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This paper is based on a question by Puk asked Jun 28, 2020 on the Stack Exchange Physics site, Newton's 2nd law for rolling motion with changing moment of inertia, about a long, hollow, rigid but infinitely thin cylinder with a point-like mass in the form of a rod with mass m attached to its wall as shown in the drawing.

enter image description here

Puk states that conservation of energy requires that the angular velocity is

$\omega^2=\frac gR\frac{1-\cos\theta}{1+\cos\theta}$

He does not show any details as to how he gets that result, which implies that the angular velocity becomes infinite at the contact point A where theta is 180 degrees.

I commented that this should be impossible but have received no reaction, so I am asking it here in the hope of a response.

For the sake of completeness I will here outline how I think the conservation of energy method gives that weird result:

The motion in the x-direction is partly translational due to the rolling to the right and partly a displacement due to the rotation around the center O, so

$x=x_0+R\int_{t_0}^t\omega dt+R(\sin\theta-\sin\theta_o)$

In the y-direction there is only a displacement due to the rotation around the center O, so

$y=R(1+\cos\theta)$

The velocities in the x- and y-directions are obtained as the time derivatives $\dot{x} $ and $\dot{y} : $

$\dot{x}=\nu_x=R\omega(1+\cos\theta)$

$\dot{y}=\nu_y=-R\omega\sin\theta$

Their vectorial sum squared becomes

$ \nu^2=2R^2\omega^2(1+\cos\theta) $

Taking the positive square root then gives

$\nu=2R\omega\cos\left(\theta/2\right)$

Using the expression for $\nu$ the total kinetic energy of the system is

$ E_{kin}=\frac12m\nu^2=mR^2\omega^2(1+\cos\theta) $

Overcoming the force of gravity to move the mass m from the contact point A at $y=0$ to the higher position at C with ordinate $y$ requires the work

$E_{pot}=mgy=mgR(1+\cos\theta)$

The total energy $E$ of the system is thus

$ E=E_{kin}+E_{pot}=mR^2(\omega^2+\frac gR)(1+\cos\theta) $

Conservation of energy implies that $E$ has some constant value. Here we let $E=2mgR$ which is the potential energy at the top where $\theta=0$ and where we have chosen to let $\omega=0\:.$

Using this value of $E$ we get

$ \omega^2=\frac gR\frac{1-\cos\theta}{1+\cos\theta} $

which becomes infinite when theta is 180 degrees ??!

With this expression for omega we get

$ E_{kin}=mgR(1-\cos\theta) $

which of course is finite for any value of theta.

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    $\begingroup$ What is "a point mass in the form of a rod"? $\endgroup$ Commented Mar 13 at 16:55
  • $\begingroup$ If you reference the previous question you might as well actually link the previous question too, as that will reinforce more context for your question. $\endgroup$
    – Triatticus
    Commented Mar 13 at 16:59
  • $\begingroup$ Does the cylinder have mass, or is it there just to establish the kinematics of the mass. $\endgroup$ Commented Mar 13 at 20:52
  • $\begingroup$ Can I assume an essentially massless hollow cylinder except for a point mass emended in its circumference and there is no actual rod where R is depicted in the diagram which is just a measurement of radius? $\endgroup$
    – KDP
    Commented Mar 13 at 22:20
  • $\begingroup$ The cylinder itself is massless. For purposes of calculation, the rods diameter is zero and may be treated as a point with mass m but zero moment of inertia. Besides the (x, y) coordinates the instantaneous position of the mass is indicated by the position vector R from O to C and the angle theta towards vertical. $\endgroup$
    – Jens
    Commented Mar 13 at 23:44

3 Answers 3

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Yes, the "point" mass has no mass moment of inertia about itself. At the bottom where the point is not moving (due to the no-slip contact) all the potential energy then would go into the rotational kinetic energy, which is finite. Since $KE = \tfrac{1}{2} I \omega^2$, and $I$ is zero, you need infinite $\omega$ to result into a finite $KE$.


So assume there is some small mass moment of inertia $I \ll m R^2$ for the rod about its center.

The equations of motion, derived from the sum of forces and torques about the rod center are:

$$ \begin{aligned} F & = m R \ddot{\theta} \cos \theta - m R \dot{\theta}^2 \sin \theta \\ N -m g & = -m R \ddot{\theta} \sin \theta - m R \dot{\theta}^2 \cos \theta \\ R ( F \cos \theta - N \sin \theta + F ) & = -I \ddot{\theta} \end{aligned}$$

Note that the rod center is defined by $$\vec{r}_C = \pmatrix{ R \sin \theta \\ R + R \cos \theta }$$ and this the acceleration of the center of mass is

$$ \vec{a}_C = \pmatrix{ R \ddot{\theta} \cos \theta - R \ddot{\theta}^2 \sin \theta \\ -R \ddot{\theta} \sin \theta - R \ddot{\theta} \cos \theta} $$

with solution

$$\begin{aligned} F &= \frac{m R \sin \theta \left( m g R \cos \theta + (I+m R^2) \dot{\theta}\right)}{I + m R^2 ( \cos \theta+1)} \\ N & = \ldots \\ \ddot{\theta} & = \frac{m R \left( g + R \dot{\theta}^2\right) \sin \theta}{I + m R^2 ( \cos \theta+1)} \end{aligned}$$

From the above, you can see the denominator $I + m R^2 ( \cos \theta+1) > 0 $ always since $I>0$, which means in real life there is no singularity in angular acceleration.

