Question 264: Centre of Mass and Rotational Motion

Question

A thin disc of mass M and radius R has mass
per unit area $$\sigma $$(r) = kr2 where r is the distance
from its centre. Its moment of inertia about an
axis going through its centre of mass and
perpendicular to its plane is :

Accepted Answer

$${I_{Disc}} = \int\limits_0^R {\left( {dm} \right)} {r^2} \Rightarrow {I_{Disc}} = \int\limits_0^R {\left( {\sigma 2\pi rdr} \right)} {r^2}$$
$${I_{Disc}} = \int\limits_0^R {\left( {k{r^2}2\pi rdr} \right)} {r^2}$$    Mass of Disc
$${I_{Disc}} = 2\pi k\int\limits_0^R {{r^2}dr} \,\,\,\,M = \int\limits_0^R {2\pi rdr\,k{r^2}} $$
$${I_{Disc}} = 2\pi k\left( {{{{r^6}} \over 6}} \right)_0^R\,\,\,\,M = 2\pi k\int\limits_0^R {{r^3}dr} $$
$${I_{Disc}} = 2\pi k{{{R^6}} \over 6} \,\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R$$
$${I_{Disc}} = {{\pi k{R^6}} \over 3} = \left( {{{\pi k{R^4}} \over 2}} \right){{{R^2}2} \over 3}\,\,\,M = 2\pi k\left. {{{{r^4}} \over 4}} \right|_0^R$$
$${I_{Disc}} = {{M2{R^2}} \over 3};{I_{Disc}} = {2 \over 3}M{R^2}$$

Answer

$${{M{R^2}} \over 3}$$

Answer

$${{M{R^2}} \over 6}$$

Answer

$${{2M{R^2}} \over 3}$$

Answer

$${{M{R^2}} \over 2}$$

Centre of Mass and Rotational Motion

A thin disc of mass M and radius R has mass per unit area $\sigma $(r) = kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is :

Choose the correct answer: