Mass liftable by a \$15 \,kg\cdot cm\$ torque stepper motor

Question

I'm working on a project for automated welding and I'm using a stepper motor. I am confused in relation to its rated \$15\, kg \cdot cm\$ torque. Can you explain what does it signify?

Answer 1

This is more a mechanical engineering question, as @AndyAka pointed out in a comment, but since it is related to a motor specs, I'll give you some clue.

A very basic, intuitive explanation goes like this: essentially the torque of a motor is the "force" the motor can exert to cause rotational motion in whatever you connect to its shaft (more precise explanation down below). Among the characteristics of a stepper motor you have the max torque that the motor can exert.

What is torque actually? It is a physical quantity analogous to force, but specifically defined to deal with rotational motion. Newton's 2nd law states that if you exert a force on a point mass in a given direction, it will be accelerated in the same direction according with the equation (I won't bother you with vectors here, just to keep it simple):

\[ F = m \cdot a \qquad\Leftrightarrow\qquad a = \dfrac F m \]

where \$F\$ is the force in newtons (\$N\$), \$m\$ is the mass in kilograms (\$kg\$) and \$a\$ is the acceleration in meters per squared seconds (\$m/s^2\$). This means that if you apply a \$10N\$ constant force to a \$1kg\$ mass it will accelerate by \$10m/s^2\$ (i.e. its speed will increase by \$10m/s\$ for each second the force is applied).

That equation has to do with linear motion, i.e. the motion in a given direction (using vectors you can generalize that, but that's not the point here). In most machines linear motion is not the norm, but rotational motion is (cogs and wheels everywhere!), since the most common source of mechanical motion are sources of rotational motions (motors and engines, for example). Therefore it would be useful to have an equation like Newton's 2nd law, but relating quantities relevant for rotational motion.

It can be shown that such an equation exists and is this:

\[ T = I \cdot \alpha \qquad\Leftrightarrow\qquad \alpha = \dfrac T I \]

where T is the torque in newtons-meter (\$N\cdot m\$) (beware, it's not newtons per meter), \$I\$ is a quantity called moment of inertia in kilogram-square meters (\$kg\cdot m^2\$) and \$\alpha\$ is the angular acceleration in radians per squared second (\$rad/s^2\$).

That equation is very simple and is valid only if the rotation is around an axis fixed in space, like the shaft of a motor, otherwise nasty vector calculus equations come into play.

More details on how to apply that equation can be found here. Another relevant question on Physics.SE.