Subfield: Mechanics

## The Mad Hatter



A frictionless hemisphere of radius $R$ is fixed on top of a flat cylinder. One end of a spring with zero relaxed length and spring constant $k$ (i.e. the force from the spring when stretched to length $\ell$ is $-k \ell$ ) is fixed to the top of the hemisphere. Its other end is attached to a point mass of mass $m$.



<img_4469>
Context 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 |
|:----------------------------------------------:|:-----------------:|:------------------------:|
|                     $0<k<?$                    |                   |                          |
|                                                |                   |                          |
|                                                |                   |                          |
|                  $?<k<\infty$                  |                   |                          |
Context 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            |


Extra Supplementary Reading Materials:

b. At $t=0$, the mass is given a speed $v$ along a line of constant latitude $\theta=\theta_{0}$.



<img_4518>
Context question:
i. Indicate which of the following trajectories the mass takes for a short time after $t=0$ and briefly explain your reasoning. The differences between the paths are exaggerated.



<img_4424>
Context answer:
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.


Context question:
ii. What is the total radial force (i.e., normal to the surface of the hemisphere) on the mass at $t=0$ ? Express your answer in terms of $m, v, R, g$, and $\theta_{0}$.
Context answer:
\boxed{$F_{r}=m g \cos \theta_{0}-\frac{m v^{2}}{R}$}


Question: c. A cylinder of radius $r \ll R \theta_{0}$ is placed on top of the sphere. Suppose the mass is launched at an angle $\alpha$ away from the direction of the spring's displacement with kinetic energy $K$, as shown.



What is the maximum angle $\alpha_{\max }$ at which the mass can be launched such that it can still hit the cylinder? Express your answer in terms of $K, m, g, \theta_{0}, r$, and $R$. You may assume $K$ is large enough for the mass to reach the cylinder for $\alpha=0$.



<img_4563>



(view from above)

Solution:
The initial energy of the system is given by



$$

m g R \cos \theta_{0}+\frac{1}{2} k R^{2} \theta_{0}^{2}+K

$$



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



$$

L=m v \sin \alpha(R \sin \theta)

$$



We now compute the distance of closest approach. If the distance of closest approach is equal to $r$ (as it does for $\alpha_{\max }$, 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:



$$

m u r=m v \sin \alpha(R \sin \theta) \Longrightarrow u=\frac{v \sin \alpha(R \sin \theta)}{r}

$$



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



$$

m g R+\frac{1}{2} m u^{2} \approx m g R+\frac{1}{2} \frac{m v^{2} \sin ^{2} \alpha R^{2} \sin ^{2} \theta}{r^{2}} \approx m g R+K\left(\frac{\alpha^{2} R^{2} \sin ^{2} \theta}{r^{2}}\right)

$$



Equating with the initial energy gives us



$$

m g R \cos \theta_{0}+\frac{1}{2} k R^{2} \theta_{0}^{2}+K=m g R+K\left(\frac{\alpha^{2} R^{2} \sin ^{2} \theta}{r^{2}}\right)

$$



Before finishing the calculation, we now compute the required $k$ for the object to be at equilibrium (since our answer cannot contain $k$ ). Setting $k R \theta_{0}=m g \sin \theta_{0}$ gives us $k=$







$\frac{m g \sin \theta_{0}}{R \theta_{0}}$. Then,



$$

m g R \cos \theta_{0}+\frac{1}{2} m g R \theta_{0} \sin \theta_{0}+K=m g R+K\left(\frac{\alpha^{2} R^{2} \sin ^{2} \theta_{0}}{r^{2}}\right)

$$



Solving for $\alpha$ gives us



$$

\alpha=\frac{r}{R \sin \theta_{0}} \sqrt{1-\frac{m g R\left(1-\cos \theta_{0}\right)-(1 / 2) m g R \theta_{0} \sin \theta_{0}}{K}}

$$



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.)

Final answer: $\alpha=\frac{r}{R \sin \theta_{0}} \sqrt{1-\frac{m g R\left(1-\cos \theta_{0}\right)-(1 / 2) m g R \theta_{0} \sin \theta_{0}}{K}}$