Question 250: 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:

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: