Year: 2006
Paper: 3
Question Number: 10
Course: zNo longer examinable
Section: Moments of inertia
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1469.6
Banger Comparisons: 2
A disc rotates freely in a horizontal plane about a vertical axis
through its centre. The moment of inertia of the disc about this axis
is $mk^2$ (where $k>0$). Along one diameter is a smooth narrow
groove in which a particle of
mass $m$ slides freely. At time $t=0\,$, the disc is rotating with angular
speed $\Omega$, and the particle is a distance $a$ from the axis and
is moving with speed~$V$ along the groove, towards
the axis, where
$k^2V^2 = \Omega^2a^2(k^2+a^2)\,$.
Show that, at a later time $t$, while the particle is still moving towards
the axis, the angular speed $\omega$ of the disc and the distance $r$ of
the particle from the axis are related by
\[
\omega = \frac{\Omega(k^2+a^2)}{k^2+r^2}
\text{ \ \ and \ \ }
\left(\frac{\d r}{\d t}\right)^{\!2} = \frac{\Omega^2r^2(k^2+a^2)^2}{k^2(k^2+r^2)}\;.
\]
Deduce that
\[
k\frac{\d r}{\d\theta} = -r(k^2+r^2)^{\frac12}\,,
\]
where $\theta $ is the angle through which the disc has turned by time $t$.
By making the substitution $u=k/r$, or otherwise, show that
$r\sinh (\theta+\alpha) = k$, where $\sinh \alpha = k/a\,$. Deduce that
the particle never reaches the axis.