Question 5.13: Conical pendulum. A small ball of mass m, suspended by a cor......
Conical pendulum. A small ball of mass m, suspended by a cord of length £, revolves in a circle of radius r = ℓ sin 0, where 0 is the angle the string makes with the vertical (Fig. 5-20). (a) In what direction is the acceler-ation of the ball, and what causes the acceleration? (b) Calculate the speed and period (time required for one revolution) of the ball in terms of \ell, \theta, g, and m.
APPROACH We can answer (a) by looking at Fig. 5-20, which shows the forces on the revolving ball at one instant: the acceleration points horizontally toward the center of the ball’s circular path (not along the cord). The force responsible for the acceleration is the net force which here is the vector sum of the forces acting on the mass m : its weight \overrightarrow{\mathbf{F}}_{\mathrm{G}} (of magnitude F_{\mathrm{G}}=m g ) and the force exerted by the tension in the cord, \overrightarrow{\mathbf{F}}_{\mathrm{T}}. The latter has horizontal and vertical components of magnitude F_{\mathrm{T}} \sin \theta and F_{\mathrm{T}} \cos \theta, respectively.
(b) We apply Newton’s second law to the horizontal and vertical directions. In the vertical direction, there is no motion, so the acceleration is zero and the net force in the vertical direction is zero:
F_{\mathrm{T}} \cos \theta-m g=0 .
In the horizontal direction there is only one force, of magnitude F_{\mathrm{T}} \sin \theta, that acts toward the center of the circle and gives rise to the acceleration v^{2} / r. Newton’s second law tells us:
F_{\mathrm{T}} \sin \theta=m \frac{v^{2}}{r} .
We solve the second equation for v, and substitute for F_{\mathrm{T}} from the first equation (and use r=\ell \sin \theta ):
\begin{aligned}v=\sqrt{\frac{r F_{\mathrm{T}} \sin \theta}{m}} & =\sqrt{\frac{r}{m}\left(\frac{m g}{\cos \theta}\right) \sin \theta} \\& =\sqrt{\frac{\ell g \sin ^{2} \theta}{\cos \theta}} .\end{aligned}
The period T is the time required to make one revolution, a distance of 2 \pi r=2 \pi \ell \sin \theta. The speed v can thus be written v=2 \pi \ell \sin \theta / T; then
\begin{aligned}T=\frac{2 \pi \ell \sin \theta}{v} & =\frac{2 \pi \ell \sin \theta}{\sqrt{\frac{\ell g \sin ^{2} \theta}{\cos \theta}}} \\& =2 \pi \sqrt{\frac{\ell \cos \theta}{g}} .\end{aligned}
NOTE The speed and period do not depend on the mass m of the ball. They do depend on \ell and \theta.