4
$\begingroup$

This question already has an answer here:

The following figure shows a cylinder resting on a rough horizontal rug that is pulled out from under it with acceleration $a$ perpendicular to the axis of the cylinder. What is the motion of the cylinder, assuming it does not slip?

enter image description here

The only horizontal force on the cylinder is that of friction at $P$. Therefore let us take the moments about $P$. The forces of gravity and reaction at the surface pass through the point $P$, as does also the friction force. Also, we know* that:

$$\dot{\mathbf{L}}=\mathbf{M}-m\mathbf{v}_P \times \mathbf{v}_O$$

Where $\mathbf{L}$ - the angular momentum of the system, $\mathbf{M}$ - the net external torque on the system, $m$ - the mass of the system, $\mathbf{v}_P$ - the velocity of point $P$ (as viewed from an inertial system - say, ground) and $\mathbf{v}_O$ - the velocity of the center of mass.

Because $\mathbf{v}_O\parallel \mathbf{v}_P$ we have $\dot{\mathbf{L}}=\mathbf{M}$. So the net torque about $P$ is $0$. Therefore:

$$\dot{\mathbf{L}}=0 \Rightarrow I_P \omega = \text{const} \Rightarrow \omega = \text{const}$$

which is wrong. Only by taking the moments about $O$ we obtain $\omega \propto a$.

Why my approach was wrong in the first place?


*Let $\mathbf{v}$ and $\mathbf{p}=m \mathbb{v}$ be the velocity and the momentum of a particle with respect to a stationary inertial system $S$, and $\mathbf{r}$ - the radius-vector of the particle with respect to the moving point $P$. The motion of $P$ can be either uniform or nonuniform (accelerating). Suppose $\mathbf{v}_P$ is the velocity of $P$. The angular momentum is $\mathbf{L}=\mathbf{r} \times \mathbf{p}$. Taking the derivative we get $\dot{\mathbf{L}}=\dot{\mathbf{r}} \times \mathbf{p} + \mathbf{r} \times \dot{\mathbf{p}}$. Because $\dot{\mathbf{r}}=\mathbf{v}-\mathbf{v}_P$ we get $\dot{\mathbf{L}}=(\mathbf{v}-\mathbf{v}_P) \times \mathbf{p} + \mathbf{r} \times \dot{\mathbf{p}}$. However $\dot{\mathbf{p}}=\mathbf{F}$ and thus we obtain:

$$\dot{\mathbf{L}}=\mathbf{M}-\mathbf{v}_P \times \mathbf{p}$$

However we can express the momentum $\mathbf{p}$ of the system in terms of $\mathbf{v}_O$ - the velocity of the center of mass: $\mathbf{p}=m\mathbf{v}_O$. Thus: $$\dot{\mathbf{L}}=\mathbf{M}-m\mathbf{v}_P \times \mathbf{v}_O$$

$\endgroup$

marked as duplicate by ja72, John Duffield, ACuriousMind, CuriousOne, John Rennie newtonian-mechanics Mar 9 '16 at 10:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Taking moments about $P$ is exactly the wrong idea, since a) the only force that will give angular acceleration to the cylinder is acting through that point so will give no torque, and b) the cylinder doesn't rotate around $P$. The cylinder is going to rotate around $O$, so you want to take moments around that point. That will give you the angular acceleration, which you can combine with the linear acceleration to describe the motion of the cylinder. $\endgroup$ – PhillS Mar 8 '16 at 14:22
  • 3
    $\begingroup$ @PhillS - what you say in essence is that I can't take the moments about $P$ because it will give the wrong result. It doesn't quite answer my question. My textbook says that I can take moments about the momentary axis of rotation (we can always view the cylinder as rotating about $P$). Many problems are easily solved once we do that. However this problem is different in that it has acceleration. But I still fail to understand why can't I take the moments about $P$ (even though it accelerates - note that the derivation at the end of my post doesn't require $\dot{\mathbf{v}_P} = 0$). $\endgroup$ – user249293 Mar 8 '16 at 14:38
  • $\begingroup$ You can't the view the cylinder as rotating about $P$. You can, by all means, take moments about any point you wish, but the physical reality of the situation is that the cylinder will only rotate around $O$, and so only moments about $O$ will be meaningful if you want to understand the rotation of the cylinder. $\endgroup$ – PhillS Mar 8 '16 at 14:41
  • 1
    $\begingroup$ @PhillS - right, we can write $\dot{\mathbf{L}}=I\dot{\mathbf{\omega}}$ only when the rigid body is purely rotating with respect to the axis. In this case, when viewed from an inertial system, the cylinder has both rotational and translational motion (even with respect to $P$). We can however analyze the motion of the cylinder in the non-inertial system of the rug itself. There the cylinder is only rotating, but there we have the fictitious force $-m\mathbf{a}$ which creates the torque. Am I correct? Thank you. $\endgroup$ – user249293 Mar 8 '16 at 14:52
  • $\begingroup$ $\dot{\mathbf{L}}=\mathbf{I}\dot{\omega}$ works fine whether the cylinder has translational motion or not. Just work out the torque in the center of mass frame of the cylinder (since in this case it will rotate about that center, which is $O$) which is only due to friction at ${P}$. That gives you the angular acceleration. The friction force also gives you a linear acceleration. BTW the cylinder does not only have rotational motion in the accelerating rug frame since the cylinder doesn't have the same acceleration as the rug. $\endgroup$ – PhillS Mar 8 '16 at 15:54
1
$\begingroup$

