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.


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 $\theta_{0}$.

Solution:
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.



<img_4348>



Here, $F_{c}=\frac{m v^{2}}{R \sin \theta_{0}}$ 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,



$$

F_{r}+F_{c}^{\perp}=m g \cos \theta_{0}

$$



The perpendicular part of the centrifugal force is $F_{c} \sin \theta_{0}$. so we get



$$

F_{r}=m g \cos \theta_{0}-\frac{m v^{2}}{R}

$$



Incidentally, there's a simple way to understand why the second term has to be exactly $m v^{2} / R$. 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 $m v^{2} / R$ inward.



Note: The phrase "total radial force" could also validly be interpreted as the net radial force. Thus, we accepted both $m g \cos \theta_{0}-m v^{2} / R$ and $-m v^{2} / R$.

Final answer: $F_{r}=m g \cos \theta_{0}-\frac{m v^{2}}{R}$