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While writing an answer to this question, I encountered a puzzling conceptual problem when applying Newton's second law for rotation about an instantaneous axis of rotation when the moment of inertia is changing. I will restate the premise of a simplified version of the original problem.

A massless cylindrical tube of radius $R$ with infinitesimally thin walls has an infinitesimally narrow rod of mass $m$ attached to its inner surface, parallel to the tube. The tube is then laid on a floor so that the rod is in the uppermost position, and then is released to roll away from that unstable equilibrium position. The floor isn't slippery, so there's no sliding. The cross section of the tube looks like the following (from ja72's answer):

enter image description here

The square of angular velocity $\omega=d\theta/dt$ can be found from conservation of energy to be $$ \omega^2 = \frac{g}{R}\frac{1 - \cos\theta}{1 + \cos\theta}.\tag{1}$$ Differentiating with respect to time yields the angular acceleration $$\alpha=\frac{g}{R}\sin\theta\frac{1}{(1+\cos\theta)^2}.\tag{2}$$

Now my actual question is on how to apply Newton's second law to the rotation about the instantaneous line of contact of the tube with the floor. This axis is instantaneously at rest since the tube is not sliding on the floor. The total moment of inertia about this axis is $$ I = 2mR^2(1 + \cos\theta)\tag{3}$$ since its squared distance to the axis of rotation is $R^2\left[(1 + \cos\theta)^2 + \sin^2\theta\right] = 2R^2(1 + \cos\theta)$.

I can think of two different ways to apply Newton's second law for rotation:

  1. During the instantaneous pure rotation about the bottom of the tube, the distance of the rod to the axis of rotation doesn't change (since it is merely rotating about this axis), so the moment of inertia doesn't change.
  2. $\theta$ changes as the tube rolls, therefore the moment of inertia changes according to $(3)$.

So we can write Newton's second law for rotation as follows: $$mgR\sin\theta = \tau = \frac{d}{dt}(I\omega) = I\alpha + \epsilon\omega \frac{dI}{dt} = I\alpha + \epsilon \omega\frac{dI}{d\theta}\frac{d\theta}{dt} = I\alpha + \epsilon \omega^2\frac{dI}{d\theta}$$ where $\epsilon = 0$ for case 1 (we are considering $I$ fixed) and $\epsilon = 1$ for case 2 ($I$ changes according to $(3)$). Substituting $(1)$, solving for $\alpha$ and simplifying yields

$$\alpha = \frac{g}{2R}\sin\theta\frac{1 + 2\epsilon+(1 - 2\epsilon)\cos\theta}{(1 + \cos\theta)^2}. $$ This yields the correct angular acceleration $(2)$ only for $\epsilon = 1/2$! What is happening here? Is there just no straightforward way to apply Newton's second law for rotation about such an axis?

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    $\begingroup$ Hmm, I don't have an answer for what is going on, but I would suggest taking the limit $\beta^{-1}=0$ where you get rid of the cylinder, since all the problems are coming from the treatment of the rod (which is just a point mass so you can do the problem without worrying about moments of inertia and rigid bodies). $\endgroup$ – octonion Jun 28 at 11:03
  • $\begingroup$ @octonion Good idea, I've made the cylinder massless to simplify equations. I'm still referring to the moment of inertia of the rod just because it is convenient and Newton's second law is often stated in terms of it. $\endgroup$ – Puk Jun 28 at 11:26
  • $\begingroup$ Is the problem like this and the outside square is irrelevant? Is there a way to update the sketch including the contact point, and any other important quantities shown? $\endgroup$ – John Alexiou Jun 28 at 16:00
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    $\begingroup$ @ja72 Yes, you are right. Thanks for updating the sketch to better illustrate the actual problem, and I hope you don't mind that I've stolen your version :) $\endgroup$ – Puk Jun 28 at 22:35
  • $\begingroup$ How is "the rod in the uppermost position unstable"? Surely if it's actually uppermost it's also as stable as it if were down-most, unless there are unspecified forces? What did I miss, please? $\endgroup$ – Robbie Goodwin Jun 28 at 22:36
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In general, you can only apply $ \tau_C = \tfrac{\rm d}{{\rm d}t} L_C $ about the center of mass C. The expression about a different point is quite more complex.

