Chapter 5 Problem 93

Problem

A small bead of mass $m$ is constrained to slide without friction inside a circular vertical hoop of radius $r$ which rotates about a vertical axis at a frequency $f$.

(a) Determine the angle $\theta$ where the bead will be in equilibrium,that is, where it will have no tendency to move up or down along the hoop.

(b) If $f=2.00$rev/s and $r=22.0$cm, what is $\theta$?

(c) Can the bead ride as high as the center of the circle ($\theta =90^{\circ }$)? Explain

Solution

The relationship between speed and frequency: $v=2\pi fr\sin \theta$

(a)

When the bead is in such an equilibrium, the net vertical force on it is zero (so that it does not move vertically)

$\sum F_{\text{vertical}}=F_{N}\cos \theta -mg=0$ $\rightarrow F_{N}={\frac {mg}{\cos \theta }}$

Using the fact that it is tracing a circular path horizontally, the angle can be calculated as:

$\sum F_{\text{radial}}=F_{N}\sin \theta =m{\frac {v^{2}}{r\sin \theta }}=m{\frac {4\pi ^{2}f^{2}r^{2}\sin ^{2}\theta }{r\sin \theta }}$ $F_{N}\sin \theta ={\frac {mg}{\cos \theta }}\sin \theta =m{\frac {4\pi ^{2}f^{2}r^{2}\sin ^{2}\theta }{r\sin \theta }}$ $\rightarrow \theta =\cos ^{-1}{\frac {g}{4\pi ^{2}f^{2}r}}$

(b)

$\theta =\cos ^{-1}{\frac {9.80{\text{m/s}}^{2}}{4\pi ^{2}(2.00{\text{Hz}})^{2}(0.220{\text{m}})}}=73.6^{\circ }$

(c)

No, the bead cannot ride as high as the center of the circle. If the bead were located there, the normal force of the wire on the bead would point horizontally. There would be no force to counteract the bead’s weight, and so it would have to slip back down below the horizontal to balance the force of gravity. From a mathematical standpoint, the expression ${\frac {g}{4\pi ^{2}f^{2}r}}$ would have to be equal to 0 and that could only happen if the frequency or the radius were infinitely large.