M3 June 2017 Q5
5.
A hollow cylinder is fixed with its axis horizontal. A particle \(P\) moves in a vertical circle, with centre \(O\) and radius \(a\), on the smooth inner surface of the cylinder. The particle moves in a vertical plane which is perpendicular to the axis of the cylinder. The particle is projected vertically downwards with speed \(\sqrt{7ag}\) from the point \(A\), where \(OA\) is horizontal and \(OA = a\). When angle \(AOP = \theta\), the speed of \(P\) is \(v\), as shown in Figure 4.
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}mv^2 - \dfrac{1}{2}m \times 7ag = mga\sin\theta\) | M1A1A1 |
| \(v^2 = 7ag + 2ag\sin\theta = ag(7 + 2\sin\theta)\) * | A1 |
| (4) |
Notes
M1 Energy equation from the point of projection to a general point. Must have 3 terms and the PE term must include a trig function.
A1 Correct difference of KE terms.
A1 Correct PE term and all signs correct.
A1cso Obtain correct given expression for \(v^2\) with no errors in the solution.
| Scheme | Marks |
|---|---|
| At top \(v^2 = 5ag\) | M1A1 |
| \(R + mg = m\dfrac{v^2}{a}\) or \(m\dfrac{v^2}{a} > mg\) | M1A1 |
| \(R = 4mg\) or substitute for \(v^2\) | dM1 |
| \(R > 0 \quad \therefore\) complete circles | A1 cso |
| (6) |
Notes
M1 Use the result given in (a) with \(\theta = 270^\circ\) to obtain \(v^2\) at the top. Substitution for \(\theta\) may occur later.
A1 Correct expression for \(v^2\). May be implied by correct work later.
M1 Attempt NL2 at the top. This mark cannot be awarded if a general position is used but can be awarded later when the motion at the highest point is considered.
A1 Correct NL2 at the top with \(R + mg\)
dM1 Eliminate \(v^2\) between the 2 equations. Depends on the 2 previous M marks in (b).
A1cso Correct result for \(R\) (at the top) seen and the conclusion stated. (Do not need to see \(R > 0\)). If working with the resultant, resultant > \(mg\) must be seen.
Full marks can be awarded if it is stated that \(v^2 > 0\) and \(R > 0\) at the top - mark the work relevant to \(R\).
ALT Last 4 marks:
If \(m\dfrac{v^2}{a} > mg\) is seen, give M1A1. M1 substitute for \(v^2\); \(5mg > mg \quad \therefore\) complete circles
| Scheme | Marks |
|---|---|
| Max \(v\) at lowest point | |
| \(\sin\theta = 1 \ \Rightarrow v^2 = 9ag\) | M1 |
| \(v = 3\sqrt{ag}\) | A1 |
| (2) | |
| (12 marks) |
Notes
M1 Using \(\sin\theta = 1\) in the result given in (a) to obtain \(v^2\) at the lowest point. Any other complete method may be used, eg an energy equation provided it leads to the speed at the lowest point.
A1 \(v = 3\sqrt{ag}\) or \(\sqrt{9ag}\) (Watch square root covers all necessary letters.)