Tetherball Physics Problem Find Velocity

Consider a tetherball setup (pole stuck in basis with string tied betwixt it and an otherwise freely moving brawl) without gravity. That is, the string'due south stretched out perpendicular to the pole, with the ball at the stop, say at an initial distance $R_0$ from the eye of the pole. And now give the ball an initial tangential velocity $v_0$ then that it starts swinging around the pole (without whatever $z$-centrality motion since at that place's no gravity).

Now, as the ball swings around the pole, the string winds effectually the pole, so the ball'due south distance from the center decreases -- if the pole's bore is $d$ we tin easily figure $R(\theta)<R_0$ after the ball swings through a total bending $\theta$, just the details aren't important.

What is important (what I was trying to effigy) is the ball'southward subsequent tangential velocity $v$ equally 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, in that location'due south some $\frac{mv^2}{R(\theta)}\Delta R$ piece of work done on the ball, and that'south got to be coming from the brawl'south initial $\frac12I\omega_0^two=\frac12mv_0^ii$, which I hadn't accounted for.

Is that right? It'southward 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), and so the tangential motion is arguably contained. Simply and so, where'southward that "radial work" (so to speak) coming from? And then which is it? And if the work's coming from the ball's energy, then how practise you get a radial strength from the tangential move? And if not coming from the brawl's energy, then coming from where?

Edit...
Regarding @zhutchens1'southward comment: there's conspicuously work done due to move in the radial direction, only past strength$\times$distance definition. My very simple work that led me to say "goofy" in a higher place was every bit follows...

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

In particular, we could cut the string at any fourth dimension, and the ball would just get flying off tangentially. So if we insist on conservation of the brawl's kinetic energy, the correct magnitude tangential velocity has to be $five=v_0$ at all times, and therefore $\omega=\frac vR=\frac{v_0}R$ must be right. Not my higher up $\omega=\frac{v_0R_0}{R^two}$ (which multiplies the correct answer by $\frac{R_0}R>1$). Annotation 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 in a higher place.

Edit#2...
In his answer, @zhutchens1's original remark that "the external torque allows for kinetic energy to increase", and @BowlofRed's comment remark"...momentum to commutation between the ball and pole" both implicitly incorporate the germ of the idea illustrated below that occurred to me when I offset saw the preceding "goofy" result, and that prompted my question...

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

If the ball's kinetic energy increased as $R$ decreased, just arrange for it to "latch onto" a 2nd pole with outstretched string length $R_0$, as illustrated, at some $R<R_0$ length of the first pole. So the initial weather for that second pole are now the $v>v_0$ you originally supplied for the first pole. And now, only keep swapping and swapping, dorsum-and-forth, and get all the free free energy you desire.

Every bit @BowlofRed suggested, that energy would presumably be coming from the pole, meaning from the Earth. And in that location are lots of energy-from-the-earth devices, specially tidal power, due east.g., https://en.wikipedia.org/wiki/Tidal_power, that gets energy from the Earth-Moon organisation. The above device would -- if it actually worked -- be getting energy from the Earth'southward rotational energy. But as nosotros figured out, it doesn't work. Nevertheless, it suggests trying to design an apparatus that tin can practise that. Tidal (and geothermal, etc) devices can only be installed at a small number of sites, just a tetherball-like device could be installed anywhere at all.

Tetherball Physics Problem Find Velocity,

Source: https://physics.stackexchange.com/questions/410541/tetherball-problem

Posted by: mcguiganselse2000.blogspot.com

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