Question

Four point masses, each of mass 1.9 $kg$ are placed at the corners of a square of side 2.2 $m$. Find the moment of inertia of this system about an axis that is perpendicular to the plane of the square and passes through one of the masses.

Answer 1

I = 36.78 kg m^{2}

Explanation:\

Given data:

side of square = 2.2 m

mass =1.9 kg

The moment of inertia i is the total sum of the moments of inertia of the 4 point masses and it is given as

`I = I_1+I_2+I_3`

`= mr^(2) +mr^(2)+m(\sqrt2 r)^(2)`

`= 2mr^(2)+2mr^(2)`

`=4mr^(2)`

=
`4*1.9*2.2^(2)`

I = 36.78 kg m^{2}