Question 7.15: The spherical pendulum. One end of a thin, rigid rod of leng...
The spherical pendulum. One end of a thin, rigid rod of length l and negligible mass is fastened to a bob P of mass m, and its other end is attached to a smooth ball joint at O. In view of the constraint, the bob moves on a spherical surface of radius l, so this device is called a spherical pendulum. The bob is given an arbitrary initial velocity v_o at a point A located in the horizontal plane at the distance h below O in Fig. 7.13. Find three equations that determine the velocity of P as a function of the vertical distance z below O , and describe how the motion x(P , t) may be found from the results.
To find v(z), it proves convenient to introduce cylindrical coordinates (r, Φ, z) with origin at the ball joint O and basis directed as shown in Fig. 7.13, with e_z downward. The velocity of P is given by (see (4.59) in Volume 1)
\mathbf{v}(P, t)=\dot{r} \mathbf{e}_r + r \dot{\phi} \mathbf{e}_\phi + \dot{z} \mathbf{e}_z (7.83a)
We wish to determine \dot{r}, r \dot{\phi}, and \dot{z} as functions of z. Three equations are needed.
The first equation is obtained from the energy principle. The force s that act on P are its weight mg and the workless force T exerted by the rod. Therefore, the principle of conservation of energy (7.73), with V_o=0 at O, yields
K+V=E, \text { a constant. } (7.73)
\frac{1}{2} m v^2 – m g z=m E_0, (7.83b)
where E_0 \equiv E / m is the total energy per unit mass. The speed v of P is thus determined by (7.83a) and (7.83b):
v^2=\dot{r}^2 + r^2 \dot{\phi}^2 + \dot{z}^2=2\left(E_0 + g z\right). (7.83c)
This provides one relation among the three unknown functions. The constant E_o is fixed by the initial conditions at A, E_0=\frac{1}{2} v_0^2 – g h at z = h in (7.83b) .
Another equation may be obtained from the principle of conservation of moment of momentum. The rod tension T has no moment about O, and mg exerts no moment about the vertical OQ axis. Hence, by (7.7 1), the moment of momentum about this line is conserved : \mathbf{h}_O \cdot \mathbf{e}_z=\eta, constant. Recalling (7.83a), we see that only the component m r \dot{\phi} of the linear momentum mv has a moment about the line OQ, namely, r(m r \dot{\phi}) \mathbf{e}_z. Therefore, \mathbf{h}_O \cdot \mathbf{e}_z=m r^2 \dot{\phi}=\eta, and with \gamma \equiv \eta / m, we have
\mathbf{M}_O \cdot \mathbf{e}=0 \Longleftrightarrow \mathbf{h}_O \cdot \mathbf{e}=\text { const } (7.7 1)
r^2 \dot{\phi}=\gamma. (7.83d)
This provides another equation relating the unknown function s. The constant \gamma is determined from the initial conditions. Let \hat{\mathbf{e}}_\phi, denote \mathbf{e}_\phi, at A. Then m \mathbf{v}_0 \cdot \hat{\mathbf{e}}_\phi, is the only component of the initial linear momentum having a moment about the line OQ, and hence \gamma=r_0 \mathbf{v}_0 \cdot \hat{\mathbf{e}}_\phi=r_0 v_0 \cos \left\langle\mathbf{v}_0, \hat{\mathbf{e}}_\phi\right\rangle, wherein r_0=\left(\ell^2 – h^2\right)^{1 / 2} in Fig. 7.13.
The final equation is derived from the suspension constraint: \ell^2=r^2 + z^2. This gives r=\left(\ell^2 – z^2\right)^{1 / 2}, and hence
\dot{r}=-\frac{z \dot{z}}{\sqrt{\ell^2 – z^2}} (7.83e)
A few moments reflection reveals that \dot{r}, r \dot{\phi}, and \dot{z} are now known as functions of z. Indeed , upon substituting (7.83d) and (7.83e) into (7.83c) , we reach
\dot{z}^2=\frac{2 g}{\ell^2}\left[\left(\ell^2 – z^2\right)\left(z + \frac{E_0}{g}\right) – \frac{\gamma^2}{2 g}\right] (7.83f)
which determines \dot{z}(z). And with r(z)=\left(\ell^2 – z^2\right)^{1 / 2}, \dot{r}(z) given by (7.83e), and r \dot{\phi}=\gamma / r(z) from (7.83d), it is now a straightforward matter to write the velocity (7.83a) as a function of z alone . We omit these details.
Finally. we need to say how the motion \mathbf{x}(P, t)=r \mathbf{e}_r + z \mathbf{e}_z, may be read from the results. In principle, integration of (7.83f) determines z(t), hence r(t), and (7.83d) provides
\phi(t)=\int_0^t \frac{\gamma}{r^2} d t (7.83g)
to fix \mathbf{e}_r(\phi), which thus determines the motion. The exact solution for z(t) may be obtained from (7.83f) in terms of Jacobian elliptic function s introduced later; however, we shall not pursue this problem further. (See Synge and Griffith.)