AQA M2 2011 June — Question 8 10 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2011
SessionJune
Marks10
PaperDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle: complete revolution conditions
DifficultyStandard +0.3 This is a standard M2 vertical circle problem requiring application of energy conservation and circular motion equations. Part (a) is a routine 'show that' derivation for complete revolution conditions (finding minimum speed at top). Part (b) involves standard energy methods and Newton's second law in circular motion. While multi-step, it follows well-established textbook procedures with no novel insight required, making it slightly easier than average.
Spec6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration

8 A smooth wire is fixed in a vertical plane so that it forms a circle of radius \(a\) metres and centre \(O\). A bead, \(B\), of mass 0.3 kg , is threaded on the wire and is set in motion with a speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at the lowest point of its circular path, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{31ba38f7-38a8-4e4e-96a3-19e819fabfb0-6_364_378_466_845}
  1. Show that, if the bead is going to make complete revolutions around the wire, $$u > 2 \sqrt { a g }$$
  2. At time \(t\) seconds, the angle between \(O B\) and the horizontal is \(\theta\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{31ba38f7-38a8-4e4e-96a3-19e819fabfb0-6_330_328_1231_858} It is given that \(u = \sqrt { \frac { 9 } { 2 } a g }\).
    1. Find the reaction of the bead on the wire, giving your answer in terms of \(g\) and \(\theta\).
    2. Find \(\theta\) when this reaction is zero.

8 A smooth wire is fixed in a vertical plane so that it forms a circle of radius $a$ metres and centre $O$. A bead, $B$, of mass 0.3 kg , is threaded on the wire and is set in motion with a speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the lowest point of its circular path, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{31ba38f7-38a8-4e4e-96a3-19e819fabfb0-6_364_378_466_845}
\begin{enumerate}[label=(\alph*)]
\item Show that, if the bead is going to make complete revolutions around the wire,

$$u > 2 \sqrt { a g }$$
\item At time $t$ seconds, the angle between $O B$ and the horizontal is $\theta$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{31ba38f7-38a8-4e4e-96a3-19e819fabfb0-6_330_328_1231_858}

It is given that $u = \sqrt { \frac { 9 } { 2 } a g }$.
\begin{enumerate}[label=(\roman*)]
\item Find the reaction of the bead on the wire, giving your answer in terms of $g$ and $\theta$.
\item Find $\theta$ when this reaction is zero.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA M2 2011 Q8 [10]}}