Edexcel M3 2014 June — Question 2 13 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2014
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on cone surface – with string attached to vertex or fixed point
DifficultyStandard +0.8 This is a challenging M3 circular motion problem requiring careful 3D geometry with the cone constraint, resolution of forces in multiple directions, and an inequality proof involving the normal reaction condition. The 7+6 mark allocation and need to handle the cone contact condition (R≥0) to derive a time inequality elevates this above standard circular motion exercises.
Spec6.05c Horizontal circles: conical pendulum, banked tracks

\includegraphics{figure_1} A cone of semi-vertical angle \(60°\) is fixed with its axis vertical and vertex upwards. A particle of mass \(m\) is attached to one end of a light inextensible string of length \(l\). The other end of the string is attached to a fixed point vertically above the vertex of the cone. The particle moves in a horizontal circle on the smooth outer surface of the cone with constant angular speed \(\omega\), with the string making a constant angle \(60°\) with the horizontal, as shown in Figure 1.
  1. Find the tension in the string, in terms of \(m\), \(l\), \(\omega\) and \(g\). [7]
The particle remains on the surface of the cone.
  1. Show that the time for the particle to make one complete revolution is greater than $$2\pi\sqrt{\frac{l\sqrt{3}}{2g}}$$ [6]

2(a)
AnswerMarks Guidance
\(T \sin 60° + R \sin 60° = mg\)M1 A1
\(T \cos 60° - R \cos 60° = ml \cos 60°\omega^2\)M1 A1 A1
\(T = \frac{1}{3}m(l\omega^2 + \frac{2}{\sqrt{3}}g)\)DM1 A1 Solve for \(T\)
(7)
2(b)
AnswerMarks Guidance
\(R = \frac{1}{3}m(\frac{2}{\sqrt{3}}g - l\omega^2)\)M1 A1
\(\frac{1}{3}m(\frac{2}{\sqrt{3}}g - l\omega^2) > 0\)DM1
\(\omega < \sqrt{\frac{2g}{l\sqrt{3}}}\)A1
\(t > 2\pi\sqrt{\frac{l\sqrt{3}}{2g}}\) **DM1 A1 Obtaining an inequality for \(t\)
(6)
13
## 2(a)
$T \sin 60° + R \sin 60° = mg$ | M1 A1 |
$T \cos 60° - R \cos 60° = ml \cos 60°\omega^2$ | M1 A1 A1 |
$T = \frac{1}{3}m(l\omega^2 + \frac{2}{\sqrt{3}}g)$ | DM1 A1 | Solve for $T$
| (7) |

## 2(b)
$R = \frac{1}{3}m(\frac{2}{\sqrt{3}}g - l\omega^2)$ | M1 A1 |
$\frac{1}{3}m(\frac{2}{\sqrt{3}}g - l\omega^2) > 0$ | DM1 |
$\omega < \sqrt{\frac{2g}{l\sqrt{3}}}$ | A1 |
$t > 2\pi\sqrt{\frac{l\sqrt{3}}{2g}}$ ** | DM1 A1 | Obtaining an inequality for $t$
| (6) |
| 13 |

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\includegraphics{figure_1}

A cone of semi-vertical angle $60°$ is fixed with its axis vertical and vertex upwards. A particle of mass $m$ is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point vertically above the vertex of the cone. The particle moves in a horizontal circle on the smooth outer surface of the cone with constant angular speed $\omega$, with the string making a constant angle $60°$ with the horizontal, as shown in Figure 1.

\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string, in terms of $m$, $l$, $\omega$ and $g$. [7]
\end{enumerate}

The particle remains on the surface of the cone.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the time for the particle to make one complete revolution is greater than
$$2\pi\sqrt{\frac{l\sqrt{3}}{2g}}$$ [6]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2014 Q2 [13]}}