| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on cone surface – no string (normal reaction only) |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics conical pendulum problem requiring resolution of forces (weight, normal reaction, friction) in two perpendicular directions, application of F=mrω² for circular motion, and use of limiting friction F=μR. While it involves multiple steps and careful geometry (finding the cone's semi-vertical angle from given dimensions), it follows a well-established template taught in FM2 with no novel insight required beyond systematic application of standard techniques. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete overall strategy | M1 | Form equation in \(\omega\) only |
| Resolve vertically: \(mg + F\cos\theta = R\sin\theta\) | M1, A1 | Needs all 3 terms; condone sign errors and sin/cos confusion |
| Horizontal equation of motion: \(mr\omega^2 = R\cos\theta + F\sin\theta\) | M1, A1 | Needs all 3 terms; condone sign errors and sin/cos confusion |
| Use of limiting friction since maximum \(\omega\) | M1 | Seen or implied |
| Substitute for trig ratios: \(\frac{3a\omega^2}{2g} = \frac{9}{2}\) | M1 | Substitute to achieve equation in \(a\), \(\omega\) and \(g\) only |
| Maximum \(\omega = \sqrt{\frac{3g}{a}}\) | A1 | Or equivalent exact form |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete overall strategy | M1 | Form equation in $\omega$ only |
| Resolve vertically: $mg + F\cos\theta = R\sin\theta$ | M1, A1 | Needs all 3 terms; condone sign errors and sin/cos confusion |
| Horizontal equation of motion: $mr\omega^2 = R\cos\theta + F\sin\theta$ | M1, A1 | Needs all 3 terms; condone sign errors and sin/cos confusion |
| Use of limiting friction since maximum $\omega$ | M1 | Seen or implied |
| Substitute for trig ratios: $\frac{3a\omega^2}{2g} = \frac{9}{2}$ | M1 | Substitute to achieve equation in $a$, $\omega$ and $g$ only |
| Maximum $\omega = \sqrt{\frac{3g}{a}}$ | A1 | Or equivalent exact form |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f06704e5-454c-41c1-9577-b1210f60480d-04_655_643_207_639}
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\caption{Figure 1}
\end{center}
\end{figure}
A hollow right circular cone, of base diameter $4 a$ and height $4 a$ is fixed with its axis vertical and vertex $V$ downwards, as shown in Figure 1.
A particle of mass $m$ moves in a horizontal circle with centre $C$ on the rough inner surface of the cone with constant angular speed $\omega$.
The height of $C$ above $V$ is $3 a$.\\
The coefficient of friction between the particle and the inner surface of the cone is $\frac { 1 } { 4 }$. Find, in terms of $a$ and $g$, the greatest possible value of $\omega$.
\hfill \mbox{\textit{Edexcel FM2 Q2 [8]}}