How Do You Calculate the Moment of Inertia for Combined Square Laminas?

[hms.tech]

Homework Statement

Two uniform square laminas are combined into a single body. One lamina ABCD has mass 5m and the other lamina PQRS has mass m. The lamina PQRS has side 2a, and its vertices are at the mid-points of the sides of ABCD, with P on AB and S on AD. The line PS meets AC at K, and the body rotates in a vertical plane about a horizontal axis k through K (see diagram). Find the moment of inertia of the body about k.

Homework Equations

parallel and the perpendicular axes theorems

The Attempt at a Solution

If i understand the question correctly, is the "dashed"(broken)line, the axis about which we need to find the moment of inertia ? If that is the case, here is what i did :Using the perpendicular axes theorem : (let I_{A} be the moment of inertia about the required axis for ABCD only)
2I_{A} = \frac{5m(2a^{2}+2a^{2})}{3}
I_{A} = \frac{10ma^{2}}{3}

Similarly, let I_{B} be the moment of inertia about the required axis for PQRS only :
Using the perpendicular axes theorem,
I_{B} = \frac{ma^{2}}{3}

The total moment of inertia = I_{A} + I_{B}
Total = \frac{11ma^{2}}{3}

The problem is that the solution notes state the answer to be \frac{40ma^{2}}{3}

 

[rude man]

Since k was not indicated on the diagram it's a bit of a puzzle all right as to what the axis of rotation is. It's supposed to be a "horizontal" axis which would seem to eliminate the dashed line though. I think the intended axis is a "hozizontal" line passing thru point K. In other words, the axis of rotation is parallel to line AB, and also to a line drawn thru SQ, and situated inbetween those two and passing thru point K.

Happy integration!

 

[haruspex]

I would interpret rotating "in a ... plane" as meaning the axis is perpendicular to the lamina.

 

[rude man]

I would interpret rotating "in a ... plane" as meaning the axis is perpendicular to the lamina.

That sounds sound. I agree.