Thanks a lot for the help, I really appreciate it. I correctly got (angular velocity of rotor around its axis) = (gravitational force of entire system)*(radius around pivot)/((given moment of interia)*(angular velocity around pivot)). My final answer turned out to be 2074.55 rev/min.
Thank you...
I have torque=I*alpha and L=I*omega
Therefore I can say that torque/alpha = L/omega
Solving for omega, i get omega=(L*alpha)/torque
and since L=r*p and p=m*v, I get L=r*m*v
Also, I know that torque=r*F
Plugging the L and torque into the omega equation, I get omega = r^2*m*v*a/(r*F) which...
Well I know that torque = (lever arm) * (force)
So I set up my equation (for part a) and have torque = .04 * (.065+.13) * 9.8
Once I found my torque, though, I'm not sure how that helps me find the upward force exerted by the pivot.
"Lagrangian" no I don't know. I'm taking AP Physics C in high school right now. If you don't know what it is, it's basic physics with a bit of calculus.
Homework Statement
The rotor (flywheel) of a toy gyroscope has mass 0.130 kg. Its moment of inertia about its axis is 1.40 * 10^-4 kg·m^2. The mass of the frame is 0.0650 kg. The gyroscope is supported on a single pivot (Diagram: http://www.webassign.net/yf10/10-43.gif) with its center of mass...