Year: 2011
Paper: 3
Question Number: 11
Course: UFM Mechanics
Section: Circular Motion 2
No solution available for this problem.
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
A thin uniform circular disc of radius $a$ and mass $m$ is held
in equilibrium in a horizontal plane a distance $b$ below a horizontal
ceiling, where $b>2a$. It is held in this way by $n$ light
inextensible vertical strings, each of length $b$; one end of
each string is attached to the edge of the disc and the other end is
attached to a point on the ceiling. The strings are equally
spaced around the edge of the disc.
One of the strings is attached to the point $P$ on the disc
which has coordinates
$(a,0,-b)$ with respect to
cartesian axes with origin on the ceiling directly above the
centre of the disc.
The disc is then rotated through an angle $\theta$ (where $\theta<\pi$)
about its vertical axis of symmetry and held at rest by a couple
acting in the plane of the disc.
Show that the string attached to~$P$ now makes an angle $\phi$
with the vertical,
where
\[
b\sin\phi = 2a \sin\tfrac12 \theta\,.
\]
Show further that the magnitude of the couple is
\[
\frac {mga^2\sin\theta}{\sqrt{b^2-4a^2\sin^2 \frac12\theta \ } \ }\,.
\]
The disc is now released from rest. Show that
its angular speed, $\omega$, when the strings
are vertical is given by
\[
\frac{a^2\omega^2}{4g} = b-\sqrt{b^2 - 4a^2\sin^2 \tfrac12\theta \;}\,.
\]
The least popular question on the paper, attempted by about 4%, but with similar mean score to question 2 (and several others). Mostly, they did pretty well in finding the couple, and using the initial trigonometric relation and its consequences to do so. At that point they tended not to know how to proceed to the last part, though there were some very good and simple solutions from considering energy.