| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces (normal reaction and weight) and application of F=mrω². The geometry is straightforward (30-60-90 triangle gives angle), and the method is routine for this topic. Slightly easier than average due to being a 'show that' question with clear target and standard technique. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R\sin 60° = mg\) | M1A1 | Resolve vertically, allow with \(60°\) or \(\theta\); correct equation with \(60°\) or \(\theta\) |
| \(R\cos 60° = mr\omega^2\) | M1A1 | Equation of motion along radius, acceleration in either form, \(60°\) or \(\theta\); correct equation with \(60°\) or \(\theta\), acceleration to be \(r\omega^2\) |
| \(\tan 60° = \dfrac{g}{r\omega^2}\) | ddM1 | Eliminate \(R\), substitute numerical value for \(\theta\) and solve to \(\omega = \ldots\) or \(\omega^2 = \ldots\); depends on both previous M marks |
| \(\omega = \sqrt{\dfrac{g}{r\sqrt{3}}} = \sqrt{\dfrac{g\sqrt{3}}{3r}}\) | A1cso | Complete to the given answer |
**Question 1:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R\sin 60° = mg$ | M1A1 | Resolve vertically, allow with $60°$ or $\theta$; correct equation with $60°$ or $\theta$ |
| $R\cos 60° = mr\omega^2$ | M1A1 | Equation of motion along radius, acceleration in either form, $60°$ or $\theta$; correct equation with $60°$ or $\theta$, acceleration to be $r\omega^2$ |
| $\tan 60° = \dfrac{g}{r\omega^2}$ | ddM1 | Eliminate $R$, substitute numerical value for $\theta$ and solve to $\omega = \ldots$ or $\omega^2 = \ldots$; depends on both previous M marks |
| $\omega = \sqrt{\dfrac{g}{r\sqrt{3}}} = \sqrt{\dfrac{g\sqrt{3}}{3r}}$ | A1cso | Complete to the given answer |1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ffe0bc72-3136-48d9-9d5b-4a364d134070-02_503_524_121_712}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A hemispherical bowl of internal radius $2 r$ is fixed with its circular rim horizontal. A particle $P$ is moving in a horizontal circle of radius $r$ on the smooth inner surface of the bowl, as shown in Figure 1. Particle $P$ is moving with constant angular speed $\omega$.
Show that $\omega = \sqrt { \frac { g \sqrt { 3 } } { 3 r } }$\\
\hfill \mbox{\textit{Edexcel M3 2016 Q1 [6]}}