Recent content by Malor

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150 views, no replies... did I put this in the wrong section? ;)

No... though both objects have the same potential Energy at t = 0s, they have different "moments of inertia" and since they are both rolling they have both kinetic and rotational Energy when they reach the bottom of the inclined plane. The hoop will be slower. Erot = 1/2 J ω²

I deleted something that I thought made no sense and replaced it with what I think the answer to be ;p h = cos 60 * r Because at point C, the object has kinetic, potential and rotational Energy.

With Erad = 1/2 Jω², ω=v/b, b=2/6∏r EkinB - Erad - EpotC = EkinC => vB²-2r²(v²/((2/6)∏r)²-2gh=v² => v=1.32m/s says my calculator

For the speed at point B try the following: EpotA = EkinB mgh = m/2*v² v = sqrt(2gh) (h=0,5m) Check what you did differently, and that's where you went wrong. EDIT: Ah... Energy at point C... oops.

Hi, it's my first time on these forum, so I hope that I made this post correctly. I have an example solution to this problem using tensors and matrices ^^ but I wanted to solve the problem in a different way and would like some feedback on whether or not this solution is correct. Homework...