Classical Mechanics (Kibble & Berkshire) · Problem 11.8

Lagrange's equations and conservation principles — T.W.B. Kibble, F.H. Berkshire

Physics — Mechanics · particle-sliding-on-cone

Problem

Particle sliding on a cone A particle PP slides on the smooth inner surface of a circular cone of semi-angle α\alpha. The axis of symmetry of the cone is vertical with the vertex OO pointing downwards. Show that the vertical component of angular momentum about OO is conserved in the motion. State a second dynamical quantity that is conserved. Initially PP is a distance aa from OO when it is projected horizontally along the inside surface of the cone with speed uu. Show that, in the subsequent motion, the distance rr of PP from OO satisfies the equation r˙2=(r−a)[u2(r+a)r2−2gcos⁡α].\dot{r}^2 = (r-a)\left[\frac{u^2(r+a)}{r^2} - 2g\cos\alpha\right]. Case A For the case in which gravity is absent, find rr and the azimuthal angle ϕ\phi explicitly as functions of tt. Make a sketch of the path of PP (as seen from `above') when α=π/6\alpha = \pi/6. Case B For the case in which α=π/3\alpha = \pi/3, find the value of uu such that rr oscillates between aa and 2a2a in the subsequent motion. With this value of uu, show that rr will first return to the value r=ar = a after a time 23(ag)1/2∫12ξ dξ[(ξ−1)(2−ξ)(2+3ξ)]1/2.2\sqrt{3}\left(\frac{a}{g}\right)^{1/2}\int_1^2 \frac{\xi\,d\xi}{[(\xi-1)(2-\xi)(2+3\xi)]^{1/2}}.
Intuitive explanation

The key to this problem is recognizing that the cone's smooth surface and vertical symmetry create a constraint: forces perpendicular to the axis (gravity and normal force) produce no torque about OO, so the vertical component of angular momentum Lz=mr2sin⁡2α ϕ˙L_z = mr^2\sin^2\alpha\,\dot{\phi} is conserved. Energy is your second conserved quantity. Use these two conservation laws together with the constraint that the particle stays on the cone surface (θ=α\theta = \alpha in spherical coordinates) to derive the radial equation of motion—this is the core insight that unlocks the given r˙2\dot{r}^2 equation without needing to solve forces explicitly.

Intuitive explanation

The key to this problem is recognizing that the cone's smooth surface and vertical symmetry create a constraint: forces perpendicular to the axis (gravity and normal force) produce no torque about OO, so the vertical component of angular momentum Lz=mr2sin⁡2α ϕ˙L_z = mr^2\sin^2\alpha\,\dot{\phi} is conserved. Energy is your second conserved quantity. Use these two conservation laws together with the constraint that the particle stays on the cone surface (θ=α\theta = \alpha in spherical coordinates) to derive the radial equation of motion—this is the core insight that unlocks the given r˙2\dot{r}^2 equation without needing to solve forces explicitly.

Related theory

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