- #1
fahraynk
- 186
- 6
I was trying to figure out the velocity a 6mm diameter shaft would spin given a torque of 4000 N-mm.
M =mass = .0227 Kg
R=radius = .003
i=moment of inertia = 1/2*MR^2 = 1.04E-7
T/R=Force = 1300N <---Am I correct that this is force experienced by shaft?
Bearing friction coefficient = .0015
iX'' + .0015X'=1300
X''+14423X'=1.25E10
X=e^st
S^2+14423S=0
S=14423
X=C1e^-14423t + c2 -> t=0, v=0, C1+C2 = 0. I just set both = 0 for homogeneous solution.
X=C3t
14423C3=1.25E10
X = 86671t
So, the velocity can NOT be 86671 m/s, because its ridiculous. Also after I did this I realized that the coeffiecient of friction was supposed to be for the force normal to the bearing. Still, it SORT of makes sense though, since the friction is ideally 0 for a bearing. So... was I correct here? Is most of the friction just aerodynamic drag on a rotating shaft for a motor? Otherwise motor speed would be ridiculously fast? Or did I make some fatal mistake?
M =mass = .0227 Kg
R=radius = .003
i=moment of inertia = 1/2*MR^2 = 1.04E-7
T/R=Force = 1300N <---Am I correct that this is force experienced by shaft?
Bearing friction coefficient = .0015
iX'' + .0015X'=1300
X''+14423X'=1.25E10
X=e^st
S^2+14423S=0
S=14423
X=C1e^-14423t + c2 -> t=0, v=0, C1+C2 = 0. I just set both = 0 for homogeneous solution.
X=C3t
14423C3=1.25E10
X = 86671t
So, the velocity can NOT be 86671 m/s, because its ridiculous. Also after I did this I realized that the coeffiecient of friction was supposed to be for the force normal to the bearing. Still, it SORT of makes sense though, since the friction is ideally 0 for a bearing. So... was I correct here? Is most of the friction just aerodynamic drag on a rotating shaft for a motor? Otherwise motor speed would be ridiculously fast? Or did I make some fatal mistake?