Year: 1995
Paper: 3
Question Number: 9
Course: zNo longer examinable
Section: Moments of inertia
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1470.3
Banger Comparisons: 2
A thin circular disc of mass $m$, radius $r$ and with its centre
of mass at its centre $C$ can rotate freely in a vertical plane about
a fixed horizontal axis through a point $O$ of its circumference.
A particle $P$, also of mass $m,$ is attached to the circumference
of the disc so that the angle $OCP$ is $2\alpha,$ where $\alpha\leqslant\pi/2$.
\begin{questionparts}
\item In the position of stable equilibrium $OC$ makes an angle $\beta$
with the vertical. Prove that
\[
\tan\beta=\frac{\sin2\alpha}{2-\cos2\alpha}.
\]
\item The density of the disc at a point distant $x$ from $C$ is $\rho x/r.$
Show that its moment of inertia about the horizontal axis through
$O$ is $8mr^{2}/5$.
\item The mid-point of $CP$ is $Q$. The disc is held at rest with $OQ$
horizontal and $C$ lower than $P$ and it is then released. Show
that the speed $v$ with which $C$ is moving when $P$ passes vertically
below $O$ is given by
\[
v^{2}=\frac{15gr\sin\alpha}{2(2+5\sin^{2}\alpha)}.
\]
Find the maximum value of $v^{2}$ as $\alpha$ is varied.
\end{questionparts}