1994 Paper 1 Q10

Year: 1994
Paper: 1
Question Number: 10

Course: UFM Mechanics
Section: Work, energy and Power 2

Difficulty: 1516.0 Banger: 1484.0

circular motion conical pendulum elastic string Hooke's law inequality angular velocity horizontal circle tension

Problem

One end \(A\) of a light elastic string of natural length \(l\) and modulus of elasticity \(\lambda\) is fixed and a particle of mass \(m\) is attached to the other end \(B\). The particle moves in a horizontal circle with centre on the vertical through \(A\) with angular velocity \(\omega.\) If \(\theta\) is the angle \(AB\) makes with the downward vertical, find an expression for \(\cos\theta\) in terms of \(m,g,l,\lambda\) and \(\omega.\) Show that the motion described is possible only if \[ \frac{g\lambda}{l(\lambda+mg)}<\omega^{2}<\frac{\lambda}{ml}. \]

No solution available for this problem.

Rating Information

Difficulty Rating: 1516.0

Difficulty Comparisons: 1

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source

One end $A$ of a light elastic string of natural length $l$ and
modulus of elasticity $\lambda$ is fixed and a particle of mass $m$
is attached to the other end $B$. The particle moves in a horizontal
circle with centre on the vertical through $A$ with angular velocity
$\omega.$ If $\theta$ is the angle $AB$ makes with the downward
vertical, find an expression for $\cos\theta$ in terms of $m,g,l,\lambda$
and $\omega.$

Show that the motion described is possible only if 
\[
\frac{g\lambda}{l(\lambda+mg)}<\omega^{2}<\frac{\lambda}{ml}.
\]

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