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...
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.
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...
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...