| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 9 |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem requiring resolution of forces and circular motion equations. Part (i) involves straightforward limiting arguments, parts (ii) and (iii) are routine applications of F=ma in circular motion with vertical equilibrium. The algebra is simple and the problem follows a well-established template, making it slightly easier than average for A-level Further Maths. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
**(i)** $T\cos\theta = mg$ (Must be seen to score in part **(i)**) **B1**
$\theta = 90° \Rightarrow mg = 0$ $\therefore AP$ can never be horizontal. **B1**
$\therefore 0 < \cos\theta < 1 \Rightarrow T\left(= \frac{mg}{\cos\theta}\right) > mg$ **B1**
**(ii)** $T\sin\theta = ml\sin\theta\varpi^2 \Rightarrow T = ml\varpi^2$ **M1 A1**
**(iii)** $T\frac{h}{l} = mg \Rightarrow T = \frac{mgl}{h}$ **B1**
$\Rightarrow ml\varpi^2 = \frac{mgl}{h}$ **B1**
$\Rightarrow \varpi^2 h = g$ **B1**
$h = 0.5 \Rightarrow \varpi = \sqrt{20} = 4.47$ Angular speed is 4.47 rad/s. **B1**9\\
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A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point $A$. The particle moves with constant angular speed $\omega$ in a horizontal circle whose centre is at a distance $h$ vertically below $A$ (see diagram).\\
(i) Show that however fast the particle travels $A P$ will never become horizontal, and that the tension in the string is always greater than the weight of the particle.\\
(ii) Find the tension in the string in terms of $m , l$ and $\omega$.\\
(iii) Show that $\omega ^ { 2 } h = g$ and calculate $\omega$ when $h$ is 0.5 m .
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q9 [9]}}