In the idealized situation of $I=0$, then the denominator $I + m R^2 ( \cos \theta+1) = 0$ when $\cos \theta = -1$, which means when $\theta = 180°$.

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  • $\begingroup$ @Alexiou The mass is “welded” onto the cylinder wall so how do you see it rotating about itself? The kinetic energy at point A is $2mgR= \frac 12m\nu^2$and $\nu= 2\sqrt {gR}$ so the moment of inertia should not come into play. $\endgroup$
    – Jens
    Commented Mar 14 at 0:41
  • $\begingroup$ @Jens - The cylinder has no mass, so the MMOI about the center of mass (the rod itself) is zero. $\endgroup$ Commented Mar 14 at 4:08
  • $\begingroup$ @Jens - I updated my answer to put some math behind my assessment. $\endgroup$ Commented Mar 14 at 5:59
  • $\begingroup$ @ Alexiou I assume there is a typo because there should be a factor of 1/2 in front of your equation for the angular acceleration in order to match the correct result when I =0 $\endgroup$
    – Jens
    Commented Mar 14 at 13:13
  • $\begingroup$ @Jens - No there is no typo. You can derive the equations yourself and get this solution. With $I=0$ and $\cos \pi = -1$ the denominator is $0 + m R^2 (-1 + 1) = 0$ which means $\ddot{\theta} \rightarrow \infty$. $\endgroup$ Commented Mar 14 at 16:17
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If you look at the motion of this it is a cycloid

The instantaneous acceleration at the moment the point touches the ground is infinite, as the motion is purely vertical and the direction changes from downwards to upwards instantaneously.
As this is, at this moment, purely radial acceleration, (point is directly vertically beneath the center of rotation), then "r" for the calculation of the moment of inertia is zero. So $\omega^2$ can be infinite as the radius of rotation is zero.
However, you will find that the $\lim_{\theta\to 180} \omega^2r$ is finite.
Here's one way to think about it:
A point mass p is moving along the x-axis, there is another point mass q at (1,0), as p passes q at (1,0) its radial velocity wrt q is infinite. No big deal.

You do have to be careful as the Newtonian expression for this motion with y=f(x) is not smooth at the point when the mass touches the ground and is not differentiable at this point. The first derivative is not continuous at this point, and the second derivative therefore does not exist.

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  • $\begingroup$ I think this is the true explanation. The y-acceleration is bounded to -2g but the x-acceleration becomes infinite and changes sign at 180 degrees and this must be reflected in omega but really has nothing to do with the moment of inertia except that the zero value allows infinite omega. In fact the rotational version of Newton 2, which involves omega and a moment of inertia, doesn't even need to be applied. $\endgroup$
    – Jens
    Commented Mar 15 at 23:03
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Since I assume comments should be rather short, I write this as an answer since it is actually just a comment on how to easily get an expression for the angular velocity when the rod has some small moment of inertia I.

In that case we can just add its kinetic energy $\frac12I\omega^2$to my previous expression for the total energy which then becomes $E=E_{_{kin}}+E_{_{pot}}=\frac12I\omega^2+mR^2(\omega^2+\frac gR)(1+\cos\theta)=2mgR$

From this we get

$ \omega^2=\frac gR\frac{1-\cos\theta}{1+\cos\theta+\frac I{2mR^2}} $

If the rod has radius $\rho$ then $I=\frac12m\rho^2$ giving

$\omega^2=\frac gR\frac{1-\cos\theta}{1+\cos\theta+\frac14\frac{\rho^2}{R^2}}$

At the critical angle $\theta=\pi$ this gives $\omega=2\sqrt[]{\frac{2g}R}\frac R\rho$ which of course is bounded.

However, I still don’t understand how the whole cylinder would be able to pivot around A with an infinite $\omega$ when I=0, since I would think that any such rotation would somehow be blocked by the ground.

So maybe the question can be resolved by somehow including other ground reaction forces than N and F!

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  • $\begingroup$ The angular velocity is $$\omega=\frac{d\theta}{dt}=\sqrt{\frac gR}tg(\theta/2)$$ Integration of this gives the theta versus time relation $$\sin{(\theta/2)}=C\exp(\frac12\sqrt{\frac gR}t)$$ where C is some integration constant. This makes no sense since we require $\theta=0$ when $t=0$ which is not possible unless C is zero. So either it is not a valid solution and/or the expression for $\omega$ must be wrong. All this just adds to the paradox but maybe it has to do with the top position being "stable" since we have set both the omega and its acceleration to 0 at that point. $\endgroup$
    – Jens
    Commented Mar 15 at 16:22

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