| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.8 This is a challenging M3 circular motion problem requiring 3D spatial reasoning to resolve forces in a rotating system with constraints. Students must handle the geometry of the rotating triangle, apply Newton's second law in both vertical and horizontal directions with two different tensions, and work with the constraint that the ring moves on a pole. The geometry setup (finding angles from the 6l/5 constraint) and coordinating multiple force equations is significantly harder than standard conical pendulum questions. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\cos\alpha = \frac{3}{5}\), where angle \(ARP = \alpha\) | B1 | May be seen in (b) |
| For \(R\), \((\uparrow)\) \(T_2\cos\alpha = mg\) | M1 | Resolving vertically for ring \(R\), with usual rules |
| \(T_2 = \frac{5mg}{3}\) | A1* (3) | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| For \(P\), \((\uparrow)\) \(T_1\cos\alpha - \frac{5mg}{3}\cos\alpha = mg\) or \(T_1\cos\alpha = 2mg\) | M1A1 | M1 for resolving vertically for \(P\) with usual rules |
| Equation of motion: \(T_1\sin\alpha + \frac{5mg}{3}\sin\alpha = m(l\sin\alpha)\omega^2\) | M1A2,1,0 | M1 for horizontal equation of motion for \(P\); A2 for correct equation; A1A0 for equation with at most one error |
| \(\omega = \sqrt{\frac{5g}{l}}\) | A1 | cao |
| Time \(= \frac{2\pi}{\omega}\) | M1 | Correct method |
| \(= 2\pi\sqrt{\frac{l}{5g}}\) oe | A1 (8) | cao |
# Question 5:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\cos\alpha = \frac{3}{5}$, where angle $ARP = \alpha$ | B1 | May be seen in (b) |
| For $R$, $(\uparrow)$ $T_2\cos\alpha = mg$ | M1 | Resolving vertically for ring $R$, with usual rules |
| $T_2 = \frac{5mg}{3}$ | A1* (3) | Given answer correctly obtained |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| For $P$, $(\uparrow)$ $T_1\cos\alpha - \frac{5mg}{3}\cos\alpha = mg$ or $T_1\cos\alpha = 2mg$ | M1A1 | M1 for resolving vertically for $P$ with usual rules |
| Equation of motion: $T_1\sin\alpha + \frac{5mg}{3}\sin\alpha = m(l\sin\alpha)\omega^2$ | M1A2,1,0 | M1 for horizontal equation of motion for $P$; A2 for correct equation; A1A0 for equation with at most one error |
| $\omega = \sqrt{\frac{5g}{l}}$ | A1 | cao |
| Time $= \frac{2\pi}{\omega}$ | M1 | Correct method |
| $= 2\pi\sqrt{\frac{l}{5g}}$ oe | A1 (8) | cao |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9777abb8-a564-40d5-8d96-d5649913737b-16_730_634_246_657}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A small smooth ring $R$ of mass $m$ is threaded on to a thin smooth fixed vertical pole. One end of a light inextensible string of length $2 l$ is attached to a point $A$ on the pole. The other end of the string is attached to $R$. A particle $P$ of mass $m$ is attached to the midpoint of the string. The particle $P$ moves with constant angular speed in a horizontal circle, with both halves of the string taut, and $A R = \frac { 6 l } { 5 }$, as shown in Figure 2.
It may be assumed that in this motion the string does not wrap itself around the pole and that at any instant, the triangle $A P R$ lies in a vertical plane.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the lower half of the string is $\frac { 5 m g } { 3 }$
\item Find, in terms of $l$ and $g$, the time for $P$ to complete one revolution.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q5 [11]}}