Challenging +1.8 This is a challenging vertical circular motion problem requiring energy conservation and Newton's second law at two positions, combined with the constraint T_A = 7T_B. Students must set up equations for tensions at symmetric positions, apply energy between them, then solve a system involving trigonometric terms. The multi-step algebraic manipulation and the non-standard constraint make this harder than typical circular motion questions, but it follows established methods without requiring novel geometric insight.
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 \(O\). The particle completes vertical circles with centre \(O\). The points A and B are on the path of \(P\), both on the same side of the vertical through \(O\). \(OA\) makes an angle \(\theta\) with the downward vertical through \(O\) and \(OB\) makes an angle \(\theta\) with the upward vertical through \(O\).
The speed of \(P\) when it is at A is \(u\) and the speed of \(P\) when it is at B is \(\sqrt{ag}\). The tensions in the string at A and B are \(T_A\) and \(T_B\) respectively. It is given that \(T_A = 7T_B\).
Find the value of \(\theta\) and find an expression for \(u\) in terms of \(a\) and \(g\). [8]
Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
Answer
Marks
5
mu2
T −mgcosθ=
Answer
Marks
A a
B1
mag
T +mgcosθ=
Answer
Marks
B a
B1
mu2 mag
mgcosθ+ =7 −mgcosθ+
T A = 7T B so
a a
Answer
Marks
Guidance
u2 =ag ( 7−8cosθ)
M1
Use given relationship and combine.
Energy: 1 mu2 − 1 mag =mg ( acosθ+acosθ)
2 2
( )
u2=ag 4cosθ+1
Answer
Marks
Guidance
So
M1 A1
Energy equation.
Equate expressions for u2
M1
1
cosθ= , θ =60°
Answer
Marks
Guidance
2
A1
CAO
u= 3ga
A1
CAO
8
Answer
Marks
Guidance
Question
Answer
Marks
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
5 | mu2
T −mgcosθ=
A a | B1
mag
T +mgcosθ=
B a | B1
mu2 mag
mgcosθ+ =7 −mgcosθ+
T A = 7T B so
a a
u2 =ag ( 7−8cosθ) | M1 | Use given relationship and combine.
Energy: 1 mu2 − 1 mag =mg ( acosθ+acosθ)
2 2
( )
u2=ag 4cosθ+1
So | M1 A1 | Energy equation.
Equate expressions for u2 | M1
1
cosθ= , θ =60°
2 | A1 | CAO
u= 3ga | A1 | CAO
8
Question | Answer | Marks | Guidance
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 $O$. The particle completes vertical circles with centre $O$. The points A and B are on the path of $P$, both on the same side of the vertical through $O$. $OA$ makes an angle $\theta$ with the downward vertical through $O$ and $OB$ makes an angle $\theta$ with the upward vertical through $O$.
The speed of $P$ when it is at A is $u$ and the speed of $P$ when it is at B is $\sqrt{ag}$. The tensions in the string at A and B are $T_A$ and $T_B$ respectively. It is given that $T_A = 7T_B$.
Find the value of $\theta$ and find an expression for $u$ in terms of $a$ and $g$. [8]
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q5 [8]}}