| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.8 This is a multi-part M3 circular motion problem requiring resolution of forces in two directions, geometric relationships, and algebraic manipulation to derive a given result, followed by an inequality proof and tension calculation. While systematic, it demands careful setup of the force diagram, understanding of conical pendulum geometry with two string sections, and non-trivial algebra to reach the specified forms—significantly above average difficulty but standard for M3. |
| Spec | 3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Dist of centre of mass from \(O = r\tan 30 = \frac{r}{\sqrt{3}}\) | M1A1 | Must use tan to find distance of c of m from \(O\) |
| Ratio of masses: \(M\), \(kM\), \((1+k)M\) (ratio \(1\), \(k\), \(1+k\)); Dist from \(O\): \(-\frac{1}{4}h\), \(\frac{kh}{4}\), \(\frac{r}{\sqrt{3}}\) | — | — |
| \(\text{M}(O): -\frac{1}{4}h + \frac{k^2h}{4} = (1+k)\frac{r}{\sqrt{3}}\) | M1A1A1ft | M1: forming moments equation; A1: LHS correct; A1ft: RHS correct for their distance |
| \(\frac{h}{4}(k^2-1) = (k+1)\frac{r}{\sqrt{3}}\) | — | — |
| \(k = \frac{4r}{h\sqrt{3}}+1\) ✳ | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(kMg\!\left(\frac{1}{4}kh\cos30 - r\sin30\right)\), \(Mg\!\left(\frac{1}{4}h\cos30 + r\sin30\right)\) | M1A1, M1A1 | M1: attempting LHS with two terms inc resolution; M1: attempting RHS with two terms inc resolution |
| \(h\cos30(k^2-1) = 4r\sin30(k+1)\) | A1ft | Collecting terms and cancelling \(Mg\) |
| \((k-1) = \frac{4r}{h}\tan30\) | — | — |
| \(k = \frac{4r}{h\sqrt{3}}+1\) ✳ | A1 | Completing to the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{M}(O): -\frac{1}{4}h + \frac{k^2h}{4} = (1+k)\bar{x}\) | M1 A1 | Attempting equation to find \(\bar{x}\) in terms of \(h\) and \(k\) |
| \(\bar{x} = \frac{h(k-1)}{4}\) | A1 | Correct expression for \(\bar{x}\) |
| Then suspend: \(\frac{\bar{x}}{r} = \tan30\) | M1 | Using \(\frac{\bar{x}}{r} = \tan30\) (LHS either way up) |
| \(\frac{h(k-1)}{4r} = \frac{1}{\sqrt{3}}\) (or \(\tan30\)) | A1ft | Substitute their \(\bar{x}\); LHS must be correct way up |
| \(k = \frac{4r}{h\sqrt{3}}+1\) ✳ | A1 | Obtaining the given answer |
## Question 5:
**Main Method:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Dist of centre of mass from $O = r\tan 30 = \frac{r}{\sqrt{3}}$ | M1A1 | Must use tan to find distance of c of m from $O$ |
| Ratio of masses: $M$, $kM$, $(1+k)M$ (ratio $1$, $k$, $1+k$); Dist from $O$: $-\frac{1}{4}h$, $\frac{kh}{4}$, $\frac{r}{\sqrt{3}}$ | — | — |
| $\text{M}(O): -\frac{1}{4}h + \frac{k^2h}{4} = (1+k)\frac{r}{\sqrt{3}}$ | M1A1A1ft | M1: forming moments equation; A1: LHS correct; A1ft: RHS correct for their distance |
| $\frac{h}{4}(k^2-1) = (k+1)\frac{r}{\sqrt{3}}$ | — | — |
| $k = \frac{4r}{h\sqrt{3}}+1$ ✳ | A1 | [6] | Obtaining the given answer |
**Alt 1 – Taking moments about A:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $kMg\!\left(\frac{1}{4}kh\cos30 - r\sin30\right)$, $Mg\!\left(\frac{1}{4}h\cos30 + r\sin30\right)$ | M1A1, M1A1 | M1: attempting LHS with two terms inc resolution; M1: attempting RHS with two terms inc resolution |
| $h\cos30(k^2-1) = 4r\sin30(k+1)$ | A1ft | Collecting terms and cancelling $Mg$ |
| $(k-1) = \frac{4r}{h}\tan30$ | — | — |
| $k = \frac{4r}{h\sqrt{3}}+1$ ✳ | A1 | Completing to the given answer |
**Alt 2 – Find $\bar{x}$ first:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{M}(O): -\frac{1}{4}h + \frac{k^2h}{4} = (1+k)\bar{x}$ | M1 A1 | Attempting equation to find $\bar{x}$ in terms of $h$ and $k$ |
| $\bar{x} = \frac{h(k-1)}{4}$ | A1 | Correct expression for $\bar{x}$ |
| Then suspend: $\frac{\bar{x}}{r} = \tan30$ | M1 | Using $\frac{\bar{x}}{r} = \tan30$ (LHS either way up) |
| $\frac{h(k-1)}{4r} = \frac{1}{\sqrt{3}}$ (or $\tan30$) | A1ft | Substitute their $\bar{x}$; LHS must be correct way up |
| $k = \frac{4r}{h\sqrt{3}}+1$ ✳ | A1 | Obtaining the given answer |
---5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{ab85ec29-b1fc-45a9-9343-09feb33ab6c5-008_531_691_299_657}
\end{center}
\end{figure}
One end of a light inextensible string is attached to a fixed point $A$. The other end of the string is attached to a fixed point $B$, vertically below $A$, where $A B = h$. A small smooth ring $R$ of mass $m$ is threaded on the string. The ring $R$ moves in a horizontal circle with centre $B$, as shown in Figure 3. The upper section of the string makes a constant angle $\theta$ with the downward vertical and $R$ moves with constant angular speed $\omega$. The ring is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Show that $\omega ^ { 2 } = \frac { g } { h } \left( \frac { 1 + \sin \theta } { \sin \theta } \right)$.
\item Deduce that $\omega > \sqrt { \frac { 2 g } { h } }$.
Given that $\omega = \sqrt { \frac { 3 g } { h } }$,
\item find, in terms of $m$ and $g$, the tension in the string.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5}}