Principal Moments of Inertia

Could someone explain how the answer for b) is obtained. Am I correct saying that the centre of mass lies on the z-plane (vertical direction) at (m2h/(2m1+m2))? where mew = total mass = 2m1+m2?

For the y-plane (horizontal direction) I obtain (m1a/(2m1+m2)), and x=0. I'm not sure what I need to do next and any help would be grateful! (

Yes.



Not sure what you're saying there. The centre of mass is on the axis of symmetry.

What does the (m1a/(2m1+m2)) refer to?

Where does the x=0 come from?


I'd be inclined to stick to the axes as given x1, x2.

Then apply the formula sigma mr^2 about each of the principal axes.

The x2 axis is easy.

We have $\displaystyle m_1\left(\frac{a}{2}\right)^2+m_1\left(\frac{a}{2}\right)^2= \frac{1}{2}m_1a^2$

For the x1 axis you need your distances relative to the horizontal axis through the centre of mass. And it's somewhat more involved, but does work out to the value given.

Sorry just ignore the first part, I'm over complicating things. I've managed to get the answer for x2 by $\displaystyle 2m_1\left(\frac{m_2 h}{2m_1 + m_2}\right)^2+m_2\left(\frac{-2m_1 h}{2m_1 +m_2}\right)^2$ Is this correct?