The moment of inertia for a set of point masses is given by the equation, i=∑nimir2i. what is the moment of inertia for a frame of four point masses positioned at the corners of an essentially massless rigid frame if the frame rotates around one side like a door rotates on its hinges. each mass is 2.00 kg and the sides of the square frame all have a length of 0.68 m.

Explanation:

It is given that,

Mass of each mass, m = 2 kg

Side of the square frame, l = 0.68 m

It is clearly written in the question that the four point masses positioned at the corners of an essentially massless rigid frame if the frame rotates around one side like a door rotates on its hinges. Let ABCD are the vertices of square and AB acts like a door.

The moment of inertia for a set of point masses is given by :

So,

The sum of inertia at A and B is 0 because l = 0

So,the moment of inertia for a frame of four point masses positioned at the corners is . Hence, this is the required solution.

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