OCR M3 2013 January — Question 4 11 marks

Exam BoardOCR
ModuleM3 (Mechanics 3)
Year2013
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeParticle in circular tube or on wire
DifficultyChallenging +1.8 This M3 question requires energy conservation with geometric constraints (string over cylinder), finding normal force using circular motion dynamics, and solving a transcendental equation. It demands careful geometric analysis of the string configuration, multi-step energy calculations, and understanding of contact forces. The transcendental equation in part (ii) is sophisticated but the overall approach follows standard M3 techniques for connected particles over pulleys/cylinders.
Spec3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

4 A smooth cylinder of radius \(a \mathrm {~m}\) is fixed with its axis horizontal and \(O\) is the centre of a cross-section. Particle \(P\), of mass 0.4 kg , and particle \(Q\), of mass 0.6 kg , are connected by a light inextensible string of length \(\pi a \mathrm {~m}\). The string is held at rest with \(P\) and \(Q\) at opposite ends of the horizontal diameter of the crosssection through \(O\) (see Fig. 1). The string is released and \(Q\) begins to descend. When \(O P\) has rotated through \(\theta\) radians, with \(P\) remaining in contact with the cylinder, the speed of each particle is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see Fig. 2). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-3_365_433_520_424} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-3_396_643_484_1000} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Show that \(v ^ { 2 } = 3.92 a ( 3 \theta - 2 \sin \theta )\) and find an expression in terms of \(\theta\) for the normal force of the cylinder on \(P\) at this time.
  2. Given that \(P\) leaves the surface of the cylinder when \(\theta = \alpha\), show that \(\sin \alpha = k \alpha\) where \(k\) is a constant to be found.

Question 4:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
Conservation of energyM1, M1 Need 4 terms; allow sign & trig errors; Both KE or both PE correct
\(\frac{1}{2}(0.4)v^2 + \frac{1}{2}(0.6)v^2 + 0.4ga\sin\theta - 0.6ga\theta = 0\)A1 Completely correct
\(v^2 = 3.92a(3\theta - 2\sin\theta)\)M1, A1 Attempt to find \(v^2\) dep both earlier M1s AG; Allow with sign and trig errors; No errors
\(F = ma\) radially for \(P\)M1* Allow sign and trig errors
\(0.4g\sin\theta - R = \frac{0.4v^2}{a}\)A1
\(R = -4.704\theta + 7.056\sin\theta\)*M1, A1 [9] Manipulation attempted, leading to \(a\theta + b\sin\theta\); Allow sign and trig errors; \(2.352(-2\theta + 3\sin\theta)\)
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
Using \(R = 0\)M1 \(0 = -4.704\theta + 7.056\sin\theta\)
\((k =)\ \frac{2}{3}\)A1 [2] Must be from correct expression in (i)
# Question 4:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Conservation of energy | M1, M1 | Need 4 terms; allow sign & trig errors; Both KE or both PE correct |
| $\frac{1}{2}(0.4)v^2 + \frac{1}{2}(0.6)v^2 + 0.4ga\sin\theta - 0.6ga\theta = 0$ | A1 | Completely correct |
| $v^2 = 3.92a(3\theta - 2\sin\theta)$ | M1, A1 | Attempt to find $v^2$ dep both earlier M1s **AG**; Allow with sign and trig errors; No errors |
| $F = ma$ radially for $P$ | M1* | Allow sign and trig errors |
| $0.4g\sin\theta - R = \frac{0.4v^2}{a}$ | A1 | |
| $R = -4.704\theta + 7.056\sin\theta$ | *M1, A1 **[9]** | Manipulation attempted, leading to $a\theta + b\sin\theta$; Allow sign and trig errors; $2.352(-2\theta + 3\sin\theta)$ |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $R = 0$ | M1 | $0 = -4.704\theta + 7.056\sin\theta$ |
| $(k =)\ \frac{2}{3}$ | A1 **[2]** | Must be from correct expression in (i) |

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4 A smooth cylinder of radius $a \mathrm {~m}$ is fixed with its axis horizontal and $O$ is the centre of a cross-section. Particle $P$, of mass 0.4 kg , and particle $Q$, of mass 0.6 kg , are connected by a light inextensible string of length $\pi a \mathrm {~m}$. The string is held at rest with $P$ and $Q$ at opposite ends of the horizontal diameter of the crosssection through $O$ (see Fig. 1). The string is released and $Q$ begins to descend. When $O P$ has rotated through $\theta$ radians, with $P$ remaining in contact with the cylinder, the speed of each particle is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see Fig. 2).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-3_365_433_520_424}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{dfe477d4-eae6-40e1-b704-1a97485f4c7e-3_396_643_484_1000}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

(i) Show that $v ^ { 2 } = 3.92 a ( 3 \theta - 2 \sin \theta )$ and find an expression in terms of $\theta$ for the normal force of the cylinder on $P$ at this time.\\
(ii) Given that $P$ leaves the surface of the cylinder when $\theta = \alpha$, show that $\sin \alpha = k \alpha$ where $k$ is a constant to be found.

\hfill \mbox{\textit{OCR M3 2013 Q4 [11]}}