A2 June 2022 Q4
4.

A small smooth ring \(R\) of mass \(m\) is threaded onto a light inextensible string. One end of the string is attached to a fixed point \(A\) and the other end of the string is attached to the fixed point \(B\) such that \(B\) is vertically above \(A\) and \(AB = 6a\)
The ring moves with constant angular speed \(\omega\) in a horizontal circle with centre \(A\). The string is taut and \(BR\) makes a constant angle \(\theta\) with the downward vertical, as shown in Figure 2.
The ring is modelled as a particle.
Given that \(\tan\theta = \dfrac{8}{15}\)
| Scheme | Marks | AO |
|---|---|---|
![]() | ||
| Resolve vertically | M1 | 3.4 |
| \(T\cos\theta = mg\) | A1 | 1.1b |
| \(T = \left(\dfrac{mg}{\cos\theta} = \dfrac{6.8mg}{6}\right) = \dfrac{17mg}{15}\) | A1 | 1.1b |
| (3) |
Notes
M1: Need all terms. Condone sin/cos confusion
A1: Correct unsimplified equation.
A1: Correct answer only
\(1.1mg\) or better \((1.13\ldots mg)\)
Do not ignore subsequent working if they try to combine this with a tension in \(AR\)
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion | M1 | 3.1b |
| \(mr\omega^2 = T + T\sin\theta \qquad \left(m\times 3.2a\omega^2 = \text{their } T\left(1 + \dfrac{8}{17}\right)\right)\) | A1 A1 | 1.1b 1.1b |
| Solves for \(\omega\) or \(\omega^2\) | M1 | 1.1b |
| \(\left(\dfrac{r\omega^2}{g} = \dfrac{1 + \sin\theta}{\cos\theta} = \dfrac{6.8 + 3.2}{6},\quad \omega^2 = \dfrac{10g}{6\times 3.2a}\right) \qquad \omega = \sqrt{\dfrac{25g}{48a}} = \dfrac{5}{4}\sqrt{\dfrac{g}{3a}}\) | A1 | 1.1b |
| (5) | ||
| (8 marks) |
Notes
M1: Equation for circular motion. Need all terms and dimensionally correct. Condone sin/cos confusion and sign errors.
Any correct form for acceleration
A1: Unsimplified equation with at most one error
A1: Correct unsimplified equation
Allow M1A1A0 for \(mr\omega^2 = T' + (\text{their (a)})\sin\theta\)
M1: Clear attempt to substitute for trig and tension or divide their two equations to solve for \(\omega\) or \(\omega^2\) in terms of \(a\) and \(g\)
Independent M mark but requires an equation using tension and trig.
A1: Any equivalent form
\(0.72\sqrt{\dfrac{g}{a}}\) or better \((0.7216\ldots)\)
