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.


1 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) Motor ramp

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.

  • $\begingroup$ This wheel is being used to launch ping pong balls, so it will not be under too much stress beyond just spinning the wheel. $\endgroup$ Aug 26, 2015 at 23:01
  • $\begingroup$ Yeah I figured it was something like that or a large gyro when you had asked specifically about a wheel with no reference to a vehicle. $\endgroup$
    – Chuck
    Aug 27, 2015 at 2:36
  • $\begingroup$ Reading through your response, I noticed that you said the motor should work. However, when I tried this out, the motor got very hot and started to smell bad. Is this normal for a motor to do? According to the information given, this motor should be able to handle 190g.cm with no load. I just assumed the "no load" detail didn't matter because it doesn't seem logical to give a "no load" torque, but does that affect the math at all? I think I might try to reduce the weight of the wheel by reducing the size of it, but in the end, I am looking for the highest linear speed. $\endgroup$ Aug 27, 2015 at 13:05
  • $\begingroup$ Yes, this does affect the math, but how specifically it impacts the math depends on the exact motor torque-speed curves. You stated in your question, "MOTOR: 190 g.cm min", which I took to mean that the minimum torque was 190 g-cm. If you look at the image I linked above, that means I thought the value you gave was the lower-end torque on the far right of the plot. What you actually have is a no-load torque, so what you have called "min" is actually the max torque. $\endgroup$
    – Chuck
    Aug 27, 2015 at 14:07
  • $\begingroup$ I still don't understand what changes I have to make to the math. As I am trying to find the maximum linear speed, I suppose I should start with the ideal mass - is there any way to do this? By the way, here is the motor (I am running it in a 12V config) - comingsoon.radioshack.com/… $\endgroup$ Aug 28, 2015 at 3:02

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