Find the Rotational Inertia of a Square Rigid Body

[VitaX]

Homework Statement

Four identical particles of mass 0.717 kg each are placed at the vertices of a 3.11 m x 3.11 m square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

Homework Equations

I think I = 1/12M(a^2 + b^2)

The Attempt at a Solution

I don't really know how to go about solving this, the way its worded is confusing me.

 

[VitaX]

Anyone can give me a hint or something on how to do this?

 

[Redbelly98]

Homework Equations

I think I = 1/12M(a^2 + b^2)

That doesn't look right.

The Attempt at a Solution

I don't really know how to go about solving this, the way its worded is confusing me.

Look up the moment of inertia for a point mass (particle) that is a distance r from the rotation axis.

 

[VitaX]

I = mr^2

I thought I had to use I = Icom + Mh^2

The hard part for me is drawing these out and finding what r would be each part.

 

[Redbelly98]

I = mr^2

I thought I had to use I = Icom + Mh^2

You could use Icom + Mh², but for a few point masses it's probably easiest to calculate ∑mr² directly.

The hard part for me is drawing these out and finding what r would be each part.

I would start by simply drawing a square first. Then figure out where the axis is in relation to the square.

If the axis is perpendicular to the plane of the square, you can just draw a dot or "x" in the figure showing where it passes through the plane.

 

[VitaX]

You could use Icom + Mh², but for a few point masses it's probably easiest to calculate ∑mr² directly.I would start by simply drawing a square first. Then figure out where the axis is in relation to the square.

If the axis is perpendicular to the plane of the square, you can just draw a dot or "x" in the figure showing where it passes through the plane.

I understand that part a is just I=4M(L/2)^2

Part b is giving me trouble but I think it is I = 2M(L/2)^2 + 2M(sqroot((L/2)^2 + L^2))^2

Part c is 2M(sqroot((L/2)^2 + (L/2)^2))^2 which goes down to ML^2 which is same as part a