Moment of inertia rigid body problem

[haruspex]

Yeah because its a rigid body of many particles. And adding them up, I-total=1/3mL^2+0+mL^2=4/3mL^2

In post #33 you came to understand that the moment of inertia of a rod about an axis parallel to it, distance L away, is mL². Why have you gone back to 1/3mL^2+mL^2?

 

[i_hate_math]

because I think the other rod that's perpendicular to the rotation axis also needs to be taken into account for the total MoI

 

[haruspex]

because I think the other rod that's perpendicular to the rotation axis also needs to be taken into account for the total MoI

Oh, ok, I didn't realize you were including that yet. In that case, yes, that is indeed the total MoI of the H shape about the given axis.

 

[i_hate_math]

Oh, ok, I didn't realize you were including that yet. In that case, yes, that is indeed the total MoI of the H shape about the given axis.

Thanks a lot mate for all the help and detailed explanations. I got the correct answer 6.45 eventually!