Centres of Mass

An object will topple if it's center of mass doesn't lie direc

Will the length along the bottom still be equal to 3r ? If so the, (3-2^0.5)r is beneath the rectangle part of the shape, so I need to only determine if the hypotenuse length of a triangle with sides r and 27/20 r is less than (3-2^0.5) ?

I don't understand what you mean - surely the "bottom" in the diagram you've drawn has length r sqrt(2)?

Your basic plan seems reasonable however, you just need to be careful with the geometry. (And frankly I'm not going to bother checking it without you posting a decent diagram. To be clear - I'd totally understand you not wanting to bother, but if you don't, I'd have to do one if I want to check carefully, and that's more effort than I want to go to myself).

I don't understand what you mean - surely the "bottom" in the diagram you've drawn has length r sqrt(2)?

Your basic plan seems reasonable however, you just need to be careful with the geometry. (And frankly I'm not going to bother checking it without you posting a decent diagram. To be clear - I'd totally understand you not wanting to bother, but if you don't, I'd have to do one if I want to check carefully, and that's more effort than I want to go to myself).

I've just scan read this, what did you get for the location of the .com of the body?

Also what angle is the cylinder at compared to the horizontal or vertical?

x = 27r/20 , y=r

X is from base or top or ... What is the angle the cylinder makes?

As the shape is initially pictured, not my diagram. The x .com is 27r/20 along the horizontal

Your "a" is the horizontal length (height) of the cylinder, have you worked this out in terms of r? If you know the angle (not hard), it's easy.

Your "a" is the horizontal length (height) of the cylinder, have you worked this out in terms of r? If you know the angle (not hard), it's easy.

A= 2rcos45

Sure or sqrt(2)*r.
So does the (horizontal projected) com lie to the left or right of the toppling point?

It lies to the right I think