Year: 2012
Paper: 3
Question Number: 10
Course: UFM Mechanics
Section: Circular Motion 2
No solution available for this problem.
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A small ring of mass $m$
is free to slide without friction on a hoop of radius $a$.
The hoop is fixed in a vertical plane.
The ring is connected by a light elastic string of natural length $a$
to the highest point of the hoop.
The ring is initially at rest at the lowest
point of the hoop and is then slightly displaced.
In the subsequent motion the
angle of the string to the downward vertical is $\phi$.
Given that the ring first comes to rest just as the string becomes slack,
find an expression for the modulus of elasticity of the string in terms
of $m$ and $g$.
Show that, throughout the motion, the magnitude $R$ of the reaction
between the ring and the hoop is given by
\[
R = ( 12\cos^2\phi -15\cos\phi +5) mg
\]
and
that $R$ is non-zero throughout the motion.
This was the most popular of the non‐Pure questions, being attempted by about a sixth of candidates, and with more success than all but three questions on the paper. Many attempts failed to include a clear, legible, accurate diagram, and so an unclear mess of variables invariably failed to lead to satisfactory conclusions. On the other hand, the general standard of mechanics was above average, and the initial energy equation was usually correct. Many candidates came to grief with the general energy equation, confusing signs. A good number of strong candidates ploughed straight through correctly, and all who did so, then gained credit at the end for using the discriminant to demonstrate that R is non‐zero.