| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2019 |
| Session | June |
| 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 circular motion problem requiring resolution of forces (normal reaction and weight) and application of F=mrω². The geometry is straightforward (given vertical distance allows finding angle), and the method is a routine textbook exercise for FM2 students. Slightly above average difficulty due to being Further Maths content, but mechanically straightforward with no novel insight required. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\uparrow R\cos\theta = mg\) | M1 | Resolve vertically. Must be dimensionally correct. |
| \(\leftrightarrow R\sin\theta = mr\omega^2\) | M1 | Resolve horizontally and form equation for circular motion. Must be dimensionally correct. |
| Correct pair of equations | A1 | Correct pair of equations for their unknowns (any \(r\)) |
| \(\tan\theta = \dfrac{r}{\frac{a}{4}}\) \(\left(\tan\theta = \sqrt{15}\right)\) | B1 | Correct trig ratio(s) seen or implied. Allow for \(r = \dfrac{\sqrt{15}}{4}a\) |
| Complete strategy to find \(\omega\) | M1 | Complete strategy to form and solve a set of equations with \(r \neq a\) to find \(\omega\). For solving their 2 equations – not dependent |
| \(\tan\theta = \dfrac{mr\omega^2}{mg} = \dfrac{4r}{a}\), \(\Rightarrow \omega^2 = \dfrac{4g}{a}\), \(\omega = 2\sqrt{\dfrac{g}{a}}\) | A1 | Eliminate additional variables to obtain \(\omega\). Accept equivalent exact forms. |
| (6) | If \(R\) does not act through the centre of the hemisphere then the maximum available is M1M1A0B0M0A0: 2/6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\uparrow R\cos\theta = mg\) | M1 | Resolve vertically. Must be dimensionally correct. |
| \(\leftrightarrow R\sin\theta = ma\sin\theta\,\omega^2\) | M1 | Resolve horizontally for circular motion. Must be dimensionally correct. |
| Correct pair of equations | A1 | |
| \(\cos\theta = \tfrac{1}{4}\) | B1 | |
| Complete strategy to find \(\omega\) | M1 | |
| \(\Rightarrow R = 4mg,\quad 4mg = ma\omega^2 \Rightarrow \omega = 2\sqrt{\dfrac{g}{a}}\) | A1 |
# Question 1 (Mechanics - Circular Motion on Hemisphere):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\uparrow R\cos\theta = mg$ | M1 | Resolve vertically. Must be dimensionally correct. |
| $\leftrightarrow R\sin\theta = mr\omega^2$ | M1 | Resolve horizontally and form equation for circular motion. Must be dimensionally correct. |
| Correct pair of equations | A1 | Correct pair of equations for their unknowns (any $r$) |
| $\tan\theta = \dfrac{r}{\frac{a}{4}}$ $\left(\tan\theta = \sqrt{15}\right)$ | B1 | Correct trig ratio(s) seen or implied. Allow for $r = \dfrac{\sqrt{15}}{4}a$ |
| Complete strategy to find $\omega$ | M1 | Complete strategy to form and solve a set of equations with $r \neq a$ to find $\omega$. For solving their 2 equations – not dependent |
| $\tan\theta = \dfrac{mr\omega^2}{mg} = \dfrac{4r}{a}$, $\Rightarrow \omega^2 = \dfrac{4g}{a}$, $\omega = 2\sqrt{\dfrac{g}{a}}$ | A1 | Eliminate additional variables to obtain $\omega$. Accept equivalent exact forms. |
| **(6)** | | If $R$ does not act through the centre of the hemisphere then the maximum available is M1M1A0B0M0A0: 2/6 |
**Alternative Method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\uparrow R\cos\theta = mg$ | M1 | Resolve vertically. Must be dimensionally correct. |
| $\leftrightarrow R\sin\theta = ma\sin\theta\,\omega^2$ | M1 | Resolve horizontally for circular motion. Must be dimensionally correct. |
| Correct pair of equations | A1 | |
| $\cos\theta = \tfrac{1}{4}$ | B1 | |
| Complete strategy to find $\omega$ | M1 | |
| $\Rightarrow R = 4mg,\quad 4mg = ma\omega^2 \Rightarrow \omega = 2\sqrt{\dfrac{g}{a}}$ | A1 | |1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b995178-a4be-4d5a-95f8-6c2978ff01b3-02_330_662_349_753}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A hemispherical shell of radius $a$ is fixed with its rim uppermost and horizontal. A small bead, $B$, is moving with constant angular speed, $\omega$, in a horizontal circle on the smooth inner surface of the shell. The centre of the path of $B$ is at a distance $\frac { 1 } { 4 } a$ vertically below the level of the rim of the hemisphere, as shown in Figure 1.
Find the magnitude of $\omega$, giving your answer in terms of $a$ and $g$.
\hfill \mbox{\textit{Edexcel FM2 2019 Q1 [6]}}