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Consider a tetherball setup (pole stuck in ground with string tied between it and an otherwise freely moving ball) without gravity. That is, the string's stretched out perpendicular to the pole, with the ball at the end, say at an initial distance $R_0$ from the center of the pole. And now give the ball an initial tangential velocity $v_0$ so that it starts swinging around the pole (without any $z$-axis motion since there's no gravity).

Now, as the ball swings around the pole, the string winds around the pole, so the ball's distance from the center decreases -- if the pole's diameter is $d$ we can easily figure $R(\theta)<R_0$ after the ball swings through a total angle $\theta$, but the details aren't important.

What is important (what I was trying to figure) is the ball's subsequent tangential velocity $v$ as a function of $\theta$, since its $I=mR(\theta)^2$ decreases, whereas $I\omega$ is presumably conserved. But that was getting goofy results. What I ended up guessing is that as $R(\theta)$ decreases, there's some $\frac{mv^2}{R(\theta)}\Delta R$ work done on the ball, and that's got to be coming from the ball's initial $\frac12I\omega_0^2=\frac12mv_0^2$, which I hadn't accounted for.

Is that right? It's a little unclear to me since both $\frac{mv^2}{R(\theta)}$ and $\Delta R$ are in the radial direction (although their work=dot_product's a scalar), so the tangential motion is arguably independent. But then, where's that "radial work" (so to speak) coming from? So which is it? And if the work's coming from the ball's energy, then how do you get a radial force from the tangential motion? And if not coming from the ball's energy, then coming from where?

Regarding @zhutchens1's comment: there's clearly work done due to motion in the radial direction, simply by force$\times$distance definition. My very simple work that led me to say "goofy" above was as follows...

Give the ball an initial tangential velocity of $v_0$ at the initial distance $R_0$ from the pole's center. Then initially, the ball's moment of inertia is just $I_0=mR_0^2$ and angular velocity $\omega_0=v_0/R_0$. Now, naively, by conservation of angular momentum, we'd say (at least I said:), $I\omega=I_0\omega_0$ at any future point. So after the ball's swung around enough so that its distance from the pole is $R<R_0$, we'd have $$\omega = I_0\omega_0/I = (mR_0^2\frac{v_0}{R_0})/(mR^2)=\frac{v_0R_0}{R^2}$$ and that just gives us $v$, $$\omega =\frac vR = \frac{v_0R_0}{R^2}\hspace{10pt}\mbox{whereby} \hspace{10pt} v=v_0\frac{R_0}{R}>v_0$$ But $v>v_0$ means the swinging ball's gaining (tangential) kinetic energy, which presumably can't be (otherwise,... patent pending:). So I'm guessing it goes to that $\frac{mv^2}R\Delta R$ work.

In particular, we could cut the string at any time, and the ball would just go flying off tangentially. So if we insist on conservation of the ball's kinetic energy, the correct magnitude tangential velocity has to be $v=v_0$ at all times, and therefore $\omega=\frac vR=\frac{v_0}R$ must be correct. Not my above $\omega=\frac{v_0R_0}{R^2}$ (which multiplies the correct answer by $\frac{R_0}R>1$). Note that, like @zhutchens1 said, $\omega$ indeed increases as $R$ decreases. But his further remark "conserve angular momentum" leads to $\omega$ increasing too much when we insist $I\omega=I_0\omega_0$ like I demonstrated above.

In his answer, @zhutchens1's original remark that "the external torque allows for kinetic energy to increase", and @BowlofRed's comment remark"...momentum to exchange between the ball and pole" both implicitly contain the germ of the idea illustrated below that occurred to me when I first saw the preceding "goofy" result, and that prompted my question...

enter image description here

What you've got illustrated -- >>if<< the ball's kinetic energy were increasing -- is a perpetual motion machine. Actually, even better than perpetual motion -- you can bleed off the excess energy after each "pole swap" and have an endless supply of free energy.

If the ball's kinetic energy increased as $R$ decreased, just arrange for it to "latch onto" a second pole with outstretched string length $R_0$, as illustrated, at some $R<R_0$ length of the first pole. Then the initial conditions for that second pole are now the $v>v_0$ you originally supplied for the first pole. And now, just keep swapping and swapping, back-and-forth, and get all the free energy you want.

As @BowlofRed suggested, that energy would presumably be coming from the pole, meaning from the Earth. And there are lots of energy-from-the-earth devices, particularly tidal power, e.g., https://en.wikipedia.org/wiki/Tidal_power, that gets energy from the Earth-Moon system. The above device would -- if it actually worked -- be getting energy from the Earth's rotational energy. But as we figured out, it doesn't work. Nevertheless, it suggests trying to design an apparatus that can do that. Tidal (and geothermal, etc) devices can only be installed at a small number of sites, but a tetherball-like device could be installed anywhere at all.


