The device I am building is responsible for supporting 200 lbs and is required to accelerate 10 feet per second. The device will have two 8 inch diameter wheels. I have worked out what I believe to be an accurate way to determine the required torque from a motor. I have assumed no slip between the wheel and the floor, and I took that point as the instantaneous center for the basis of the calculations. I am providing my calculations and would appreciate if any one could point out flaws in my calculations, or if my final answers intuitively seem feasible. Torque Calculations Pt1 Torque Calculations Pt2Torque Calculations Pt3

  • $\begingroup$ You seem to be ignoring rolling friction. Also, are the wheels 200lbs? Or is the load supported by the wheels 200lbs? Because you need to accelerate both the mass linearly and the wheel rotationally if the wheels' moment of inertia is not negligible. It is not correct to toss the mass the wheels are supporting into the moment of inertia for a cylinder if not all of that mass is part of the wheel and actually rotating. $\endgroup$
    – DKNguyen
    Nov 14, 2021 at 21:18
  • $\begingroup$ The effective mass of the wheels are 100 pounds. Meaning the total weight of the device is 200 pounds but each wheel is essentially supporting 100 pounds. Therefore, each motors will be required to rotate 100 pounds worth of weight $\endgroup$ Nov 14, 2021 at 21:24
  • $\begingroup$ When you assume there is no slip. The contact point between the wheel and the ground becomes the instantaneous center. Therefore that is the point that torques and forces are summed relative to. Hence, the friction term is ignored because it passes through that point. There is only one force being accounted for, which is the axle force. I do account for both the linear acceleration and angular acceleration in my work $\endgroup$ Nov 15, 2021 at 1:26
  • $\begingroup$ You do not account for linear acceleration and angular acceleration in your work so much as you use the linear acceleration to derive the angular acceleration. As for rolling friction it sounds like you don't understand what rolling friction is. It's not the same as skidding. All materials deform. Two things are going to happen: Your wheel is going to flatten a bit at the contact point so it is no longer a perfect circle and your wheel is going to sink into the surface it is sitting on. Your wheel has to force it's way out of the flat spot and out of the valley. $\endgroup$
    – DKNguyen
    Nov 15, 2021 at 1:38
  • $\begingroup$ It's a real project so you can't just ignore it. If you could ignore it you could apply any small force to an object on wheels and it would start moving, even if it only accelerated very slowly. But you can't. A minimum force is required and sometimes a considerable one. It's one thing to investigate and determine the coefficient of friction is so low relative to the rotational and angular inertia that you can ignore it, but you shouldn't neglect it based on the condition that there is no slip because it has nothing to do with slip. $\endgroup$
    – DKNguyen
    Nov 15, 2021 at 1:49

2 Answers 2


You seem to be ignoring rolling friction. I don't advise you do that. Rolling friction exists even if there is no slip. It is not the same as skidding. Rolling friction cannot be neglected in a real situation just because you assume there is no slip. It's caused by deformation of both the wheel and the surface at the contact point. In a lot of cases, you should be able to get a good estimate for the coefficient of rolling friction intuition and gut feel and by taking into account the material of wheel and the surface it is rolling on. Just imagine what fraction of the force it would take to push the object as it would take to actually lift it. That will give your coefficient of rolling friction.

enter image description here https://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html

It's one thing to determine the coeficient of rolling friction is so low and the inertias involved are so high that rolling friction can be neglected, but you should not neglect it based on the no-slip condition because rolling friction has nothing to do with slip. But I'll ignore that for the rest of my answer.

I don't see a summation anywhere and thus I think your approach is incorrect. You seem to have taken the linear momentum and directly translated that into the angular momentum. The implication there is that a rolling and translating object has the same momentum and energy has a sliding object on frictionless surface (since we are ignoring rolling friction) which I do not believe to be true.

The rolling and translating wheel has both linear momentum from translating but also has angular momentum from spinning and you need to accelerate both. So you need to calculate the torque required to spool up the wheel as if it were off the ground on something like a lathe, and you also need to calculate the linear force required at the edge of the wheel to produce the linear acceleration. You can then convert the linear force to a torque and add up that torque with the spool-up torque.

This seems to verify what I suspected: http://web.mit.edu/8.01t/www/materials/Presentations/Presentation_W11D1.pdf

From the article it looks like an alternative method to determining the torque required for the linear acceleration part (which I had initially proposed as treating it as a sliding object to calculate the linear force required at the edge of the wheel and then converting that to a torque), is to directly calculate a torque by considering the rotation of the entire wheel about the contact point on the ground.

