1991 Paper 2 Q13

Year: 1991
Paper: 2
Question Number: 13

Course: zNo longer examinable
Section: Moments of inertia

Difficulty: 1600.0 Banger: 1500.0

moments of inertia non-uniform rod friction sliding condition angular motion pivot energy conservation rotational dynamics

Problem

A non-uniform rod \(AB\) of mass \(m\) is pivoted at one end \(A\) so that it can swing freely in a vertical plane. Its centre of mass is a distance \(d\) from \(A\) and its moment of inertia about any axis perpendicular to the rod through \(A\) is \(mk^{2}.\) A small ring of mass \(\alpha m\) is free to slide along the rod and the coefficient of friction between the ring and rod is \(\mu.\) The rod is initially held in a horizontal position with the ring a distance \(x\) from \(A\). If \(k^{2} > xd\), show that when the rod is released, the ring will start to slide when the rod makes an angle \(\theta\) with the downward vertical, where \[ \mu\tan\theta=\frac{3\alpha x^{2}+k^{2}+2xd}{k^{2}-xd}. \] Explain what will happen if (i) \(k^{2}=xd\) and (ii) \(k^{2} < xd\).

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

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Problem source

A non-uniform rod $AB$ of mass $m$ is pivoted at one end $A$ so
that it can swing freely in a vertical plane. Its centre of mass is
a distance $d$ from $A$ and its moment of inertia about any axis
perpendicular to the rod through $A$ is $mk^{2}.$ A small ring of
mass $\alpha m$ is free to slide along the rod and the coefficient
of friction between the ring and rod is $\mu.$ The rod is initially
held in a horizontal position with the ring a distance $x$ from $A$.
If $k^{2} > xd$, show that when the rod is released, the ring will
start to slide when the rod makes an angle $\theta$ with the downward
vertical, where 
\[
\mu\tan\theta=\frac{3\alpha x^{2}+k^{2}+2xd}{k^{2}-xd}.
\]
Explain what will happen if (i) $k^{2}=xd$ and (ii) $k^{2} < xd$.

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