| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces (normal reaction and weight) and application of F=ma in both vertical and horizontal directions. The geometry is straightforward (finding the radius of the circular path and angle from the given vertical distance), and the algebraic manipulation to reach the given answer is routine. It's slightly easier than average because it's a 'show that' question with clear geometric setup and standard M3 techniques. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(R\sin 30° = mg\) | M1A1 | M1: Resolving vertically, 30° or \(\theta\); A1: Correct equation |
| \(R\cos 30° = m(r\cos 30°)\omega^2\) | M1A1A1 | M1: Equation of motion along radius; A1: LHS correct; A1: RHS correct (30° or \(\theta\), not \(r\) for radius) |
| \(\omega^2 = \frac{R}{mr} = \frac{g}{r\sin 30}\) | DM1 | Obtaining expression for \(\omega^2\) or \(v^2\) and length of path; dependent on both previous M marks |
| \(\omega = \sqrt{\frac{2g}{r}}\) | A1 | Correct expression for \(\omega\); must have numerical value for trig function |
| \(\text{Time} = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{r}{2g}} = \pi\sqrt{\frac{2r}{g}}\) | A1cso | Deducing the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(mg\cos 30 = m \times rad \times \omega^2\cos 60 = mr\cos 30\,\omega^2\cos 60\) | M2A1(LHS) A1(RHS), A1 | |
| Obtain \(\omega\) | M1A1 | |
| Correct time | A1 |
## Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $R\sin 30° = mg$ | M1A1 | M1: Resolving vertically, 30° or $\theta$; A1: Correct equation |
| $R\cos 30° = m(r\cos 30°)\omega^2$ | M1A1A1 | M1: Equation of motion along radius; A1: LHS correct; A1: RHS correct (30° or $\theta$, not $r$ for radius) |
| $\omega^2 = \frac{R}{mr} = \frac{g}{r\sin 30}$ | DM1 | Obtaining expression for $\omega^2$ or $v^2$ and length of path; dependent on both previous M marks |
| $\omega = \sqrt{\frac{2g}{r}}$ | A1 | Correct expression for $\omega$; must have numerical value for trig function |
| $\text{Time} = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{r}{2g}} = \pi\sqrt{\frac{2r}{g}}$ | A1cso | Deducing the given answer |
**ALT:** Resolve perpendicular to reaction:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $mg\cos 30 = m \times rad \times \omega^2\cos 60 = mr\cos 30\,\omega^2\cos 60$ | M2A1(LHS) A1(RHS), A1 | |
| Obtain $\omega$ | M1A1 | |
| Correct time | A1 | |
**Total: [8]**
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b7cfcf0a-8f54-4350-8e07-a3b51d94d0f2-02_406_537_264_715}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A hemispherical bowl, of internal radius $r$, is fixed with its circular rim upwards and horizontal. A particle $P$ of mass $m$ moves on the smooth inner surface of the bowl. The particle moves with constant angular speed in a horizontal circle. The centre of the circle is at a distance $\frac { 1 } { 2 } r$ vertically below the centre of the bowl, as shown in Figure 1.\\
The time taken by $P$ to complete one revolution of its circular path is $T$.\\
Show that $T = \pi \sqrt { \frac { 2 r } { g } }$.\\
\hfill \mbox{\textit{Edexcel M3 2015 Q1 [8]}}