There might be another way to ask this question but I don't know what that would be. Say I have the following setup:

enter image description here

The shaft is turned by the motor, which itself is fixed and cannot move. The key/lever is embedded in the wheel and attached solidly to the shaft. The wheel is difficult to turn and is only connected to the motor by the shaft.

When the motor turns the shaft, does this create a lateral force on the bearings?

Additional notes:

If the wheel were removed and the shaft and lever were sat against a flat surface, turning the shaft would obviously apply a lateral force on the bearings (it'd lift the motor). Since the shaft is also embedded in the wheel I don't know whether it would still try to lift the wheel.

If the lever were duplicated on the opposite side of the shaft it would definitely not create a lateral force.


1 Answer 1


There is no lateral force on the bearings in the case of a rotating wheel, as opposed to the shaft's key hitting a solid object like the table. More precisely:

$\Sigma F = 0$ but $\Sigma \tau \ne 0$ in the case of shaft that rotates an object about it's center of mass.

$\Sigma F \ne 0$ and $\Sigma \tau \ne 0$ in the case of the shaft that strikes an object like a flat table.

To consider why this is true, please allow me to simplify your picture somewhat. Let's ignore gravity, and instead of a flat 'key', the shaft has a single peg that rotates. The single peg avoids having to consider the integration of torque. See the picture below: torque on a shaft

You can see the clock-wise rotating shaft's peg on the left striking the flat surface would produce an upward force (on the shaft). This force is unbalanced and so the motor would buck upwards as you correctly expect.

The picture on the right shows the same upward force on the shaft from the wheel at the point of the peg. However, there is also an additional force on the shaft from the center of the wheel. This force is a constraining force: because the wheel is constrained to rotation and not translation, there must be an additional force from the shaft to keep it stationary. The constraint force will guarantee $\Sigma F = 0$. Since the net force is zero, the motor feels no 'bucking' from rotation of the wheel. It should be noted the net torque is not zero because the constraint force is located at the axis of rotation.

  • $\begingroup$ I'll add that this is exactly what I believed would be the case, but I didn't know how to prove it. $\endgroup$
    – Clonkex
    Commented Feb 7, 2018 at 2:16

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