Challenging +1.2 This is a standard circular motion problem requiring energy conservation and Newton's second law in a familiar context. While it involves multiple steps (deriving tension formula, finding max/min ratio, solving for u, then finding angle), each step follows well-established techniques taught in Further Maths mechanics. The derivation is straightforward application of formulas, and the ratio calculation is algebraic manipulation without requiring novel insight.
3 A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). When \(P\) is hanging vertically below \(O\), it is given a horizontal speed \(u\). In the subsequent motion, \(P\) moves in a complete circle. When \(O P\) makes an angle \(\theta\) with the downward vertical, the tension in the string is \(T\). Show that
$$T = \frac { m u ^ { 2 } } { a } + m g ( 3 \cos \theta - 2 )$$
Given that the ratio of the maximum value of \(T\) to the minimum value of \(T\) is \(3 : 1\), find \(u\) in terms of \(a\) and \(g\).
Assuming this value of \(u\), find the value of \(\cos \theta\) when the tension is half of its maximum value.
3 A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. When $P$ is hanging vertically below $O$, it is given a horizontal speed $u$. In the subsequent motion, $P$ moves in a complete circle. When $O P$ makes an angle $\theta$ with the downward vertical, the tension in the string is $T$. Show that
$$T = \frac { m u ^ { 2 } } { a } + m g ( 3 \cos \theta - 2 )$$
Given that the ratio of the maximum value of $T$ to the minimum value of $T$ is $3 : 1$, find $u$ in terms of $a$ and $g$.
Assuming this value of $u$, find the value of $\cos \theta$ when the tension is half of its maximum value.
\hfill \mbox{\textit{CAIE FP2 2013 Q3 [12]}}