marked as duplicate by sammy gerbil, Community Jun 9 '18 at 6:43

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  • $\begingroup$ I don't think there's any work. In order to conserve angular momentum, $\omega$ has to increase with $R(\theta)$ decreasing. Could you outline your work where you found the $\frac{mv^2}{R(\theta)}\Delta R$? $\endgroup$ – Zack Hutchens Jun 7 '18 at 11:40
  • $\begingroup$ @zhutchens1 Thanks for the comment. See Edit above. $\endgroup$ – John Forkosh Jun 7 '18 at 12:04
  • $\begingroup$ I suspect that the interaction is causing angular momentum to exchange between the ball and pole. The pole is large enough that any $\Delta \omega$ it experiences is negligible, but the $\Delta L$ is still relevant. Rather like the momentum change of a wall when you bounce a ball against it. $\endgroup$ – BowlOfRed Jun 7 '18 at 20:57
  • $\begingroup$ "That is, the string's stretched out perpendicular to the pole" and "the string winds around the pole" are contradictory statements. The described system cannot exist... $\endgroup$ – DJohnM Jun 8 '18 at 2:45
  • $\begingroup$ @sammygerbil yeah, that's pretty much the same question, with JohnMCavall's answer the closest to zhutchens1's approach below. But I don't see a clear discussion answering the op's question of which approach (conservation of energy or angular momentum) is correct any why. While JMC mentions in passing that the string is tangential to the pole, he doesn't go on to mention that this results in a tangential component to the $mv^2/r$ force, thereby retarding the ball's tangential motion, whereby $I\omega$ isn't conserved as you'd usually expect. (P.S. thanks for the downvotes, ditto zh's answer:) $\endgroup$ – John Forkosh Jun 9 '18 at 5:43

Note that this might not be a full answer, so I intend to discuss this more (which hopefully will lead to a clear answer). I was interested, so I decided to attack this problem with the Lagrangian mechanics.

The first thing to note is that the mass $m$ really only has one degree of freedom, $\theta$. As you mention, its radial position depends on $\theta$ because the tether, length $L$, winds around. Let's suppose it winds around a small pole of radius $a$, where $a << L$ so I don't have to care about the pole in my coordinate system.

The radial length of the tether for any $\theta$ should be given by

$$ r = L - a\theta,$$

where I have defined that $r=L$ at $\theta(t)=0$. Then, the mass' radial velocity is

$$ \dot r \hat r = \frac{dr}{dt}\hat r = -a\dot \theta \hat r.$$

We know that velocity in polar coordinates is $\dot{\vec{r}} = \dot{r}\hat{r} + r\dot\theta\hat\theta,$ so the mass' velocity vector is

$$ \vec{v} = -a\dot \theta \hat r + \left[(L-a\theta)\dot\theta \right]\hat\theta.$$

By computing $\vec{v} \cdot \vec{v}$, you can get the needed term to compute the kinetic energy, which in this case, is the Lagrangian (no potential):

$$\mathcal{L} = T = \frac{1}{2}mv^2 = \frac{1}{2}m\dot{\theta}^2\left(a^2 + L^2 + a^2\theta^2 - 2La\theta \right).$$

We know it speeds up as it passes, so $\dot\theta$ is not constant, and thus the Lagrangian is not conserved. In other words, the kinetic energy is changing. We can easily show that the generalized force is

$$ \tau_\theta = \frac{\partial \mathcal{L}}{\partial\theta} = \frac{1}{2}m\dot{\theta}^2 \left(2a^2\theta - 2La\right).$$

Here the generalized force is a torque (note its units, $\rm N \cdot m$).

So based on this, my (perhaps incomplete/unjustified) argument is that $I\omega$ is not conserved quantity, because an external torque acts on the system. The external torque allows for kinetic energy to increase.

The source of this torque I am not certain. It's a fairly basic system, so the best idea that comes to mind is a torque from the normal force in contact with the rod -- I believe it is a constraint force due to the fact that $r$ depends explicitly on $\theta$.

Update, based on discussion

@John Forkosh realized that because the pole has some radius $a$, the string does not pointly exactly radially outward from it. We can generalize the tension force on this object as

$$ F_{T} = F_{||} + F_{\perp},$$

where $F_{||}$ and $F_{\perp}$ are the components parallel and perpendicular to the momenutum $\vec p$, respectively. The centripetal force is such that

$$F_{\perp} = \frac{mv^2}{r}. $$

Then $F_{||} = F_T - \frac{mv^2}{r}$.

  • $\begingroup$ Your emphasis on the pole's radius $a$ suggested the following to me: while $a<<L$, it's not entirely ignorable. Because $a>0$, the force exerted by the string on the ball isn't entirely central (towards the pole's exact center). Rather, there's the usual right triangle with legs $L$ and $a$, with the string force directed along the hypoteneuse. And therefore, that string force has a (small but non-negligible) tangential component retarding the ball's motion. And I'm guessing that if you work it out, that's what keeps $|v|$ constant. $\endgroup$ – John Forkosh Jun 8 '18 at 1:27
  • $\begingroup$ @John Forkosh Yes, you are totally right. I knew that giving the pole a radius would affect the geometry of my coordinate system, for the reason you state, which I was I tried to ignore it -- not realizing that was the answer. I'll formalize my answer to include this. $\endgroup$ – Zack Hutchens Jun 8 '18 at 1:39
  • 1
    $\begingroup$ Thanks. There's a second edit to my question, above, illustrating why I was pretty sure your original remark "the external torque allows for kinetic energy to increase" had to be wrong. But, if you can conjure up another device design that actually accomplishes that, let's share the patent 50-50:) $\endgroup$ – John Forkosh Jun 8 '18 at 2:40

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