Question

Three particles of masses 1 g, 2g and 3 g are kept at points (2cm,0), (0.6 cm), (4cm, 3cm) find moment of inertia of all three particles (in gm-cm(2)) about (a) x-axis (b). Y-axis (c). Z-axis.

Answer 1

Given:  

1. Masses of all particles

2. Distance of particles

To find:

Moment of Inertia

Solution :

1. About X axis

Total Inertia = Sum of individual Inertias along axis

I=$I_{1}$+$I_{2}$+$I_{3}$

According to the formula of inertia we get:

$I_{1}$=$m_{1}r_{1}^{2}$    

$I_{2}$= $m_{2}r_{2}^{2}$

$I_{3}$= $m_{3}r_{3}^{2}$

I= $m_{1}r_{1}^{2}$+ $m_{2}r_{2}^{2}$+$m_{3}r_{3}^{2}$

As we know:

r is perpendicular distance from x axis

I=$1\times(0)^2+2\times(6)^2+3\times(3)^2$

I=99 $gcm^2$

2. About Y axis , we get:

I=$I_{1}$+$I_{2}$+$I_{3}$

$I_{1}$=$m_{1}r_{1}^{2}$    

$I_{2}$= $m_{2}r_{2}^{2}$

$I_{3}$= $m_{3}r_{3}^{2}$

I= $m_{1}r_{1}^{2}$+ $m_{2}r_{2}^{2}$+$m_{3}r_{3}^{2}$

Puttting the values we get:

$I=1\times(2)^2+2\times(0)^2+3\times(4)^2$

I=52  $gcm^2$

3. About z axis

 I=$I_{1}$+$I_{2}$+$I_{3}$

$I_{1}$=$m_{1}r_{1}^{2}$    

$I_{2}$= $m_{2}r_{2}^{2}$

$I_{3}$= $m_{3}r_{3}^{2}$

I= $m_{1}r_{1}^{2}$+ $m_{2}r_{2}^{2}$+$m_{3}r_{3}^{2}$

Now:

R=$\sqrt( 3^2+4^2)$ =5

I=$1\times(2)^2+2\times(6)^2+3\times(5)^2$ =151$gcm^2$

Thus we get the moment of inertia.