Question

Four masses are arranged in square one meter on a side as follows: (clockwise starting in the upper lefthand corner): 7 kg, 2 kg, 9.1 kg, 1 kg. What is the rotational inertia, to the nearest tenth of a kg*m^2, about an axis perpendicular to the square and through the 1 kg mass?

Answer 1

Here ,

For the square , side , a = 1 m

as 2 kg mass is on diagonally oposite end from axis

distance , d = a*sqrt(2)

d = 1.414 m

Now ,

Moment of inertia of a mass , I = m*r^2

Here ,

Inet = 7 * 1^2 + 9.1 * 1^2 + 2*1.141^2

Inet = 20.1 Kg.m^2

the moment of inertia is 20.1 Kg.m^2