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

Answer 1

Answer: (D)

$\sqrt{\frac{a^2+b^2+ab}{3}}$

$dI=\left(dm\right)r^2$
$=\left(\sigma dA\right)r^2$
$\left(\frac{{\sigma }_0}{r}2\pi rdr\right)r^2$
$=\left({\sigma }_{0}2\pi \right)r^{2}dr$
$I=\int dI={\int }_a^b{\sigma }_{0}2\pi r^{2}dr$
$={\sigma }_{0}2\pi \left(\frac{b^3-a^3}{3}\right)$
$m=\int dm=\int \sigma dA$
$={\sigma }_{0}2\pi {\int }_a^{b}dr$
$m={\sigma }_{0}2\pi (b-a)$
Radius of gyration
$k=\sqrt{\frac{I}{m}}=\sqrt{\frac{(b^3-a^3)}{3(b-a)}}$
$=\sqrt{\left(\frac{a^3+b^3+ab}{3}\right)}$