,Calculating the Imbalance Force
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where F = the imbalance force, Im = the mass, r = its distance from the pivot, and w (omega) is the angular frequency, equal to 2p times the frequency in Hz..
From this, it is seen that the force on the pivot is proportional to its distance from the center of rotation and to the speed squared.
A rotor containing a heavy spot is not exactly equivalent to the stone on a string. In the case of the stone, the center of gravity of the system is the center of the stone itself, whereas the CG of a rotor with imbalance is outside the imbalance mass and is near the axis of rotation of the rotor.
If the structure holding the bearings in such a system is infinitely rigid, the center of rotation is constrained from moving, and the centripetal force resulting from the imbalance mass can be found from the above formula. This force is borne by the bearings. Now, consider a hypothetical machine where the bearings are not rigidly supported, but are suspended on springs.
Under these conditions the shaft centerline is not constrained, and the rotor will rotate around its center of gravity. The 1 x RPM force on the bearings will be very small because it is only required to accelerate the bearings to the above mentioned amplitude. The double amplitude of vibration of the bearings will be equal to twice the distance between the CG and the centerline of the rotor. Moreover, the amplitude of bearing vibration is constant regardless of the rotor speed, provided the speed is higher than the natural frequency of the spring-rotor system. It is seen here that the vibration amplitude has nothing to do with the above centripetal force formula.
At speeds well below the natural frequency, the system is said to be "spring controlled", and the centripetal force formula holds. Speeds above the natural frequency are in the "mass-controlled" region where the amplitude is constant, and the bearing forces are not so easily predictable, be dependent on the equivalent mass of the bearings and springs.
Static Imbalance