Answer:
The moment of inertia of these particles is 3.82 kg-m².
Explanation:
Given that,
Edge of the cube = 0.905 m
Mass of particle = 0.584 kg
We need to calculate the moment of inertia of these particles
Using formula of moment of inertia
[tex]I=\sum m_{i}r_{i}^2[/tex]
[tex]I=4ml^2+2m(l\sqrt{2})^2[/tex]
[tex]I=4ml^+4ml^2[/tex]
[tex]I=8ml^2[/tex]
Put the value into the formula
[tex]I=8\times0.584\times(0.905)^2[/tex]
[tex]I=3.82\ kg m^2[/tex]
Hence, The moment of inertia of these particles is 3.82 kg-m².
Answer:
moment of inertia = 3.826 kg m^2
Explanation:
moment of inertia for given imaginary cube condition is given as
[tex]I = \sum m_i r_i^2[/tex]
[tex]= 4\times m \times l^2 + 2\times m\times (l\sqrt{2}^2)[/tex]
[tex]4\times m \times l^2[/tex] is due to four masses at distance l
[tex]2\times m\times (l\sqrt{2}^2)[/tex] is due to two masses at distance l√2
[tex]= 4ml^2 + 4ml^2[/tex]
[tex]= 8ml^2[/tex]
[tex]= 8\times 0.584 \times 0.905^2[/tex]
= 3.826 kg m^2
