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Suppose I have wheels driven by a motor. The wheels make contact with a horizontal surface, so that the wheels accelerate in translational and rotational motion. If I want to know the linear acceleration of the wheels I use: $a=\alpha r$.

The torque on the wheel is given by:

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

The moment of inertia for a solid wheel rotating on its axis is given by $I=\frac{1}{2}mr^2$.

We get: $\alpha=\frac{\tau}{\frac{1}{2}mr^2}$ and so $a=\frac{2r\tau}{mr^2}=\frac{2\tau}{mr}$.

From this I conclude that the linear acceleration gets bigger when I take a smaller radius. But what are the limits of the linear acceleration at a constant torque when the radius gets smaller?

Lets say the torque of the motor is 1Nm, the mass is 0.1kg and the radius is 0.01m. In this case the linear acceleration is: $a=\frac{2}{0.001}=2000 m/s^2$. This seems to be way too high. Is the limit the maximum angular velocity of the motor?

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  • $\begingroup$ You say you have a wheel driven by a motor. But then you talk about linear acceleration. Is this wheel on the ground as well? Is the ground providing a torque also? $\endgroup$
    – BowlOfRed
    Nov 2, 2020 at 23:01

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Naturally infinite in newtonian mechanics? If you have constant force and if mass goes to zero acceleration clearly goes to infinite(of course, in the realm of classical newtonian mech.)

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  • $\begingroup$ and as a result, if we don't consider relativistic things, it's not weird to have such a big number. To maintain constant torque while radius goes smaller and smaller, the force is going infinitely large too. $\endgroup$
    – hwang
    Nov 3, 2020 at 6:21

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