USAPhO 2021, Полуфинал — Задача 6. B3 · The Mad Hatter

• a. 0 < k < 2mg/(πR): 1 equilibrium, 0 stable; 2mg/(πR) < k < mg/R: 3 equilibria, 2 stable; mg/R < k < ∞: 1 equilibrium, 1 stable.

• b. i. C

• b. ii. $F_r = mg\cos\theta_0 - \frac{mv^2}{R}$

  The net radial force −mv²/R was also accepted.

• c. $\alpha = \frac{r}{R\sin\theta_0}\sqrt{1 - \frac{mgR(1 - \cos\theta_0) - (1/2)mgR\theta_0\sin\theta_0}{K}}$

a.

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 $0 < |\theta^{\star}| < \pi/2$. 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

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

We thus get the following table:

Notice that the system is symmetric under flipping $\theta \to -\theta$. 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.

b. i.

The correct path is option $C$. Several explanations would work here. Here are two.

• If we go to the rotating frame of reference, there is an outward centrifugal force that the mass experiences, pushing it down the sphere.
• For the mass to go in a circle around the sphere, the spring force not only has to compensate for the gravitational force but also must provide centripetal acceleration. Therefore, the spring must get longer.

b. ii.

We draw a free-body diagram. It is helpful to draw the diagram in the noninertial reference frame that revolves around the central axis of the hemisphere with speed $v$ at the location of the mass.

Here, $F_c = \frac{mv^2}{R\sin\theta_0}$ is the centrifugal force, and $F_r$ is the radial force from the hemisphere. The forces in the radial direction must balance for the mass to be constrained to the surface of the sphere. Thus,

The perpendicular part of the centrifugal force is $F_c\sin\theta_0$. so we get

Incidentally, there’s a simple way to understand why the second term has to be exactly $mv^2/R$. Consider decomposing the total force on the mass into radial and tangential parts. The radial part simply keeps the mass on the hemisphere; in the absence of a tangential force, the mass would travel in a great circle of radius $R$. Adding a tangential force deflects the mass away from this great circle trajectory, but doesn’t change the radial force required, so the net radial force always has to be $mv^2/R$ inward.

Note: The phrase “total radial force” could also validly be interpreted as the net radial force. Thus, we accepted both $mg\cos\theta_0 - mv^2/R$ and $-mv^2/R$.

c.

The initial energy of the system is given by

Suppose the mass is launched with speed $v$. Then, the speed in the $\theta$ direction is $v\cos\alpha$ and the speed in the $\phi$ direction is $v\sin\alpha$, and therefore, the $z$-component of the angular momentum of the mass is

We now compute the distance of closest approach. If the distance of closest approach is equal to $r$ (as it does for $\alpha_{\mathrm{max}}$, then at $r$, the motion of the mass has no inward component, and the speed of the object at $r$ is given by conservation of angular momentum:

Because $r \ll R$, at the point of closest approach, the energy of the system is roughly

Equating with the initial energy gives us

Before finishing the calculation, we now compute the required $k$ for the object to be at equilibrium (since our answer cannot contain $k$). Setting $kR\theta_0 = mg\sin\theta_0$ gives us $k = \frac{mg\sin\theta_0}{R\theta_0}$. Then,

Solving for $\alpha$ gives us

Solving for $\alpha$ without using the small angle approximation for $\alpha$ also earned full credit. (The answer $\alpha = \pi$, which is technically also correct, earned partial credit.)