The other night, a former graduate student of mine posted the problem of the dynamics of motion of a tethered ball, undoubtedly leading many readers to a sleepless night. It's called nerd sniping; us nerds have to drop everything in a compulsive search for a solution.
The figure shows the problem, where we ignore gravity and search for a solution for the motion of the ball as it wraps around the pole of radius $R$, whose size I have exaggerated for clarity. The ball of mass $m$ is assumed to be a point and is attached to a massless rope of length $L$. The rope is attached at point $A$ and the angle $\theta$ to point $B$ determines the part of the rope in contact with the pole. We chose the center of the pole as the origin of the inertial frame from which we will calculate the equations of motion. Physical Reasoning
A physicist can often solve a problem quickly using intuition based on experience. For example, in this case, we can imagine that for an infinitesimal time, the rope can be viewed as pivoting about point $B$, so the tension in the rope $T$, which pulls the mass toward point $B$, is always perpendicular to the velocity $v$. As such, the work done on the mass must vanish so the kinetic energy is constant. Then, the velocity remains constant. On the other hand, the angular momentum decreases. If the initial velocity of the mass is $v_0$ at point $A$, the angular momentum is $m v_0 L$. When the rope has gone around once so that its length is now $L - 2 \pi R$, if the velocity is the same, the angular momentum is $m v_0 (L - 2 \pi R)$. Upon each wind, the mass loses angular momentum $2 \pi m v_0 R$.Where does this angular momentum go? If the pole were were free to rotate upon it's axis, the pull of the rope at contact point $B$ would apply a torque, so the angular momentum lost by the mass would be transferred to the pole. Since the pole is attached to a massive earth, it is the earth that absorbs the angular momentum.
There are many things about this problem that make me uneasy about such reasoning, and some of it stems from those times that I have been fooled by a subtlety. For example, you might have falsely argued that angular momentum must be conserved since the pole is obviously fixed and only action and reaction forces are acting. Alternatively, you might be concerned with the energy conservation argument because the contact point is moving around the surface of the pole, and is therefore not an inertial frame. The key to why no work is done by the rope is the that the velocity and tension are always perpendicular in any frame.
If you are uncomfortable with using intuition, the Lagrangian method is for you. Not only is it fool proof, but this problem is easy to solve and has some neat properties.
Lagrangian Method
From the figure, the coordinates of the mass are $x = R \cos \theta - (L-R \theta) \sin \theta $ and $y = R \sin \theta + (L-R \theta) \cos \theta $ which gives $r^2 = R^2 + (L - R \theta)^2$, as expected.The Lagrangian contains only the kinetic energy terms and after some simple trigonometric identities leads to the beautiful result $$ {\cal L} = \frac {1} {2} m (\dot{x}^2 + \dot{y}^2 )= \frac {1} {2} m (L-R \theta)^2 \dot{\theta}^2.$$ The Euler Lagrange equation yields $$ (L - R \theta)^2 \ddot{\theta} = R(L-R \theta) \dot{\theta}^2,$$ and with the simple substitution $\ell = L - R \theta$, with $\ell$ being the unwound part of the rope, gives $$ \ell \ddot{\ell} + \dot{\ell}^2 =0,$$ which gives the exact differential $$ \frac {d} {dt} (\ell \dot{\ell}) = 0.$$ Integration gives $\ell \dot{\ell} = c$, yielding the general solution $$ \ell^2 = 2 c t + b,$$ where $b$ and $c$ are integration constants that we fix by setting $\ell = L$ and $\dot{\ell} = \dot{\ell}_0$ at $t=0$ to give $$ \ell = \sqrt{L^2 + 2L \dot{\ell}_0 t}.\hspace{3em} (1)$$
At this point, we are done but it is fun to recast the results in other ways. We can re-express the Lagrangian in terms of the length of unwound string, which gives $$ {\cal L} = \frac {1} {2} m \frac {\ell^2} {R^2} \dot{\ell}^2, \hspace{3em} (2)$$ so the velocity is given by $$ v = \frac {\ell} {R} \dot{\ell} \hspace{3em} (3)$$ yielding an initial velocity of $$ v_0 = \frac {L} {R} \dot{\ell}_0. \hspace{3em} (4).$$ Combining Equations 1 and 4 we get $$\ell = \sqrt{L^2 + 2 R v_0 t}, $$ and Equation (3) gives $v = v_0$, confirming our intuition that energy is conserved.
