Year: 1992
Paper: 3
Question Number: 14
Course: UFM Mechanics
Section: Circular Motion 2
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
$\,$
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A horizontal circular disc of radius $a$ and centre $O$ lies on
a horizontal table and is fixed to it so that it cannot rotate. A
light inextensible string of negligible thickness is wrapped round
the disc and attached at its free end to a particle $P$ of mass $m$.
When the string is all in contact with the disc, $P$ is at $A$.
The string is unwound so that the part not in contact with the disc
is taut and parallel to $OA$. $P$ is then at $B$. The particle
is projected along the table from $B$ with speed $V$ perpendicular
to and away from $OA$. In the general position, the string is tangential
to the disc at $Q$ and $\angle AOQ=\theta.$ Show that, in the general
position, the $x$-coordinate of $P$ with respect to the axes shown
in the figure is $a\cos\theta+a\theta\sin\theta,$ and find $y$-coordinate
of $P$. Hence, or otherwise, show that the acceleration of $P$ has
components $a\theta\dot{\theta}^{2}$ and $a\dot{\theta}^{2}+a\theta\ddot{\theta}$
along and perpendicular to $PQ,$ respectively.
The friction force between $P$ and the table is $2\lambda mv^{2}/a,$
where $v$ is the speed of $P$ and $\lambda$ is a constant. Show
that
\[
\frac{\ddot{\theta}}{\dot{\theta}}=-\left(\frac{1}{\theta}+2\lambda\theta\right)\dot{\theta}
\]
and find $\dot{\theta}$ in terms of $\theta,\lambda$ and $a$. Find
also the tension in the string when $\theta=\pi.$