You can see that taking the torque about another point A (not the center of mass C), and the derivative of angular momentum about A isn't enough to solve the problem.

Using the standard transformations rules I calculated the torque about a point A away from the center of mass, as well as the angular momentum about A.

$$\begin{aligned}\boldsymbol{\tau}_{A} & =\boldsymbol{\tau}_{C}+\left(\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\right)\times\boldsymbol{F}\\ \boldsymbol{L}_{A} & =\boldsymbol{L}_{C}+\left(\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\right)\times\boldsymbol{p} \end{aligned}$$

The derivative of angular momentum about C is $$\begin{aligned}\boldsymbol{\tau}_{C} & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{C}\\ \boldsymbol{\tau}_{A}-\left(\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\right)\times\boldsymbol{F} & =\tfrac{{\rm d}}{{\rm d}t}\left(\boldsymbol{L}_{A}-\left(\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\right)\times\boldsymbol{p}\right)\\ \boldsymbol{\tau}_{A}-\left(\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\right)\times\boldsymbol{F} & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{A}-\left(\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\right)\times\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{p}-\tfrac{{\rm d}}{{\rm d}t}\left(\boldsymbol{r}_{C}-\boldsymbol{r}_{A}\right)\times\boldsymbol{p}\\ \boldsymbol{\tau}_{A} & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{A}-\left(\boldsymbol{v}_{C}-\boldsymbol{v}_{A}\right)\times\boldsymbol{p}\\ \boldsymbol{\tau}_{A} & =\tfrac{{\rm d}}{{\rm d}t}\boldsymbol{L}_{A}+\boldsymbol{v}_{A}\times\boldsymbol{p} \end{aligned}$$

Where $\tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = \boldsymbol{F}$.

Newton's 2nd law (Or more precisely Euler's law of rotation) when applied not at the center of mass is $$\boxed{ \boldsymbol{\tau}_A = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_A + \boldsymbol{v}_{A} \times \boldsymbol{p} } \tag{1}$$ where $\boldsymbol{\tau}_A$ is the net torque about the point of reference, $\boldsymbol{L}_A$ is the angular momentum about the point of reference, $\boldsymbol{p}$ is momentum of the body and $\boldsymbol{v}_{A}$ is the velocity of the point of reference.

You can see that when the reference point is not moving $\boldsymbol{v}_A = \boldsymbol{0}$, or it is co-moving with the center of mass $\boldsymbol{v}_A = \boldsymbol{v}_C$ then the second terms drop out since it is parallel to momentum.

Solve using the center of mass

fig1

Put the origin at the contact point, and describe the position of the rod as $$\boldsymbol{r}_C = \pmatrix{x + R \sin \theta \\ R + R \cos \theta \\ 0}$$ where $x$ is the horizontal distance the center of the tube moves. The no slip condition has $x = R \theta$, as well as $\dot{x} = R \dot \theta$ and $\ddot{x} = R\ddot{\theta}$.

In this case, the MMOI about the center of mass is 0, and therefore the angular momentum about the center of mass is also zero. So the torque about the center of mass should be zero.

$$ \boldsymbol{\tau}_C = (0-\boldsymbol{r}_C) \times \boldsymbol{A} = \boldsymbol{0} \tag{2}$$

where $\boldsymbol{A} = \pmatrix{F_A \\ N_A \\ 0}$ is the contact force vector. This leads to the expression of $$F_A = N_A \tan \left( \frac{\theta}{2} \right)$$

Take the 2nd derivative of the position vector to form the equations of motion

$$ \boldsymbol{A} + \boldsymbol{W} = m\, \boldsymbol{\ddot{r}}_C \tag{3} $$

where $\boldsymbol{W} = \pmatrix{0\\-m g\\0}$ is the weight vector of the rod and

$$ \boldsymbol{\ddot{r}}_C = \pmatrix{(R+R \cos \theta) \ddot{\theta} + (-R \sin \theta) \dot{\theta}^2 \\ (-R\sin \theta)\ddot{\theta} + (-R \cos \theta)\dot{\theta}^2 \\ 0 }$$

