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202209281521 Problem 5.37

A bead can slide without friction on a circular hoop of radius R in a vertical plane. The hoop rotates at a constant rate of \omega about a vertical diameter, as shown in the figure below.

(a) Find the angle \theta at which the bead is in vertical equilibrium. (Of course it has a radial acceleration toward the axis.)
(b) Is it possible for the bead to “ride” at the same elevation as the centre of the hoop?
(c) What will happen if the hoop rotates at a slower rate \omega' = \omega /2?

Modified from H. D. Young. (1989). University Physics.

Roughwork.

(a)

Kinetic energy T:

\begin{aligned} T & = \frac{1}{2}mv^2 \\ \dots\enspace v & = r\omega \enspace\dots \\ T & = \frac{1}{2}mr^2\omega^2 \\ \dots\enspace r & = R\sin\theta \enspace\dots \\ T & = \frac{1}{2}mR^2\omega^2\sin^2\theta \\ \end{aligned}

Potential energy V:

\begin{aligned} V & = mg(R-R\cos\theta ) \\ & = mgR(1-\cos\theta ) \\ \end{aligned}

Lagrangian \mathcal{L}=T-V:

\displaystyle{\mathcal{L}=\frac{1}{2}mR^2\omega^2\sin^2\theta-mgR(1-\cos\theta )}

Euler–Lagrange equation:

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial \dot{\theta}}\bigg)-\frac{\partial\mathcal{L}}{\partial \theta}=0}

Thereby

\begin{aligned} \frac{\partial\mathcal{L}}{\partial\dot{\theta}} & = 0 \quad \textrm{and}\quad \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial\dot{\theta}}\bigg) = 0 \\ \frac{\partial\mathcal{L}}{\partial\theta} & = mR^2\omega^2\sin\theta\cos\theta -mgR\sin\theta \\ \end{aligned}

The bead is in vertical equilibrium when:

mR^2\omega^2\sin\theta\cos\theta -mgR\sin\theta =0

solving for \theta then :

\begin{aligned} 0 & = (mR\sin\theta )(R\omega^2\cos\theta -g) \\ \theta & = 0\enspace\textrm{ (rej.)}\quad\textrm{\scriptsize{OR}}\quad \cos^{-1}\bigg(\frac{g}{R\omega^2}\bigg) \\ \end{aligned}

(b)

\theta\to 90^\circ iff R\omega^2\gg g.

(c) This is left as an exercise to the reader.