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The massless inextensible string has a common tension magnitude $T$. Since ${\mathbf{r}}_1=r_1({\mathbf{e}}_r-{\mathbf{e}}_z)$, the unit vector from the first particle towards the apex is $(-{\mathbf{e}}_r+{\mathbf{e}}_z)/\sqrt{2}$. The second particle lies below the apex, soThe string length constraint isThe velocity of the first particle in cylindrical coordinates is ${\dot{r}}_1{\mathbf{e}}_r+r_1\dot{\theta }{\mathbf{e}}_{\theta }+{\dot{z}}_1{\mathbf{e}}_z$. Therefore the powers delivered by the two tension forces areThusThe tension can transfer energy between the particles, but its total power on the pair is zero.

All forces on the first particle lie in the plane spanned by ${\mathbf{e}}_r$ and ${\mathbf{e}}_z$: this holds for gravity, tension, and the normal reaction. Hence the azimuthal component of Newton's second law vanishes:Using the product rule givesStrictly, $h$ is the axial specific angular momentum; the actual axial angular momentum is $m_{1}h$, which is conserved as well.

Differentiate the stated total energy using the chain rule and then substitute the two Newton's second law equations:The gravitational potential energy terms have canceled the power of gravity. Also $\mathbf{N}$ is proportional to ${\mathbf{e}}_r+{\mathbf{e}}_z$, whose dot product with ${\dot{\mathbf{r}}}_1$ is ${\dot{r}}_1+{\dot{z}}_1=0$. Thus the stationary smooth mechanical constraint does no work. The remaining tension powers cancel by part (b), proving $\dot{E}=0$ and conservation of total energy while the massless inextensible string remains taut and contact is maintained.

The massless inextensible string has hanging length $y=l-r$, measured downward from the table. Thus $\dot{y}=-\dot{r}$. In polar coordinates, the tabletop mass has squared speed ${\dot{r}}^2+r^2{\dot{\theta }}^2$. Adding the two kinetic energies and setting the hanging gravitational potential energy to $-Mgy$ givesand the Lagrangian isThe canonical momenta are $p_r=(M+m)\dot{r}$ and $p_{\theta }=m{r}^2\dot{\theta }$. Performing the Legendre transform gives the HamiltonianIt equals the total mechanical energy and is conserved because $L$ has no explicit time dependence. Also $\theta$ is a cyclic coordinate, so $J=p_{\theta }=m{r}^2\dot{\theta }$ is conserved angular momentum about the vertical through the hole. The table and the string guide supply external forces; conservation of the system's total linear momentum does not follow. These are the two independent generic integrals needed for the reduced tabletop mass coupled to a hanging mass.

The holonomic constraint obtained by summing the straight and curved segment lengths of a taut massless inextensible string and setting the result equal to its fixed length $l$. Differentiation gives ${\sum }_i(\partial L/\partial q_i){\dot{q}}_i=0$, a relation between the permitted velocities. A slack string does not impose this equality on endpoint separation.

A tabletop mass $m$ joined through a hole by a massless inextensible string of length $l$ to a hanging mass $M$ has radial inertia $M+m$ and conserved angular momentum $J=m{r}^2\dot{\theta }$. The radial energy is $\frac{1}{2}(M+m){\dot{r}}^2+U_J(r)$, with $0<r<l$. Its effective potential diverges at the hole for $J\ne 0$ and has a unique minimum at $r_0^3=J^2/(mMg)$. Since $U_J^{\prime \prime }(r_0)=3Mg/r_0$, small radial oscillations at fixed $J$ have squared angular frequency $3Mg/[(M+m)r_0]$.