Question 245: Centre of Mass and Rotational Motion

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 $\left( {{{{\sigma _0}} \over r}} \right)$
, then the radius of gyration of the disc about its axis passing through the centre is:
![](https://imagex.cdn.examgoal.net/3rsa0w2w9jx3ajhyi/d95c13d8-1802-45e6-9392-c9f548b64051/c1aaa0a0-9298-11e9-86c1-73f7f991b766/file-3rsa0w2w9jx3ajhyj.png)

Accepted Answer

dI = (dm)r
2
= ($\sigma $ dA)r
2
= $\left( {{{{\sigma _0}} \over r}2\pi dr} \right){r^2} = {\sigma _0}2\pi {r^2}dr$
$I = \int {DI}  = \int\limits_a^b {{\sigma _0}2\pi {r^2}dr} $
$ = {\sigma _0}2\pi \left( {{{{b^3} - {a^3}} \over 3}} \right)$
m = $\int {dm = \int {\sigma dA = {\sigma _0}2\pi } } \int\limits_a^b {dr} $

Answer

$\sqrt {{{{a^2} + {b^2} + ab} \over 2}} $

Answer

$\sqrt {{{a + b} \over 3}} $

Answer

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

Answer

$\sqrt {{{a + b} \over 2}} $

Centre of Mass and Rotational Motion

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( {{{{\sigma _0}} \over r}} \right) , then the radius of gyration of the disc about its axis passing through the centre is:

Choose the correct answer: