Question

4 solid sphere each of radius r and mass m are placed with their centres on the four corners of a square with side b . Find the moment of Inertia of system about 1 side of the square taken as axis?

Answer 1

Given: 4 solid sphere each of radius r and mass m are placed with their centres on the four corners of a square with side b .

To find the moment of Inertia of system about 1 side of the square taken as axis

Solution:

Moment of inertia of the sphere at axis of rotation is given by

$\frac{2}{5}m{r}^2$

Let AB be the axis as shown in the above fig. Then,

Moment of inertia of sphere A about the side AB=Moment of inertia of sphere B about the side AB,

$I_{AB}=I_A+I_B=2\frac{2}{5}m{r}^2=\frac{4}{5}m{r}^2.......\left(i\right)$

Similarly

Moment of inertia of sphere C and D about the CD is

$I_{CD}=I_C+I_D=2\times \frac{2}{5}m{r}^2=\frac{4}{5}m{r}^2)$

Since AB is one fixed side for rotation, hence using the theorem of parallel axes, the moment of inertia of C and D aphere about the axis AB is

$I_{AB\left(CD\right)}=I_{CD}+\sum m_i{r}_i^2$

where $r$ is the perpendicular distance between CD and AB

$I_{AB\left(CD\right)}=\frac{4}{5}m{r}^2+\left(2m\right)b^2...........\left(ii\right)$

From eqn(i) and eqn(ii), we get

Moment of inertia of the four speres about the axis AB as

$=\frac{4}{5}m{r}^2+\frac{4}{5}m{r}^2+2m{b}^2⟹=\frac{8}{5}m{r}^2+2m{b}^2$

is the required moment of Inertia of system about 1 side of the square taken as axis