Recent content by Rheegeaux

I got it! But could you exlpain why that is the answer? What I did is I treated the first one as a point particle and applied the parallel axis theorem, then what I did is I added that to the moment of inertia of a rod hence I got 17.222. I got it but I want to understand how and why I got it...

1/12ML^2 + 2/3*M(2/3L)^2 + 1/12 ML^2 = and I get 6.29. Please explain the question to me because I don't seem to grasp it yet even make a proper solution. thanks

Homework Statement Two identical rods, each with mass 5.00 kg and length 2.00m are welded as shown in the figure. If the moment of inertia of a rod about an axis passing through its center is Icm = 1/12ML2, what is the moment of inertia of the system of rods about an axis perpendicular to the...

I got it already *ZOINKS* I just needed to use the right hand rule

how did it yield a negative answer? What I got was letter D. my solution is: (5kg)(7.50m/s)(4m)sin(35) + 1/12(5kg)(3m)^2(2.50 rad) = 95.411 Thanks for the reply, I really need to learn this before tomorrow. Cheers :)

[Note: Post moved to homework forum by mentor] So I stumbled upon a reviewer for my physics exam tomorrow and I was wondering how the equation was formulated. Your help is very much appreciated :) ! Normally I would consult my professor for this but it's Sunday in my country today so I can't...