# How to calculate resisting torque due to moment of inertia

I'm trying to determine whether a motor is suitable for an experiment. I know the motor's torque and the moment of inertia of the disk it will be turning. I was able to find the angular acceleration. The motor has a stall torque. Intuitively, if a large enough mass is attached to the motor, the motor won't be able to turn. But isn't the torque exerted on the axle by the disk just the same as the torque exerted on the disk by the axle? If so, then the motor could spin any free mass, but this doesn't make sense. What am I doing wrong?

• are you able to expand on how you got from 'isn't the torque exerted on the axle by the disk just the same as the torque exerted on the disk by the axle? If so, then the motor could spin any free mass.'? Indeed, every torque has an equal and opposite torque – L. Maynard Sep 14 '17 at 3:46
• Mathematically, the motor could spin any mass. It would just accelerate much larger masses much more slowly. Physically though, if a counter torque too great is exerted, the motor will stall. If the opposite torque is equal to the motor's torque, then it will never exceed the stall torque because that value is higher than the motor's output torque. If a large mass were attached, the motor would stall (this is an intuitive assumption), meaning the mass is applying more torque on the axle than just the opposite of the output torque. Is this correct? If so, how do I compute this counter torque. – Rafael Sep 14 '17 at 12:07

Consider the typical load curve of a motor. At any speed $\omega$, the motor produces a torque $T(\omega)$. Now attached a rotating mass of MMOI of $I$ and some frictional torque $H$. How much is the acceleration?

$$\dot{\omega} = \frac{T(\omega)-H}{I}$$

What is the maximum speed of the motor? Find the speed that makes $T(\omega)=H$.

The torque above is given by $$T(\omega) = \begin{cases} T_1 & \omega<\omega_1 \\ T_{max} \left( 1 - \frac{\omega}{\omega_{max}} \right) & \omega \ge \omega_1 \end{cases}$$

So the maximum speed under load is

$$\begin{cases} \omega = \omega_{max} \left( 1 - \frac{H}{T_{max}} \right) & H \le T_1 \\ \omega =0 & H > T_1 \end{cases}$$

As you can see if the load $H$ is high enough the motor will stall.

• So the point is, it comes down to the friction. If starting friction is greater than stall torque, the disk never moves. Otherwise it does. In the real world, starting friction will tend to scale up with the mass of the disk. – Ben51 Oct 2 '18 at 1:48

If your motor has a stall torque, then it means that it has a magnetic field rotating at a certain angular velocity, providing the torque. If therefore, the motor cannot accelerate the mass to a certain speed fast enough, it will stall and likely overheat up as it dissipates the electrical energy as heat.

The motor - disk system turning friction plus the disk's moment of inertia determine how much torque the motor must exert in order to achieve the required minimum angular acceleration.

One way to calculate the motor torque at stall, over and above the system friction torque, is to load the motor with a hanging variable mass via a pulley to find the limiting mass or cause the motor torque to push on a weigh scale via a stick-type attachment, and calculate the torque by multiplying the mass on end of pulley (or the scale indication) by gravitational acceleration and the stick length from the motor rotational centre.