma }_0}{r}\right)$ , then the radius of gyration of the disc about its axis passing through the centre is

• A. $\frac{a+b}{2}$
• B. $\sqrt{\frac{a^2+b^2+ab}{3}}$
• C. $\sqrt{\frac{a^2+b^2+ab}{2}}$
• D. $\frac{a+b}{3}$

Answer 1

The correct option is B $\sqrt{\frac{a^2+b^2+ab}{3}}$

Step-1:Draw a diagram of the small mass element.
Considering a circular ring of radius $r$ and thickness $dr$

Step-2: Find the moment of inertia and mass of the disc.

Moment of inertia of the elemental ring,
$dI=\left(dm\right)r^2=\left(\sigma dA\right)r^2$

$=\left(\frac{{\sigma }_0}{r}2\pi rdr\right)r^2=\left({\sigma }_{0}2\pi \right)r^{2}dr$

$I=\int dI$

$={\int }_a^b{\sigma }_{0}2\pi r^{2}dr$

$={\sigma }_{0}2\pi \left(\frac{b^3-a^3}{3}\right)$

$m=\int dm=\int \sigma dA={\sigma }_{0}2\pi {\int }_a^{b}dr$

$m={\sigma }_{0}2\pi (b-a)$

Find the radius of gyration of the disc.
Radius of gyration,

$k=\sqrt{\frac{I}{m}}=\sqrt{\frac{(b^3-a^3)}{3(b-a)}}=\sqrt{\left(\frac{b^2+a^2+ab}{3}\right)}$
Final answer is (d)