Don’t ground your motors; instead, ensure the motor installation is properly bonded.
One solution that is sometimes put forth to prevent a repeat episode of catastrophic failure of motor bearings is to ensure the motor is “properly grounded.” So a ground rod is driven next to the motor, and the monthly PM is updated to include visually inspecting the connections.
The motor is not a separately derived source (see Art. 100 definition in the National Electrical Code). It’s a load. So right away, we see this “solution” is not following the requirements of NEC Art. 250 because this portion of the Code has you ground sources but bond loads [Sec. 250.4].
Section 250.4 can be a bit confusing because for grounded systems, we’re told to connect such things as metallic raceway to earth [250.4(A)(2)]; but this is achieved via the equipment grounding conductor (EGC), which is actually a bonding conductor that is ultimately grounded. The metallic raceway itself is part of the EGC.
Let’s stop to think about what this ground connection at the motor does. That ground rod is going to provide a high-impedance path for stray current to get back to the source. The path through the motor bearings is of a much lower impedance. If you calculate the parallel currents involved, you’ll find that ground rod is pointless. And it is probably a tripping hazard.
Don’t ground your motors (see Art. 100 definition of “ground”). Instead, ensure the motor installation is properly bonded (see Art. 100 definition of “bonding”).
Your goal is to reduce the amount of undesired current going through the bearings. Although it is not true that electricity “takes the path of least resistance,” it is true that much more of it will flow through a low-impedance path than through a parallel high-impedance path.
By creating a metallic path (via bonding jumpers) to the EGC (which is actually bonding the equipment), you provide the undesired electrical current a very low-impedance path back to its source.
To understand this concept, just draw a simple parallel circuit with two resistors. Label one resistor “1 ohm” and the other resistor “2,000 ohms.” Now assuming a 100V power supply, calculate the current flowing through each resistor. In the real world, there is a much greater difference in impedances, but this simple exercise helps illustrate why you bond (low impedance) your motors rather than ground (high impedance) them.