Recent content by duffman1278

I cannot seem to figure out how to start this problem when I eliminate the 12V source. I don't seem to see any resistors that are in parrallel or series. I'm just completely blank on this. Hopefully someone can put me in the right direction for this problem already...

It's online. I'll ask my professor about it then. This online thing has issues sometimes.

You're waaaaaay too paranoid. Don't even sweat it.

Any help?

Okay so I went back and tried it again. For Ix'x'=\frac{mr^2}{2}+md2 m=(7830kg/m3)(\frac{volume of cylinder}{2})=5.95kg r=.110m d=.07m+\(frac{4*.11}{3\pi})=.116m Ix'x'=.117kg*m2 This ended up being wrong.

oye vay! Of course wrong axis. fml lol I'll try that then.

Hmm I rechecked the equation and it seems to be the correct one to use, unless I'm missing somethings here or am looking at it wrong? Because the line Xo is under the center of the half circle. As for the distance that makes sense so I used (4r/3pi) for the centroid and added that to the 70mm.

Homework Statement specific volume of steel = 489 lb/ft^3 http://ultraxs.com/image-5D1E_4AFD177E.jpg Homework Equations I tried using Ixo=1/4mr2+1/12ml2+md2 The Attempt at a Solution I tried using d=70mm because of the parallel axis therm, but that was wrong. my... r=110mm...

I love you! omfg this stupid problem was so easy the entire time.

The 4 degree's I put on here was just a type-o this is what I got when I redid it. \phi= .069813 T(shaft)= 52,521.1312 lb-in G= 12x106 \tau= 12,000psi L= 120" I solved for T as you said in which I got T=\stackrel{\tau*J}{c} I then plugged that T into the \phi=\stackrel{TL}{JG}...

Homework Statement Homework Equations tao=Tc/J theta=TL/JG The Attempt at a Solution I got the torque of the motor as 210,085lb-in Then I converted the 4* to .698 rad From there I plugged it into the equations. I solved for T from tao=Tc/J then plugged that T into the...