# Pure rolling of cylinder with arbitrary inertia

I'm having trouble understanding how pure rolling is possible for cylinders with arbitrary inertia and arbitrary torques applied to them, because it seems that the reaction force doesn't necessarily coincide with the force required for pure rolling.

If a cylinder of radius $$r$$, mass $$m$$ (not necessarily uniform density) and some inertia $$I$$ is subject to a torque $$\tau$$ about its geometric center, then the force at the contact point, and the reaction force because there is no slip, is $$F=\frac{\tau}{r}$$. However, for no slip we also require that the bulk angular and linear acceleration be related by $$r\alpha=a$$. Given these two constraints, $$r\frac{\tau}{I}=\frac{F}{m}=\frac{\tau}{rm}\implies I=mr^2$$, which is not true in general.

Was it wrong to assume that the resultant forces on the cylinder comprise only of a pure torque at the geometric center and a reaction force from friction?

• Your assumption $F\,r=\tau$ Is just for static case for dynamic , you have to take this equation $I\alpha =\tau -Fr$ and this one $m\,a=F$ together with the $a=r\,\alpha$
– Eli
Mar 4 '19 at 21:55

If a cylinder of radius 𝑟, mass 𝑚 (not necessarily uniform density) and some inertia 𝐼 is subject to a torque 𝜏 about its geometric center, then the force at the contact point, and the reaction force because there is no slip, is $$F = \frac{\tau}{r}$$.
If the cylinder is accelerating, then the applied torque must exceed the reaction torque from the ground. Otherwise the rotational velocity would be constant. So you cannot assume them both to be equal to $$\tau$$.