Question

A particle is located at each corner of an imaginary cube. Each edge of the cube is 0.905 m long, and each particle has a mass of 0.584 kg. What is the moment of inertia of these particles with respect to an axis that lies along one edge of the cube?

Answer 1

moment of inertia = 3.826 kg m^2

Step-by-step explanation:

moment of inertia for given imaginary cube condition is given as

`I = \sum m_i r_i^2`

`= 4* m * l^2 +  2* m* (l√(2)^2)`

`4* m * l^2` is due to four masses at distance l

`2* m* (l√(2)^2)` is due to two masses at distance l√2

`= 4ml^2 + 4ml^2`

`= 8ml^2`

`= 8* 0.584 * 0.905^2`

= 3.826 kg m^2