M3 June 2013 Q3
3.
Two particles \(P\) and \(Q\), of mass \(m\) and \(2m\) respectively, are attached to the ends of a light inextensible string of length \(6l\). The string passes through a small smooth fixed ring at the point \(A\). The particle \(Q\) is hanging freely at a distance \(l\) vertically below \(A\). The particle \(P\) is moving in a horizontal circle with constant angular speed \(\omega\). The centre \(O\) of the circle is vertically below \(A\). The particle \(Q\) does not move and \(AP\) makes a constant angle \(\theta\) with the downward vertical, as shown in Figure 2.
Show that
| Scheme | Marks |
|---|---|
| (i) For \(Q\) \(T = 2mg\) | B1 |
| For \(P\) \(T\cos\theta = mg\) | M1 |
| \(\cos\theta = \dfrac{1}{2}\ \ \ \theta = 60^\circ\) * | A1cso |
| (ii) For \(P\ \rightarrow\) \(T\sin\theta = mr\omega^2\) | M1A1 |
| \(2mg\sin\theta = m \times 5l\sin\theta \times \omega^2\) | M1depA1 |
| \(\omega^2 = \dfrac{2g}{5l}\ \ \ \ \ \omega = \sqrt{\dfrac{2g}{5l}}\) * | A1cso |
| (8 marks) |
Notes
In this question, award marks as though the question is not divided into two parts - ie give marks for equations wherever seen.
(i)
B1 for using \(Q\) (no need to state Q being used) to state that \(T = 2mg\) or \(T_Q = 2mg\) with \(T_P = T_Q\) seen or implied later.
M1 for attempting to resolve vertically for \(P\) \(T\) must be resolved but sin/cos interchange or omission of \(g\) are accuracy errors. \(mg + 2mg = T + T\cos\theta\) gets M0
A1cso for combining the two equations to obtain \(\theta = 60^\circ\) *
NB: This is a "show" question, so if no expression is seen for \(T\) and just \(2mg\cos\theta = mg\) shown, award 0/3 as this equation could have been produced from the required result, so insufficient working.
(ii)
M1 for attempting NL2 for \(P\) along the radius. The mass used must be \(m\) if the particle is not stated to be \(P\); a mass of \(2m\) would imply use of \(Q\). \(T\) must be resolved. Acceleration can be in either form.
A1 for \(T\sin\theta = mr\omega^2\) or \(T\dfrac{\sqrt{3}}{2} = mr\omega^2\)
M1 dep for eliminating \(T\) between the two equations for \(P\) and substituting for \(r\) in terms of \(l\) and \(\theta\) dependent on the second but not the first M mark.
A1 for \(2mg\sin\theta = m \times 5l\sin\theta \times \omega^2\) or \(\dfrac{T\sin\theta}{T\cos\theta} = \tan\theta = 5l\sin\theta\left(\dfrac{\omega^2}{g}\right)\) \(\theta\) or 60°
A1cso for re-arranging to obtain \(\omega = \sqrt{\dfrac{2g}{5l}}\) * ensure the square root is correctly placed
Alternatives: Some candidates "cancel" the \(\sin\theta\) without ever showing it.
M1A1 for \(T = m \times 5l\omega^2\)
M1A1 for \(2mg = 5ml\omega^2\)
A1cso as above
Vector Triangle method: Triangle must be seen
| \(T = 2mg\) | B1 |
| \(\cos\theta = \dfrac{mg}{2mg}\) | M1 |
| \(\theta = 60^\circ\) | A1 |
| Correct triangle | M1A1 |
| \(\sin\theta = \dfrac{5ml\sin\theta\omega^2}{2mg}\) | M1A1 |
| \(\omega = \ldots\) | A1cso (as above) |