You may have over-complicated the problem. Take the momentum definition at the center of mass, differentiate it and equate it with the applied loading to find the equations of motion.

$$\require{cancel} \begin{align} \boldsymbol{p} & = m \boldsymbol{v}_O & \boldsymbol{L}_O & = \mathcal{I}_O \boldsymbol{\omega} \\ \boldsymbol{F}_{fric} & =\dot{\boldsymbol{p}} = m \dot{\boldsymbol{v}}_O & (\boldsymbol{r}_P-\boldsymbol{r}_O) \times \boldsymbol{F}_{fric}& = \dot{\boldsymbol{L}}_O = \mathcal{I}_O \dot{\boldsymbol{\omega}}+\cancel{\boldsymbol{\omega} \times \mathcal{I}_O \boldsymbol{\omega}} \\ F_{fric} & = m \dot{v} & R\; F_{fric} & = \mathtt{I}_O \dot{\omega} \end{align} $$

Now for the no-slip condition

$$ \hat{\boldsymbol{i}} \cdot (\boldsymbol{v}_{rug} - \boldsymbol{v}_P) = 0$$ $$ \hat{\boldsymbol{i}} \cdot (\boldsymbol{v}_{rug} - \boldsymbol{v}_O+\boldsymbol{\omega}\times(\boldsymbol{r}_O-\boldsymbol{r}_P)) = 0$$ $$\begin{align} \omega &= \frac{v_{rug}-v}{R} & \dot{\omega} & = \frac{a-\dot{v}}{R} \end{align}$$

Now you have three equations with three unknowns $F_{fric}$, $\dot{v}$ and $\dot{\omega}$:

$$\begin{align} F_{fric} & = m \dot{v} & R\;F_{fric} &= \mathtt{I}_O \dot{\omega} & \dot{\omega} & = \frac{a-\dot{v}}{R} \end{align} $$

$\endgroup$
  • 1
    $\begingroup$ Sorry, but that does not answer my question. I know how to solve the problem with respect to the center of mass. I asked why it is problematic to solve it with respect to $P$. $\endgroup$ – user249293 Mar 8 '16 at 15:58
  • $\begingroup$ See physics.stackexchange.com/a/80449/392 on the equations of motion not on the CM. Also physics.stackexchange.com/a/186591/392 $\endgroup$ – ja72 Mar 8 '16 at 18:23
  • 1
    $\begingroup$ By trying to solve this not on the center of mass you are over-complicating the problem because you have to account for a lot of cross terms. $\endgroup$ – ja72 Mar 8 '16 at 18:26
  • 2
    $\begingroup$ I know that it is "over-complication". I do this because I want to understand the principle. Not because I can't solve this particular question. I'll read the links, thanks. $\endgroup$ – user249293 Mar 8 '16 at 18:33
  • 1
    $\begingroup$ I wish you had summarized this in the OP. The way the question is written it reads like help for this particular problem. $\endgroup$ – ja72 Mar 8 '16 at 18:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.