A standard textbook question is to ask about some rigid body (say, a 2D disk) rolling down an incline without slipping (cf. John Taylor's Classical Mechanics, Problem 3.35).
The standard approach is either to analyze the motion from the centre of mass (CM) frame, or from the point of contact $P$ at which there is no slipping. I'd like to focus on the validity of the latter approach. In particular, one uses "Newton's rotational second law" (effectively, just a statement about the derivative of the total angular momentum $\bf{L}$ of the system which uses Newton's second law in the derivation to connect acceleration and force). However, it is invariably noted during the derivation of
$$\dot{\bf{L}}=\bf{\Gamma}$$
that the result holds only in an inertial frame precisely because Newton's second law has been invoked (with the standard "coincidence"/exception of the result holding in the CM frame being noted).
At any rate, and if this is the case, why can we use Newton's rotational second law in $P$'s frame? The point $P$ of contact moves (indeed, accelerates, with respect to time). I suppose that, at an instant of time, we can consider $P$ as the point on the fixed ground which is in contact with the disk. Is this how we get around the difficulty?
Best Answer
The validity in $p$'s frame is yet another interesting coincidence. We just have to derive Newton's 2nd law for rotation relative to an inertial frame, but keeping track of the intermediate non-inertial frame. First, let $s$ be any particle in the object, and $g$ be any inertial frame. I'll use subscript pairs to indicate properties of one frame/point relative to another, and I'll use $\vec\tau$ for torque. Then we have
$\vec F_{s} = m_{s} \vec a_{sg} = m_{s}(\vec a_{sp} + \vec a_{pg}) \\ \vec r_{sp} \times \vec F_{s} = m_{s}(\vec r_{sp} \times \vec a_{sp} + \vec r_{sp} \times \vec a_{pg}) \\ \vec\tau_{sp} = m_{s}(\vec r_{sp} \times \frac{d}{dt}\vec v_{sp}) + m_{s}\vec r_{sp} \times \vec a_{pg} \\ \vec\tau_{sp} = m_{s}(\frac{d}{dt}(\vec r_{sp} \times \vec v_{sp}) - (\frac{d}{dt}\vec r_{sp}) \times \vec v_{sp}) + m_{s}\vec r_{sp} \times \vec a_{pg} \\ \vec\tau_{sp} = \frac{d}{dt}(\vec r_{sp} \times m_{s}\vec v_{sp}) + m_{s}\vec r_{sp} \times \vec a_{pg} \\ \vec\tau_{sp} = \frac{d}{dt}\vec L_{sp} + m_{s}\vec r_{sp} \times \vec a_{pg}$
Since $p$ is in general non-inertial, we now have the corrective term on the right. Now, sum over all particles:
$\Sigma_{s} \vec\tau_{sp} = \Sigma_{s}\frac{d}{dt}\vec L_{sp} + \Sigma_{s} (m_{s}\vec r_{sp} \times \vec a_{pg}) \\ \vec\tau_{net,p} = \frac{d}{dt} \vec L_{net,p} + (\Sigma_{s} m_{s}\vec r_{sp}) \times \vec a_{pg} \\ \vec\tau_{net,p} = \frac{d}{dt} \vec L_{net,p} + M_{tot}\vec r_{com,p} \times \vec a_{pg}$
Finally, rearranging so that the corrective term appears as the "fictitious torque" as mentioned in Claudio's answer:
$\frac{d}{dt} \vec L_{net,p} = \vec\tau_{net,p} + (\vec r_{p,com} \times M_{tot} \vec a_{pg})$
where $\vec r_{com,p}$ is the position of the center of mass relative to $p$. That's why it coincidentally disappears when you choose $p$ as the center of mass. But what if $p$ is the contact point on the object? Then, $\vec r_{com,p}$ is normal to the incline. The trickier part is $\vec a_{pg}$, but we do know that $\vec a_{pg} = \vec a_{p,com} + \vec a_{com,g}$. The motion relative to the center of mass is circular, so $\vec a_{p,com}$ has a tangential component up the incline, and a centripetal part normal to it. But the tangential part has magnitude $\frac{dv_{p,com}}{dt}$ where $v_{p,com}$ is the speed of $p$ relative to the COM. Meanwhile, $\vec a_{com,g}$ has magnitude $\frac{dv_{com,g}}{dt}$ down the slope. But since we are rolling without slipping, the speeds $v_{p,com}$ and $v_{com,g}$ are always equal (not just at this instant). Hence the tangential part of $\vec a_{p,com}$ cancels $\vec a_{com,g}$, leaving $\vec a_{pg}$ with only the centripetal part normal to the incline. This is parallel to $\vec r_{com,p}$, zeroing out their cross product. So the corrective term disappears in this frame, as you were hoping for.