Subfield: Mechanics


Question: a. The number and nature of the equilibrium points on the hemisphere depends on the value of the spring constant $k$. Consider the semicircular arc shown above as a dashed line, which is parameterized by angles in the range $-\pi / 2 \leq \theta \leq \pi / 2$. Make a table indicating the number of equilibrium points on the arc, and the number that are stable, for each range of $k$ values. A blank table for your reference is given below. (You may need more or fewer rows than shown.)



| Range of $k\left(k_{\min }<k<k_{\max }\right)$ | \#  of Equilibria | \#  of Stable Equilibria |
|:----------------------------------------------:|:-----------------:|:------------------------:|
|                     $0<k<?$                    |                   |                          |
|                                                |                   |                          |
|                                                |                   |                          |
|                  $?<k<\infty$                  |                   |                          |

Solution:
The spring force attracts the mass toward the top of the hemisphere, whereas the gravitational force tends to pull it away.



For very small $k$, the spring force is negligible compared to the gravitational force for $|\theta|<\pi / 2$. For these $k$ values, we only have one unstable equilibrium at the top of the hemisphere due to the gravitational force being zero there.



For some large enough value of $k$, the spring force at $\pi / 2$ exactly compensates for the gravitational force - this marks the end of the first regime. In this next regime, we still have an unstable equilibrium at the top of the hemisphere - because the spring force only compensates for the gravitational force for larger $\theta$. Additionally, we have two equilibria at $0<\left|\theta^{\star}\right|<\pi / 2$. These equilibria must be stable because for $\theta<\theta^{\star}$, the gravitational force is stronger than the spring force and forces the mass toward $\theta^{\star}$, whereas the opposite holds for $\theta>\theta^{\star}$, and the mass is again forced toward $\theta^{\star}$. Thus, we have three total equilibria (two stable, one unstable).



Finally, for sufficiently large $k$ the spring force is stronger than the gravitational force even for arbitrarily small $\theta$, and so we have just one stable equilibrium point at the top of the hemisphere in this third regime.



We now compute the two critical points for us to fill out the table. The first critical point







is given by the balancing of the two forces at $\theta=\pi / 2$, so we compute



$$

m g \sin \pi / 2=k_{1} R(\pi / 2) \Longrightarrow k_{1}=\frac{2 m g}{\pi R}

$$



The second critical point is given by the balancing of the two forces for $\theta \ll 1$, so we compute



$$

m g \sin \theta \approx m g \theta=k_{2} R \theta \Longrightarrow k_{2}=m g / R

$$



We thus get the following table:



| Range of $k$ values $\left(k_{\min }<k<k_{\max }\right)$ | \#  of Equilibria | \#  of Stable Equilibria |
|:--------------------------------------------------------:|:-----------------:|:------------------------:|
|                   $0<k<2 m g /(\pi R)$                   |         1         |             0            |
|                $2 m g /(\pi R)<k<m g / R$                |         3         |             2            |
|                    $m g / R<k<\infty$                    |         1         |             1            |



Notice that the system is symmetric under flipping $\theta \rightarrow-\theta$. Thus, one could incorrectly guess that the only possible equilibrium point is $\theta=0$ by symmetry. In fact, when $k$ is in the right range, we get a pair of new equilibrium points at opposite $\theta$, which map to each other under symmetry. This kind of situation, where the overall setup is still symmetric but the individual equilibrium points are not, is called spontaneous symmetry breaking.



For the rest of the problem, suppose the value of $k$ is such that the mass begins at stable equilibrium on the surface of the hemisphere at angle $\theta_{0}$. The mass can move on the two-dimensional surface of the hemisphere, but a radially-inward external force prevents it from jumping off the surface.

Final answer: | Range of $k$ values $\left(k_{\min }<k<k_{\max }\right)$ | \#  of Equilibria | \#  of Stable Equilibria |
|:--------------------------------------------------------:|:-----------------:|:------------------------:|
|                   $0<k<2 m g /(\pi R)$                   |         1         |             0            |
|                $2 m g /(\pi R)<k<m g / R$                |         3         |             2            |
|                    $m g / R<k<\infty$                    |         1         |             1            |