ii. What is the total radial force (i. · OlympiadBench #1405

Physics · mechanics

Problem

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}

. Given constants: ## 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

-k \ell

) is fixed to the top of the hemisphere. Its other end is attached to a point mass of mass

m

. <image_1> 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}

. <image_2> 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. <image_3> 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.

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