| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem with a smooth ring on a string. Part (a) requires resolving forces and applying F=mrω², which is routine for this module. Part (b) is a straightforward inequality using calculus or AM-GM. Part (c) substitutes a given value back into the tension equation. While it requires careful geometry and force resolution, it follows a well-established template for conical pendulum-type problems, making it slightly easier than average. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\uparrow T \cos \theta = mg\) | B1 | |
| \(\leftrightarrow T + T \sin \theta = mr\omega^2\) (3 terms) | M1 A1 | |
| \(r = h \tan \theta\) | B1 | |
| \(\frac{mg(1 + \sin \theta)}{\cos \theta} = \frac{m\omega^2 h \sin \theta}{\cos \theta}\) (eliminate r) | ↓ M1 | |
| \(\omega^2 = \frac{g}{h}\left(\frac{1 + \sin \theta}{\sin \theta}\right)\) (*) (solve for \(\omega^2\)) | ↓ M1 A1 (7) | |
| (b) \(\omega^2 = \frac{g}{h}\left(\frac{1}{\sin \theta} + 1\right) > \frac{2g}{h}\) (\(\sin \theta < 1\)) \(\Rightarrow \omega > \sqrt{\frac{2g}{h}}\) (*) | M1 A1 (2) | |
| (c) \(\frac{3g}{h} = \frac{g}{h}\left(\frac{1 + \sin \theta}{\sin \theta}\right) \Rightarrow \sin \theta = \frac{1}{2}\) | M1 A1 | |
| \(T \cos \theta = mg \Rightarrow T = \frac{2\sqrt{3}}{3}mg\) or 1.15mg (awrt) | ↓ M1 A1 (4) | |
| (a) Allow first B1 M1 A1 if assume different tensions (so next M1 is effectively for eliminating r and T. | Guidance note | |
| (b) M1 requires a valid attempt to derive an inequality for \(\omega\). (Hence putting \(\sin \theta = 1\) immediately into expression of \(\omega^2\) [assuming this is the critical value] is M0.) | Guidance note |
(a) $\uparrow T \cos \theta = mg$ | B1 |
$\leftrightarrow T + T \sin \theta = mr\omega^2$ (3 terms) | M1 A1 |
$r = h \tan \theta$ | B1 |
$\frac{mg(1 + \sin \theta)}{\cos \theta} = \frac{m\omega^2 h \sin \theta}{\cos \theta}$ (eliminate r) | ↓ M1 |
$\omega^2 = \frac{g}{h}\left(\frac{1 + \sin \theta}{\sin \theta}\right)$ (*) (solve for $\omega^2$) | ↓ M1 A1 (7) |
(b) $\omega^2 = \frac{g}{h}\left(\frac{1}{\sin \theta} + 1\right) > \frac{2g}{h}$ ($\sin \theta < 1$) $\Rightarrow \omega > \sqrt{\frac{2g}{h}}$ (*) | M1 A1 (2) |
(c) $\frac{3g}{h} = \frac{g}{h}\left(\frac{1 + \sin \theta}{\sin \theta}\right) \Rightarrow \sin \theta = \frac{1}{2}$ | M1 A1 |
$T \cos \theta = mg \Rightarrow T = \frac{2\sqrt{3}}{3}mg$ or 1.15mg (awrt) | ↓ M1 A1 (4) |
(a) Allow first B1 M1 A1 if assume different tensions (so next M1 is effectively for eliminating r **and** T. | | Guidance note
(b) M1 requires a **valid** attempt to derive an inequality for $\omega$. (Hence putting $\sin \theta = 1$ immediately into expression of $\omega^2$ [assuming this is the critical value] is M0.) | | Guidance note
---5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{25b3ece7-69ed-4ec4-a6c7-4cd83ec2cc5e-07_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 2007 Q5 [13]}}