# What is this torque I am measuring?

I have a small brushless motor running at 3 V and a cardboard disc (about 5.5 cm radius, 3 mm thick) which can be attached and detached from the shaft of the motor. When the disc is not attached the current drawn is 0.015 A and 0.11 A when the disc is attached.

I agree that the angular velocity of the system is lower with the disc attached because of the conservation of angular momentum. However, I am unsure what the torques values I have calculated actually represent. By using

$$\tau=\frac{P}{\omega}$$

after measuring the angular velocity with and without the disk to be $$\ \omega_{disc}=105 \ rad/s\$$ and $$\ \omega=418 \ rad/s\$$ I calculated the torque with the disk to be $$\\tau_{disc}=0.003 \ Nm\$$ and without the disc to be $$\\tau=0.0001 \ {Nm}\$$ . But what do these torque values represent?

I prefer to think about these rotational system as if they were linear systems as I find them easier to think about conceptually. Let's say I have a box on the ground and I push on it to accelerate it. My force must be larger than the frictional force. Once I have accelerated it to a desired speed my force decreases to the point where my force equals the frictional force and the box is travelling at this desired speed. Let's say the system uses 100 W of power when travelling at constant velocity. Using

$$F=\frac{P}{v}$$

will only tell you the force I am exerting to counteract friction. It does not tell you the force I exerted to accelerate the box in the first place.

Is this the same principle as the torque of the motor? Is it telling you the torque required to maintain a certain angular velocity? If so, how can we calculate the max torque output of the motor?

• 0.0001 Nm is the torque needed to spin the unloaded shaft at 418 radians per second. You might assume it to be linearly increasing with angular speed. Jan 10, 2022 at 13:37

When nothing is attached to the motor shaft, the measured power is entirely losses in the motor. One part of those losses is the power lost due to the resistance of the copper windings. That loss is called the copper loss or I squared R loss. There is also power lost in the iron due to hysteresis and eddy currents, but that might be negligible in some motors. That is called iron loss. A third loss is mechanical loss, That is due to friction between the moving parts and aerodynamic drag. Aerodynamic drag is also called windage.

When you attach a disk to the shaft, there is aerodynamic drag associated with that. Aerodynamic drag is far from linear. That power is approximately proportional to speed cubed. Note that the power increase due to adding the drag of the disc also increases the current and thus the copper loss.

I agree that the angular velocity of the system is lower with the disc attached because of the conservation of angular momentum.

No. Conservation of angular momentum has nothing to do with it.

Mechanical power = torque * angular speed in rad/s

The 'torque' you are calculating by (presumably) motor power / speed, assuming the motor to be 100% efficient, is the torque required to keep the motor and load rotating against all losses. These losses include air resistance on the disc, air resistance on the rotor, bearing friction, and most importantly if you measure power at the electrical input of the motor, electrical losses in the motor and the drive.

As the motor will be less than 100% efficient, and some losses are internal to the motor, the torque the motor can deliver to a load at that speed will be somewhat less than your calculated torque.

There are two principal ways to measure the torque of a motor properly. One is to mount the motor on a sprung arm and measure the torque required to keep it still when driving a load. The other it to load the shaft with a brake that produces a constant or known torque. A common load for this purpose is another similar motor or generator on a spring arm which can measure the torque, and a variable electrical load to change the torque required to turn it.