Physics - SHM Question with Solution | TestHub

Teeter Toy\\ The toy shown in the figure, with two heavy but equal masses embedded at the lower end of its wings remains in perfectly stable equilibrium, even though it is supported by just a point (the beak of the bird in this case.) We shall investigate the making of this toy.\\

\\ The simplest model of this toy consists of two identical weights hanging from a peg with drooping arms as shown. The arrangement is unexpectedly stable. The toy can, be spun or rocked with little danger of toppling over.\\

\\ We calculate the gravitational potential energy of the toy when it has been swayed by an angle $\theta$.

$$(\mathrm{U}=0)-\cdots -\frac{1}{\mathrm{m}}{e}_{\mathrm{HCD}\theta }{\int }_{\mathrm{m}}^{\alpha -\theta }\ell \cos (\alpha -\theta )$$

Taking datum (zero) of potential energy at the pivot,

$$\begin{array}{rl}& \mathrm{U}(\theta )=\mathrm{mg}[\mathrm{H}\cos \theta -\ell \cos (\alpha +\theta )]\\ & +\mathrm{mg}[\mathrm{H}\cos \theta -\ell \cos (\alpha -\theta )]\\ & =2\mathrm{mg}\cos \theta [\mathrm{H}-\ell \cos \alpha ]\end{array}$$

Equilibrium occurs when

$$\frac{\mathrm{dU}}{\mathrm{d}\theta }=-2\mathrm{mg}\sin \theta [\mathrm{H}-\ell \cos \alpha ]=0$$

The solution is $\theta =0$ as we expect from symmetry.\\ To investigate the stability of equilibrium, we examine the $2^{\text{nd }}$ derivative of potential energy.

$$\begin{array}{rl}{\frac{{\mathrm{d}}^2\mathrm{U}}{\mathrm{d}{\theta }^2}|}_{\theta =0}& =[-2\mathrm{mg}\cos \theta (\mathrm{H}-\ell \cos \alpha ){]}_{\theta =0}\\ & =-2\mathrm{mg}(\mathrm{H}-\ell \cos \alpha )\end{array}$$

for equation to be stable i.e. for the teeter toy to be stable, the weights must hang below the pivot.

$${\frac{{\mathrm{d}}^2\mathrm{U}}{\mathrm{d}{\theta }^2}|}_{\theta =0}>0\Rightarrow \mathrm{H}-\ell \cos \alpha <0$$

Weight is the only force providing torque about the pivot and hence torque about the pivot would be given by :

$$\tau =-\frac{\mathrm{dU}}{\mathrm{d}\theta }$$