Torque Required From Motor For a Project

Question

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.

Answer 1

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.

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.

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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,

Answer 2

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.

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 $m = m_{wheel} + m_{axel}$.