Consider a pendulum (see first drawing in figure) built up with an homogeneous bar of length $L$ and mass $M$. The moment of inertia with respect to the joint (black point in the figure) is $I = \frac{ML^2}{3}.$ Its center of gravity (CoG) is placed on the middle of the bar. It is well known that the total energy of the system is:

$$E(\theta) = Mg\frac{L}{2}(1 - \cos(\theta)) + \frac{1}{2}I\omega^2(\theta) ,$$

where $\omega(\theta)$ is the angular velocity of the pendulum as a function of the angle $\theta$. For $\theta = 0$ (i.e. the pendulum is in the stable position), the pendulum has an angular velocity $\omega_0$, and hence the total energy is:

$$E(0) = \frac{1}{2}I\omega_0^2.$$

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Here I need to find the maximum angle $\overline{\theta}$ reached by the pendulum. Of course, $\omega(\overline{\theta}) = 0$, and hence:

$$E(\overline{\theta}) = Mg\frac{L}{2}(1 - \cos(\overline\theta)) = \frac{1}{2}I\omega_0^2 \Rightarrow \overline\theta = \arccos\left(1 - \frac{L\omega_0^2}{3g}\right).$$

Notice that this is possible only when $\omega_0^2 < \frac{6g}{L}$, otherwise the pendulum will perform complete rotations (is this right?).

Now, we move to the second drawing in figure, where the pendulum is tied to a movable sleeve (red object) with mass $m$, which is free to move without friction along the fixed grey horizontal bar. Indicating with $v(\theta)$ the velocity of the sleeve, we have that:

$$E(\theta) = Mg\frac{L}{2}(1 - \cos(\theta)) + \frac{1}{2}I\omega^2(\theta) + \frac{1}{2}mv^2(\theta).$$

For $\overline\theta$ such that $\omega(\overline{\theta}) = 0$, I get:

$$E(\overline{\theta}) = Mg\frac{L}{2}(1 - \cos(\overline\theta)) + \frac{1}{2}mv^2(\overline\theta) = \frac{1}{2}I\omega_0^2.$$

Here I'm stuck, since I don't know anything about $v(\theta)$. And my question arises:

does the momentum along $x$ axis is conserved?

The momentum along $x$ axis is given by:

$$p_x(\theta) = mv(\theta) + M\frac{L}{2}\omega(\theta).$$

Assuming that the sleeve is steady at the beginning (i.e. $v(0) = 0$), then:

$$p_x(\theta) = p_x(0)= M\frac{L}{2}\omega_0.$$

At $\theta = \overline\theta$, we have that $\omega(\overline\theta) = 0$, and thus:

$$v(\overline\theta) = \frac{ML}{2m}\omega_0.$$

The last equation can help me to find $\overline\theta$.

However... is the assumption on the momentum right? The only external force is the gravity which acts only along the $y$ axis. But if this is true for the second system (with sleeve), then is the momentum along $x$ conserved also in the simpler case (first system with no sleeve)?

I also know that for the presence of the gravity, the angular moment is not conserved in both cases.


Using the conservation of the momentum along $x$ and of the total energy, I come to this result:

$$\cos(\overline\theta) = 1 + \frac{L\omega_0^2}{12mg}(3M - 4m).$$

Now, if $M > \frac{4m}{3}$, then $\cos(\overline\theta) > 1$, which means that, for any initial velocity $\omega_0$, I can't find a $\overline\theta$ such that $\omega(\overline\theta)$ is $0$. That is, the pendulum will perform complete rotations. This sounds weird to me, and this is why I'm having doubt on conservation of the momentum.

closed as off-topic by sammy gerbil, Jon Custer, John Rennie, Kyle Kanos, Mostafa Sep 20 '17 at 19:09

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  • You appear to be asking two questions here : (1) is momentum conserved horizontally? and (2) what is wrong with my calculation? Philip Wood answered (1). Your comment to his answer is that your main concern is (2). Asking what is wrong with your calculation is off topic. – sammy gerbil Sep 18 '17 at 10:49
  • Sorry, but I see the whole story from a different perspective.I don't know if the momentum is conserved. If it is, then I do some calculations which lead to strange results. The strange results suggest me that the momentum is not conserved. That's not a homework. Anyway, thanks a lot for the downvote :D – the_candyman Sep 20 '17 at 14:27
  • LOL - Homework-like questions should ask about a specific physics concept (I asked) and show some effort to work through the problem (I showed). Thanks again for downvote and for the "put on hold", is fantastic! – the_candyman Sep 20 '17 at 19:40
up vote 3 down vote accepted

The x-wise momentum of the sleeve-pendulum system is conserved, but for the ordinarily-suspended pendulum the momentum of the Earth-pendulum system is conserved. In neither case is the x-wise momentum of just the pendulum conserved, because there is an external x-wise force on it.

  • +1 It bears repeating: a component of momentum of an object is conserved if and only if that component of the force on the object is zero, because $\vec{F} = m \vec{a} = d\vec{p}/dt$. So when you're talking about just the pendulum (and not the sleeve), its momentum is clearly not conserved. – Mike Sep 17 '17 at 11:26
  • Thanks a lot, but I'm missing something, and this is my fault for sure. Which is the main difference between the system "pendulum with sleeve" and "pendulum without sleeve"? Is this related to the presence of "reaction forces" between pendulum and sleeve? – the_candyman Sep 17 '17 at 11:38
  • There are forces (a Newton 3 pair) between pendulum and sleeve, but no x–wise force between the sleeve and the Earth. With the pendulum suspended as usual the forces are directly between the pendulum and the Earth (which, I'm led to believe, is much more massive than the sleeve)! – Philip Wood Sep 17 '17 at 13:01
  • I think I'm starting to understand the point. Anyway, my main concern is about the result I get on the maximum angle $\overline\theta$. Indeed, assuming also that the momentum is conserved, then I get that: $$\cos(\overline\theta) = 1 + \frac{L\omega_0^2}{12mg}(3M - 4m).$$ Apparently, this means that if $M>\frac{4m}{3}$, then for any initial velocity $\omega_0$, the pendulum will perform a full rotation... So, there must be something that I'm doing wrong. – the_candyman Sep 17 '17 at 14:51
  • I have update my main question consequently. – the_candyman Sep 17 '17 at 14:55

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