Equation (3) yields the solution of

$$\begin{aligned} N_A &= m \left( g - R \dot{\theta}^2 \right) \cos^2 \left( \tfrac{\theta}{2} \right) \\ \ddot{\theta} &= \frac{ g + R \dot{\theta}^2}{2 R} \tan\left( \tfrac{\theta}{2} \right) \end{aligned} \tag{4} $$

Solve using the Contact Point

Take equation (1) at the contact point A.

$$\boldsymbol{\tau}_A = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_A + \boldsymbol{\dot{r}}_A \times \boldsymbol{p} \tag{5} $$ with

$$\begin{aligned} \boldsymbol{\dot{r}}_A & = \pmatrix{R \dot \theta \\ 0\\0} \\ \boldsymbol{p} & = m \boldsymbol{\dot{r}}_C = \pmatrix{m R (1+\cos\theta)\dot\theta \\ -m R (\sin \theta )\dot\theta \\ 0} \\ \boldsymbol{\tau}_A &= (\boldsymbol{r}_C-\boldsymbol{r}_A) \times \boldsymbol{W} = \pmatrix{0\\0\\-m g R \sin \theta} \\ \boldsymbol{L}_A &= \boldsymbol{L}_C + (\boldsymbol{r}_C-\boldsymbol{r}_A) \times \boldsymbol{p} = \pmatrix{0\\0\\ -2m R^2 (1+\cos\theta) \dot \theta} \end{aligned}$$

Remember that angular mometum about the center of mass is zero here $\boldsymbol{L}_C =\boldsymbol{0}$.

So the derivative of angular momentum is

$$ \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_A = \pmatrix{ 0\\0\\ -2mR^2 \left( (1+\cos \theta)\ddot \theta - (\sin\theta)\dot \theta^2 \right)} $$

and the solution is

$$ \ddot \theta = \frac{g + R \dot \theta^2 }{2 R} \tan \left( \tfrac{\theta}{2} \right) \; \checkmark \tag{6}$$

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  • $\begingroup$ Fixed a typo in the explanation of equation (1). $\endgroup$ – John Alexiou Jun 29 at 1:09
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The answer is that the moment of inertia is changing not only due to the instantaneous rotation about the contact point, but also because of the horizontal motion of the cylinder.

The position of the rod (I won't write the z component of vectors) is $$r=R(\sin \theta, 1+\cos\theta)+\int_{t_0}^t (R\omega(t'),0)dt'$$ where at $t=t_0$, the cylinder is above the instantaneous contact point we are measuring torques from.

The velocity is $$v=R\omega(1+\cos\theta,-\sin\theta)$$ Note that the velocity is perpendicular to $r$ at $t=t_0$ so it does make sense to think about this as instantaneous rotation about the contact point.

At $t=t_0$ the rod feels a normal force ${N}$ perpendicular to $v$ (so no work is done) to keep it on the cylinder, and a force of $mg$ downward. $${N}+(0,-mg)=m\frac{d}{dt}v$$ Now let's cross both sides by $r$. At $t=t_0$ $r$ is perpendicular to $v$ so $N$ exerts no torque, and only the external torque due to gravity $\tau$ survives $$\tau=mr\times\left(\frac{d}{dt}v\right)=\frac{d}{dt}\left(mr\times v\right)$$ This is just the law for external torque changing angular momentum. I just wanted to show that nothing funny happened yet.

Now let's find the angular momentum. At $t=t_0$, where the second term in $r$ vanishes, we get

$$|mr\times v|_{t=t_0} = 2mR^2\omega(1+\cos \theta)=I\omega$$ which is exactly what you expect from the naive moment of inertia (which I'm calling $I$). But for later times that second term doesn't vanish and you get $$|mr\times v| = I\omega+mR^2\omega\sin\theta\left(\theta(t)-\theta(t_0)\right)$$

Now taking the time derivative and setting $t=t_0$ we get $$\tau= I\alpha + \omega \frac{d}{dt}I + mR^2\omega^2\sin\theta = I\alpha -2mR^2\omega^2\sin\theta + mR^2\omega^2\sin\theta$$ So you can see the extra piece that was missing cancels out half of the $\omega \frac{d}{dt}I$ piece.