And of course, the approach to determine torque required to spool up the wheel around its center (as if it were just free-spinning off a surface like on a lathe) remains the same as before,

  • $\begingroup$ The website in which I am ordering the wheels from show that the coefficient of friction is 1.1 for the wheel. Would you assume that is static or kinetic or would that be the rolling resistance ? $\endgroup$ Nov 15, 2021 at 2:08
  • $\begingroup$ @user15588486 That sounds like static friction coefficient for traction, not rolling since if rolling were >1 then wheels are pointless since it would require less force to just lift the damn thing and walk with it. What wheels are you using and what surface? The traction friction is useful for you to know if you can actually accelerate the way you want without skidding though. You should convert your final torque to wheel edge force to check it doesn't exceed max force of static friction. $\endgroup$
    – DKNguyen
    Nov 15, 2021 at 2:15
  • $\begingroup$ Just met with the department head of mechanical engineering at my university and we calculated the required torque for my given situation to be 14 Nm. And the required power to achieve the desired acceleration to be 0.6 horse power, just in case y’all were curious on an update $\endgroup$ Nov 16, 2021 at 0:47
  • $\begingroup$ @user15588486 I don't have a good feel for how much 14N-m is, but 3.388N-m did few a few times too small. $\endgroup$
    – DKNguyen
    Nov 16, 2021 at 0:51

Your derivation doesn't look correct to me, but maybe I am misunderstanding stuff. The way I have tried to calculate the torque you seek is by working in a stationary inertial coordinate system $O\,\vec{i}\,\vec{j}\,\vec{k}$, where the plane of the picture is the coordinate plane $O\,\vec{i}\,\vec{j}$ and the coordinate axis $O\,\vec{k}$ is perpendicular to the picture, pointing from the picture towards us.

enter image description here

According to the diagram, the dynamic equations for translational motion and rotational motion are \begin{align} & m\,\ddot{x} \,\vec{i} \, =\, F\,\vec{i}\\ & I \, \ddot{\theta} \, \vec{k} \,=\, \big(-\,R\vec{j}\,\big) \times \big(\,F\vec{i}\,\big) \, + \, \tau\,\vec{k} \end{align} The condition of rolling without slipping implies that $$x \, =\, -\,R\,\theta$$ The minus is there because when $x$ increases, the angle $\theta$ becomes more and more negative. This is due to the fact that the rotation is clockwise (i.e. negatively directed). In terms of accelerations the latter can be expressed as $$\ddot{\theta} \, =\, -\,\frac{1}{R}\,\ddot{x}$$ Hence, the equations become
\begin{align} & m\,\ddot{x} \,\vec{i} \, =\, F\,\vec{i}\\ -\,& \frac{I}{R} \, \ddot{x} \, \vec{k} \,=\, \big(-\,R\,\vec{j}\,\big) \times \big(\,F\,\vec{i}\,\big) \, + \, \tau\,\vec{k} \end{align} and after that \begin{align} & F\,\vec{i} \, =\, m\,\ddot{x} \,\vec{i} \\ -\,& \frac{I}{R} \, \ddot{x} \, \vec{k} \,=\, \big(-\,R\vec{j}\,\big) \times \big(m\,\ddot{x} \,\vec{i}\,\big) \, + \, \tau\,\vec{k} \end{align} so only one equation matter now: \begin{align} -\,& \frac{I}{R} \, \ddot{x} \, \vec{k} \,=\, -\,R\,m\,\ddot{x}\, \big(\,\vec{j} \times \vec{i}\,\big) \, + \, \tau\,\vec{k} \end{align} and since $\vec{j} \times \vec{i} \,=\, -\vec{k}$ \begin{align} -\,& \frac{I}{R} \, \ddot{x} \, \vec{k} \,=\, m\,R\,\ddot{x} \,\vec{k} \, + \, \tau\,\vec{k} \end{align} The latter vector equation reduces to the scalar equation \begin{align} -\,& \frac{I}{R} \, \ddot{x}\,=\, m\,R\,\ddot{x} \, + \, \tau \end{align} which can be rewritten as \begin{align} & \left(\frac{I}{R}\,+\,m\,R \right) \ddot{x}\,=\, -\,\tau \end{align} Observe, since the left-hand side is positive, the torque $-\tau$ is also positive, which means that $\tau$ is negative, reflecting the fact that the torque should be clock-wise (which makes sense). The formula for the required torque, in order to maintain the prescribed acceleration $\ddot{x} = a$ is $$\tau \, =\, -\,\left(\frac{I}{R}\,+\,m\,R \right)a $$ The question that remains is what is the moment of inertia $I$. Since you describe the rotating body to be a wheel of radius $R$ with axel of radius $r$ attached to it, as on your picture, then, since the moment of inertia is additive, $I$ is the sum of the moments of inertia of the two cylinders, wheel and axel, along their common longitudinal axis of rotational symmetry (parallel to the vector $\vec{k}$). Hence, $$I \,=\, \frac{1}{2}m_{wheel} R^2 \, + \, \frac{1}{2}m_{axel} r^2 $$ Hence, the torque should be something like $$\tau \, =\, -\,\left(\frac{1}{2}m_{wheel} R \, + \, \frac{1}{2}m_{axel} \frac{r^2}{R}\,+\,(m_{wheel} + m_{axel})\,R \right)a $$ because $m = m_{wheel} + m_{axel}$.


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