Standard +0.8 This is a multi-step circular motion problem requiring application of Newton's second law in circular motion at two positions, energy conservation between those positions, and careful resolution of forces. While the setup is standard for Further Mechanics, the geometric configuration (60° angles) and the need to find a ratio requires systematic work through multiple equations, making it moderately challenging but within expected Further Maths scope.
A particle \(Q\) of mass \(m\) is attached to a fixed point \(O\) by a light inextensible string of length \(a\). The particle moves in complete vertical circles about \(O\). The points \(A\) and \(B\) are on the path of \(Q\) with \(AB\) a diameter of the circle. \(OA\) makes an angle of \(60°\) with the downward vertical through \(O\) and \(OB\) makes an angle of \(60°\) with the upward vertical through \(O\). The speed of \(Q\) when it is at \(A\) is \(2\sqrt{ag}\).
Given that \(T_A\) and \(T_B\) are the tensions in the string at \(A\) and \(B\) respectively, find the ratio \(T_A : T_B\). [6]
Question 3:
3 | 4ag 9
T −mgcos60=m. (T = mg)
A a A 2 | B1
v2
T +mgcos60=m.
B a | B1
1 ( )
Energy: m 4ag−v2 =mg.2acos60
2
v2 =2ag | M1A1
3mg
T =
B 2 | A1
Ratio is 3 : 1 | A1
6
Question | Answer | Marks
A particle $Q$ of mass $m$ is attached to a fixed point $O$ by a light inextensible string of length $a$. The particle moves in complete vertical circles about $O$. The points $A$ and $B$ are on the path of $Q$ with $AB$ a diameter of the circle. $OA$ makes an angle of $60°$ with the downward vertical through $O$ and $OB$ makes an angle of $60°$ with the upward vertical through $O$. The speed of $Q$ when it is at $A$ is $2\sqrt{ag}$.
Given that $T_A$ and $T_B$ are the tensions in the string at $A$ and $B$ respectively, find the ratio $T_A : T_B$. [6]
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q3 [6]}}