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In general, for combined translation and rotational motion of rigid bodies, rotational variation of Newton's second law ($\Sigma\tau = I\alpha$) is valid (without pseudo forces) only when torques and moment of inertia are considered about an axis through the centre of mass. (Since, net torque due to pseudo forces is zero, about an axis through the centre of mass)

But for pure rolling, motion is purely rotational about a perpendicular axis through the instantaneous point of contact with the surface. Is Newton's second law ($\Sigma\tau = I\alpha$) valid here, when torques and moment of inertia are considered about this axis?

If it is valid, will the angular acceleration calculated by solving Newton's second law about this axis and about the axis through the centre of mass be the same? i.e, \begin{equation}\text{Is } \alpha=\frac{\Sigma\tau_{CoM}}{I_{CoM}}=\frac{\Sigma\tau_{PoC}}{I_{PoC}} \text{ ?} \end{equation}( where: $CoM$ - centre of mass, $PoC$ - point of contact)

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2 Answers 2

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This analysis about the point of contact is known as analyzing about the instantaneous axis of rotation(IAOR). It's a neat trick and you can read up on this using any standard textbook.

The angular accelerations in both COM frame and IAOR frame will be the same. Only the point about which angular acceleration is defined is different for the two cases $\Rightarrow$ the linear acceleration of the same point(s) in the body won't match for these two cases(in their respective frame of references) but it'll match in Laboratory frame $\therefore$ when you analyze the motion in the Laboratory frame using either of the angular accelerations, the results will be identical.

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    $\begingroup$ Wouldn't IAOR have a centripetal acceleration in the laboratory frame? (making IAOR frame different from laboratory frame and also non-inertial). With this assumption, my derivation attempts for a relation between angular accelerations (also angular velocities) as observed from IAOR frame and COM frame, ends up in them being equal. Is this expected in pure rolling? $\endgroup$
    – DarkMIR4GE
    Oct 9, 2022 at 5:36
  • $\begingroup$ Yes, I've edited the answer now. I was also trying to explain the motion in both IAOR and COM frames, but the choice of words was sloppy (if not entirely incorrect) $\endgroup$
    – user292464
    Oct 9, 2022 at 7:53
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For pure rolling, rotational variation of Newton's second law is valid about IAOR: With respect to an inertial (laboratory) frame, the instantaneous axis of rotation (IAOR) has only a centripetal acceleration ($\omega^2R$); tangential components of acceleration get cancelled out since it's pure rolling ($R\alpha=a_{com}$). Centrifugal force acts as a pseudo-force in the body when observed from the IAOR frame. However, torque about IAOR, due to centrifugal force will be zero since, $\vec{R}\parallel\vec{F}_{centrifugal}$. Hence, the rotational variation of Newton's second law is valid when taken about IAOR. $$\Sigma\tau_{iaor}=I_{iaor}\alpha$$

For pure rolling, angular velocity about CoM is the same as that about IAOR: Consider a general point P on a purely rolling body defined by $\vec{r}$ with respect to CoM. In the laboratory frame, this point will experience velocities as depicted in the figure below.

    

Velocity of point P in laboratory frame; $$\vec{v}_{p}=\vec{\omega}_{com}\times\vec{r}+\vec{v}_{com} =\vec{\omega}_{com}\times(\vec{r}-\vec{R}) $$ Velocity of IAOR with respect to the laboratory is zero: $\vec{v}_{iaor}=0$

Velocity of P in IAOR frame: $\vec{v}_{p,iaor}=\vec{v}_p-\vec{v}_{iaor}$ $$\vec{v}_{p,iaor}=\vec{\omega}_{com}\times(\vec{r}-\vec{R})$$

With respect to IAOR, the body is in pure rotation with an angular velocity $\vec{\omega}_{iaor}$. Hence, $$\vec{v}_{p,iaor}=\vec{\omega}_{iaor}\times(\vec{r}-\vec{R})$$ From the above two equations, $$\vec{\omega}_{iaor}=\vec{\omega}_{com}=\vec{\omega}$$

For pure rolling, angular acceleration about CoM is the same as that about IAOR: Consider a general point P on a purely rolling body defined by $\vec{r}$ with respect to CoM. In the laboratory frame, this point will experience accelerations as depicted in the figure below. (Position vectors $\vec{r}$ and $\vec{R}$ are the same as in last figure)

    

Acceleration of point P in laboratory frame; $$\vec{a}_p=\vec{\alpha}_{com}\times\vec{r}+\vec{a}_{com}+\vec{\omega}\times(\vec{\omega}\times\vec{r}) =\vec{\alpha}_{com}\times(\vec{r}-\vec{R})+\vec{\omega}\times(\vec{\omega}\times\vec{r})$$

Acceleration of IAOR with respect to the laboratory is only centripetal: $\vec{a}_{iaor}=\vec{\omega}\times(\vec{\omega}\times\vec{R})$

Acceleration of point P in IAOR frame: $\vec{a}_{p,iaor}=\vec{a}_p-\vec{a}_{iaor}$ $$\vec{a}_{p,iaor}=\vec{\alpha}_{com}\times(\vec{r}-\vec{R})+\vec{\omega}\times(\vec{\omega}\times(\vec{r}-\vec{R}))$$

With respect to IAOR, the body is in pure rotation with an angular acceleration $\vec{\alpha}_{iaor}$. Hence, $$\vec{a}_{p,iaor}=\vec{\alpha}_{iaor}\times(\vec{r}-\vec{R})+\vec{\omega}\times(\vec{\omega}\times(\vec{r}-\vec{R}))$$ From the above two equations, $$\vec{\alpha}_{iaor}=\vec{\alpha}_{com}=\vec{\alpha}$$

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