How to calculate if a motor is capable for a load?

Question

I am trying to figure out if a motor I have is capable of spinning a wheel. Here is all of the information I have:

MOTOR: 190 g.cm min.

WHEEL: 26 cm diameter, weighs about 59 grams

I imagine that I would have to calculate the linear speed of the wheel in order to convert it to the same units that the motor torque is listed in. However, I don't know if I am doing it correctly or what the next step would be to solve this.

Answer 1

Check out my answer to a similar question at the Robotics SE. You are asking only about a motor and a wheel, and you do not give the weight of the vehicle (assuming there is one), so I cannot comment regarding vehicular motion.

Basically, if you assume friction is negligible, the torque a motor outputs is used to accelerate a load. Then you have to ask, "What is my desired acceleration?" A [Honda Civic] can accelerate 0-60mph in about 9 seconds, while a [1968 VW bus] went the same 0-60mph in 37 seconds. Both vehicles eventually get to the same top speed, but higher torque values give you a higher acceleration rate.

You can calculate your own desired acceleration rate by giving a top speed you would like to reach and a time limit in which you would like to achieve that speed. So, for example, if you wanted to get to 10rpm in 4 seconds, you would need:

$$ a = (\mbox{Top speed} - \mbox{Starting speed})/(\mbox{Time limit}) \\ a = (10\mbox{rpm} - 0\mbox{rpm})/(4\mbox{s}) \\ a = 2.5 \frac{\mbox{rpm}}{\mbox{s}} \\ $$

To be used in an equation, you typically have to convert to radians per second, which is easy knowing that $2\pi \mbox{radians} = 1\mbox{revolution}$ and that $1\mbox{min} = 60\mbox{s}$:

$$ \mbox{rpm} = \frac{\mbox{rev}}{\mbox{min}} \\ $$ $$ a = 2.5\frac{\mbox{rpm}}{\mbox{s}} = 2.5\frac{\mbox{rev}}{\mbox{min s}}\\ $$ Convert to $\mbox{rad/s}^2$: $$ a = 2.5\frac{\mbox{rev}}{\mbox{min s}} * \frac{1\mbox{min}}{60\mbox{s}} * \frac{2\pi\mbox{rad}}{1\mbox{rpm}} \\ \boxed{a = 0.2618 \mbox{rad/s}^2} $$

You can calculate your own target acceleration using the same method.

Once you have your target acceleration, you can use the torque formula:

$$ \tau = I \alpha \\ $$

Where $\tau$ is the motor torque in Nm, $\alpha$ is the acceleration in $\mbox{rad/s}^2$, and $I$ is the moment of inertia of your load in $\mbox{kg m}^2$. For a uniform disc, the moment of inertia about the center is given by:

$$ I = mr^2/2 \\ $$

where $m$ is the mass in kg and $r$ is the wheel radius in meters. This is how you would solve for the required (minimum) torque for your target application; as the question has been asked you do not know the acceleration you can achieve, so you rearrange the torque equation to solve for it as follows:

$$ \tau = I \alpha \\ \frac{\tau}{I} = \alpha \\ $$

Substitute in the definition for $I$ and you can determine your acceleration, after you convert to standard units:

$$ \tau = 190 \mbox{g cm} \\ \tau = 190 \mbox{g cm} * \frac{1 \mbox{kg}}{1000 \mbox{g}} * \frac{1 \mbox{m}}{100 \mbox{cm}} \\ \tau = 0.019 \mbox{kg cm} \\ $$ Here the torque is tricky - to get to Nm from kg-m you need to multiply by the gravitational constant, $9.81 \mbox{m/s}^2$:

$$ \tau = 0.019 \mbox{kg cm} * 9.81\mbox{m/s}^2 \\ \boxed{\tau = 0.1864 \mbox{Nm}} $$ Divide the mass by 1000 to get to kg: $$ m = 59 \mbox{g} * \frac{1 \mbox{kg}}{1000 \mbox{g}} \\ \boxed{m = 0.059 \mbox{kg}} \\ $$ And lastly divide the wheel diameter by 100 to get to meters and by 2 to get to a radius: $$ r = d/2 \\ r = (26 \mbox{cm} * \frac{1 \mbox{m}}{100 \mbox{cm}})/(2) \\ \boxed{r = 0.13 \mbox{m}} \\ $$ Now, plug those values into the rearranged torque equation: $$ \frac{\tau}{mr^2/2} = \alpha \\ \frac{0.1864 \mbox{Nm}}{(0.059 \mbox{kg} * 0.13^2 \mbox{m}^2)/2} = \alpha \\ \frac{0.1864 \mbox{Nm}}{0.0004986 \mbox{kg m}^2} = \alpha \\ \boxed{374 \mbox{rad/s}^2 = \alpha} \\ \alpha = 374 \mbox{rad/s}^2 * \frac{1 \mbox{rev}}{2\pi \mbox{rad}} * \frac{60 \mbox{s}}{1 \mbox{min}} \\ \boxed{\alpha = 3570 \mbox{rpm/s}} \\ $$

This is a lot, but it's not so unreasonable - this article about motion control has the figure below: ("back then" - 2000 - people didn't put graphics on the page, it's linked as Figure 2)

Here you can see the motor accelerates from stopped to 2000 rpm in 0.12 seconds, for an acceleration of 16,667 rpm/s. Your motor is 3,570 rpm/s for your given load. You probably won't see rates quite that high due to friction, but it's a good ballpark.

So, in conclusion, you haven't stated what "adequate" is, but whatever your case, yes this motor will probably be good enough.

PS - again, see my answer about climbing hills for more information on how ramps affect required torque. If you're building a vehicle.