A light inextensible string of length \(l\) has one end attached to a fixed point \(A\). The other end is attached to a particle \(P\) of mass \(m\). The particle moves with constant speed \(v\) in a horizontal circle with the string taut. The centre of the circle is vertically below \(A\) and the radius of the circle is \(r\). Show that $$gr^2 = v^2\sqrt{l^2 - r^2}.$$ [9]

$T\cos\theta = mg$ | M1 A1 |

$T\sin\theta = \frac{mv^2}{r}$ | M1 A1 |

$\tan\theta = \frac{r}{\sqrt{(l^2 - r^2)}}$ or equivalent | M1 A1 |

$\tan\theta = \frac{v^2}{rg}$ Eliminating $T$ | M1 |

$\frac{r}{\sqrt{(l^2 - r^2)}} = \frac{v^2}{rg}$ Eliminating $\theta$ | M1 |

$gr^2 = v^2\sqrt{(l^2 - r^2)}$ | A1 (cso) | 9 marks [9]

---

A light inextensible string of length $l$ has one end attached to a fixed point $A$. The other end is attached to a particle $P$ of mass $m$. The particle moves with constant speed $v$ in a horizontal circle with the string taut. The centre of the circle is vertically below $A$ and the radius of the circle is $r$.

Show that
$$gr^2 = v^2\sqrt{l^2 - r^2}.$$
[9]

\hfill \mbox{\textit{Edexcel M3 2007 Q4 [9]}}