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)

$^{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^{b}_{a} \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^{b}_{a} dr $
$ m = \sigma_{0} 2\pi\left(b-a\right) $
Radius of gyration
$ k= \sqrt{\frac{I}{m}} = \sqrt{\frac{\left(b^{3} -a^{3}\right)}{3\left(b-a\right)}} $
$ = \sqrt{\left(\frac{a^{3}+b^{3}+ab}{3}\right)} $