M5 June 2014 Q5
5. A uniform rod \(AB\), of mass \(m\) and length \(2a\), is free to rotate in a vertical plane about a fixed smooth horizontal axis \(L\). The axis \(L\) is perpendicular to the rod and passes through the point \(P\) of the rod, where \(AP = \dfrac{2}{3}a\).
The rod is held at rest with \(B\) vertically above \(P\) and is slightly displaced.
| Scheme | Marks |
|---|---|
| \(I_L = \tfrac{1}{3}ma^2 + m\left(\tfrac{1}{3}a\right)^2\) | M1 A1 |
| \(= \tfrac{4}{9}ma^2\) | A1 |
| (3) |
Notes
M1 for use of parallel axes rule
First A1 for correct expression
Second A1 for answer
| Scheme | Marks |
|---|---|
| \(\tfrac{1}{2}\tfrac{4}{9}ma^2\dot{\theta}^2 = mg\tfrac{1}{3}a(1 - \cos\theta)\) | M1 A1 A1 |
| \(\dot{\theta} = \sqrt{\dfrac{3g(1 - \cos\theta)}{2a}}\) | A1 |
| (4) |
Notes
M1 for energy equation
First A1 for KE terms
Second A1 for PE terms
Third A1 for answer
| Scheme | Marks |
|---|---|
| \(mg\tfrac{1}{3}a\sin\theta = \tfrac{4}{9}ma^2\ddot{\theta}\) | M1 A1 |
| \(\dfrac{3g\sin\theta}{4a} = \ddot{\theta}\) | A1 |
| (3) |
Notes
M1 for moments about axis (or differentiate energy equation)
First A1 for a correct equation
Second A1 for answer
| Scheme | Marks |
|---|---|
| \(mg\cos\theta - X = m\tfrac{1}{3}a\dot{\theta}^2;\ X = 0\) | M1 A1 A1 |
| \(\dot{\theta}^2 = \dfrac{3g(1 - \cos\theta)}{2a}\) | |
| eliminating \(\cos\theta\) and solving, | |
| \(\dot{\theta} = \sqrt{\dfrac{g}{a}}\) | DM1 A1 |
| (5) | |
| (15 marks) |
Notes
First M1 for resolving along the rod
First A1 for forces incl. \(X = 0\)
Second A1 for mass × accln
Second M1, dependent on first M1, for eliminating \(\cos\theta\) and solving for \(\dot{\theta}\)
Third A1 for correct answer