Question 11.4: Particle sliding inside a bowl A particle P of mass m slides...
Particle sliding inside a bowl
A particle P of mass m slides on the inside surface of an axially symmetric bowl. Find its angular momentum about its axis of symmetry in terms of the coordinates ρ, Φ shown in Figure 11.3.
In order to express L _{O} \cdot k in terms of coordinates, we draw a velocity diagram for the system as explained in section 10.10. The velocities v_{\rho} \text { and } v_{\phi} , corresponding to the coordinates ρ and Φ, have the directions shown in Figure 11.3. These two velocities are perpendicular, with the v_{\phi} contribution in the azimuthal direction around the vertical axis {O, k}. It follows that v \cdot \widehat{\phi}=v_{\phi}=\rho \dot{\phi} . The required axial angular momentum is therefore
L _{O} \cdot k =m \rho( v \cdot \widehat{ \phi })=m \rho(\rho \dot{ \phi })=m \rho^{2} \dot{\phi}.
Just for the record, the velocity v_{\rho} is the (vector) sum of \dot{\rho} radially outwards and \dot{z} vertically upwards. Note that \dot{z} is not an independent quantity. If the equation of the bowl is z = f (ρ), then \dot{z}=f^{\prime}(\rho) \dot{\rho} . In particular then, the kinetic energy of P is given by
T=\frac{1}{2} m\left[\dot{\rho}^{2}+\left(f^{\prime}(\rho) \dot{\rho}\right)^{2}+(\rho \dot{\phi})^{2}\right].
and its potential energy by V = mgf (ρ).