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enter image description here To calculate the equation of motion we obtain the sum of the torques about point A, because we don't have to take care about the contact force.

first I obtain the vector u from point B to A

$$\vec{u}=R\,\begin{bmatrix} 0 \\ -1 \\ 0 \\ \end{bmatrix}- R\,\begin{bmatrix} \sin{\theta} \\ \cos(\theta) \\ 0 \\ \end{bmatrix}=-R\,\begin{bmatrix} \sin{\theta} \\ (1+\cos(\theta)) \\ 0 \\ \end{bmatrix}$$

forces at point B are the inertial forces

$$\vec{F}_I=m\,R\,\ddot{\theta}\,\vec{t}$$

where $\vec{t}$ is the tangent on the circle

$$\vec{t}=\begin{bmatrix} \cos{\theta} \\ -\sin(\theta) \\ 0 \\ \end{bmatrix}$$

and weight force

$$\vec{F}_G=\left[ \begin {array}{c} 0\\ -mg \\ 0\end {array} \right] $$

take the sum of all torques about point A you obtain:

$$\sum \tau_A=\vec{u}\times (-\vec{F}_I+\vec{F}_G)-I\,\ddot{\theta}-M\,\ddot{x}\,R=0$$

solving for $\ddot{\theta}$

$$\ddot{\theta}={\frac {R\sin \left( \theta \right) m \, g }{{R}^{2}m+{R}^{2}m\cos \left( \theta \right) +I+M{R}^{2}}} $$

with $\ddot{x}=R\,\ddot{\theta}$ the roll condition

this is your equation of motion

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All great and valid answers here, and they have helped me see the key piece of the puzzle I was missing: the subtle point that although the instantaneous axis of rotation has zero velocity, it is accelerating upward (centripetal acceleration). Therefore to apply Newton's second law (or Euler's law) about this axis, we need to take into consideration the inertial forces/torques that result.

Let $\vec{R}$ denote the position of a point on the instantaneous axis of rotation, relative to a fixed point that coincides with the axis of rotation at $t=t'$, but does not accelerate like it. This means that at $t=t'$, $\vec{R}=0$ and $\dot{\vec{R}}=0$ but $\ddot{\vec{R}}\ne 0$. Let $\vec{r}$ be the position of a particle of $M$ in the rotating system, relative to $\vec{R}$. Its angular momentum about the fixed point is $$\vec{L}=M(\vec R + \vec r)\times (\dot{\vec R} + \dot{\vec r})$$ $$\vec \tau=\frac{d\vec L}{dt}=M(\vec R + \vec r)\times (\ddot{\vec R} + \ddot{\vec r}) $$ At $t=t'$, since $\vec{R}=0$, $$\frac{d\vec{L}'}{dt}=M\vec{r}\times\ddot{\vec r}=\vec{\tau}-M\vec r\times\ddot {\vec R}$$ where $\vec{L}'$ here is the angular momentum about the instantaneous axis of rotation. The second term on the right hand side is the inertial torque that must be added to the torque due to the real forces: $$\vec{\tau}_i=-M\vec{r}\times\ddot{\vec{R}}.$$ $$\ddot{\vec R}=\omega^2R\hat{y}$$ where $\hat y$ is in the upward direction. The intertial torque on mass $m$ in this problem is $\tau_i=m\omega^2R^2\sin\theta$ (clockwise/inward).

All this so far was to derive the expression for the inertial torque. Now writing the second law, with the moment of inertia $I$ taken to be constant, as in case 1 in the question: $$mgR\sin\theta + m\omega^2R^2\sin\theta= \tau'= I\alpha $$ $$\alpha = \frac{g}{2R} \sin\theta \frac{1 + \frac{\omega^2}{gR}}{1 + \cos\theta} $$ Plugging in $\omega^2$ leads to the correct expression: $$\alpha = \frac{g}{R}\sin\theta\frac{1}{(1 + \cos\theta)^2}. $$

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