| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation, critical speed analysis, and tension equations at extreme points. Part (a) is routine energy conservation (4 marks), part (b) applies the standard condition for complete circles (4 marks), and part (c) requires setting up tension equations at top and bottom then solving simultaneously (9 marks). While multi-step with 17 total marks, it follows predictable M3 patterns without requiring novel insight—slightly above average due to the algebraic manipulation in part (c) and the need to coordinate multiple standard techniques. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks |
|---|---|
| \(mgl(\cos\alpha - \cos\theta) = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\) | M1A1=A1 |
| \(v^2 = u^2 + 2gl(\cos\alpha - \cos\theta)\) * | A1 |
| Answer | Marks |
|---|---|
| \(\cos\alpha = \frac{3}{5}\): \(v^2 = 2gl(\frac{3}{5} - \cos\theta) + u^2\) | M1A1 |
| At top \(\theta = 360°\): \(v^2 = 2gl(\frac{3}{5} - 1) + u^2\) | M1 |
| \(v^2 > 0: -2gl \times \frac{2}{5} + u^2 > 0\) | |
| \(u^2 > \frac{4gl}{5}\) | A1 |
| \(u > 2\sqrt{\frac{gl}{5}}\) * |
| Answer | Marks |
|---|---|
| Equation of motion along radius at lowest point: \(T_1 - mg = \frac{mv^2}{l}\) | M1A1 |
| \(\theta = 180°\): \(v^2 = 2gl(\frac{3}{5} + 1) + u^2\) | M1 |
| \(v^2 = \frac{16gl}{5} + u^2\) | |
| \(T_1 = \frac{m}{l}(\frac{16gl}{5} + u^2) + mg\) | A1 |
| \(= \frac{21mg}{5} + \frac{mu^2}{l}\) | |
| At highest point: \(T_2 + mg = \frac{mv^2}{l}\) | M1 |
| \(\theta = 360°\): \(T_2 = 2mg(-\frac{2}{5}) + \frac{mu^2}{l} - mg\) | M1 |
| \(T_2 = \frac{mu^2}{l} - \frac{9mg}{5}\) | A1 |
| \(T_1 = 5T_2\) | M1 |
| \(\frac{21mg}{5} + \frac{mu^2}{l} = 5(\frac{mu^2}{l} - \frac{9mg}{5})\) | |
| \(\frac{66mg}{5} = \frac{4mu^2}{l}\) | A1 |
| \(u^2 = \frac{33gl}{10}\) * | A1 |
## Part (a):
$mgl(\cos\alpha - \cos\theta) = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ | M1A1=A1 |
$v^2 = u^2 + 2gl(\cos\alpha - \cos\theta)$ * | A1 |
## Part (b):
$\cos\alpha = \frac{3}{5}$: $v^2 = 2gl(\frac{3}{5} - \cos\theta) + u^2$ | M1A1 |
At top $\theta = 360°$: $v^2 = 2gl(\frac{3}{5} - 1) + u^2$ | M1 |
$v^2 > 0: -2gl \times \frac{2}{5} + u^2 > 0$ | |
$u^2 > \frac{4gl}{5}$ | A1 |
$u > 2\sqrt{\frac{gl}{5}}$ * | |
## Part (c):
Equation of motion along radius at lowest point: $T_1 - mg = \frac{mv^2}{l}$ | M1A1 |
$\theta = 180°$: $v^2 = 2gl(\frac{3}{5} + 1) + u^2$ | M1 |
$v^2 = \frac{16gl}{5} + u^2$ | |
$T_1 = \frac{m}{l}(\frac{16gl}{5} + u^2) + mg$ | A1 |
$= \frac{21mg}{5} + \frac{mu^2}{l}$ | |
At highest point: $T_2 + mg = \frac{mv^2}{l}$ | M1 |
$\theta = 360°$: $T_2 = 2mg(-\frac{2}{5}) + \frac{mu^2}{l} - mg$ | M1 |
$T_2 = \frac{mu^2}{l} - \frac{9mg}{5}$ | A1 |
$T_1 = 5T_2$ | M1 |
$\frac{21mg}{5} + \frac{mu^2}{l} = 5(\frac{mu^2}{l} - \frac{9mg}{5})$ | |
$\frac{66mg}{5} = \frac{4mu^2}{l}$ | A1 |
$u^2 = \frac{33gl}{10}$ * | A1 |\includegraphics{figure_5}
A particle $P$ of mass $m$ is attached to one end of a light rod of length $l$. The other end of the rod is attached to a fixed point $O$. The rod can turn freely in a vertical plane about $O$. The particle is projected with speed $u$ from a point $A$, where $OA$ makes an angle $\alpha$ with the upward vertical through $O$ and $0 < \alpha < \frac{\pi}{2}$. When $OP$ makes an angle $\theta$ with the upward vertical through $O$ the speed of $P$ is $v$ as shown in Figure 5.
\begin{enumerate}[label=(\alph*)]
\item Show that $v^2 = u^2 + 2gl (\cos \alpha - \cos \theta)$.
[4]
\end{enumerate}
It is given that $\cos \alpha = \frac{3}{5}$ and that $P$ moves in a complete vertical circle.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $u > 2\sqrt{\frac{gl}{5}}$.
[4]
\end{enumerate}
As the rod rotates the least tension in the rod is $T$ and the greatest tension is $5T$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that $u^2 = \frac{33}{10}gl$.
[9]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2011 Q7 [17]}}