M2 June 2013 (R) Q4
4. A rough circular cylinder of radius \(4a\) is fixed to a rough horizontal plane with its axis horizontal. A uniform rod \(AB\), of weight \(W\) and length \(6a\sqrt{3}\), rests with its lower end \(A\) on the plane and a point \(C\) of the rod against the cylinder. The vertical plane through the rod is perpendicular to the axis of the cylinder. The rod is inclined at 60\(^\circ\) to the horizontal, as shown in Figure 1.
The coefficient of friction between the rod and the cylinder is \(\dfrac{\sqrt{3}}{3}\) and the coefficient of friction between the rod and the plane is \(\mu\). Given that friction is limiting at both \(A\) and \(C\),
| Scheme | Marks |
|---|---|
| \(AC = 4a\tan 60^\circ = 4a\sqrt{3}\). | M1 A1 |
| (2) |
Notes
M1 A1 Or \(\dfrac{4a}{\tan 30}\) or \(\sqrt{(8a)^2 - (4a)^2}\)
3 independent equations required. Award M1A1 for each in the order seen. If more than 3 relevant equations seen, award the marks for the best 3.
| Scheme | Marks |
|---|---|
| use of \(F = \mu R\) at either \(A\) or \(C\) | M1 |
| \(M(A),\ \ \ R_C.4a\sqrt{3} = W.3a\sqrt{3}\cos 60^\circ\) | M1 A1 |
| \((\uparrow),\ \ \ R_A + R_C\cos 60^\circ + F_C\cos 30^\circ = W\) | M1 A1 |
| \((\rightarrow),\ \ \ F_A - R_C\cos 30^\circ + F_C\cos 60^\circ = 0\) | M1 A1 |
| M(C) \(a\sqrt{3}\cos 60W + F_A.4a\sqrt{3}\sin 60 = R_A.4a\sqrt{3}\cos 60\) | |
| Parallel: \(F_A\cos 60 + R_A\cos 30 + F_C = W\cos 30\) | |
| Perpendicular: \(R_C + R_A\cos 60 = F_A\cos 30 + W\cos 60\) | |
| solving to give \(\mu = \dfrac{\sqrt{3}}{5}\); 0.346 or 0.35. | DM1 A1 |
| (9) | |
| (11 marks) |
Notes
M1 A1 \(R_C = \dfrac{3W}{8}\)
M1 A1 \(R_A = \dfrac{5W}{8}\)
M1 A1 \(F_A = R_C\dfrac{\sqrt{3}}{3}\)
DM1 Equation in \(\mu\) only. Dependent on 4 M marks for their equations.
Reactions in the wrong direction(s) – check carefully