M5 June 2018 Q6
6. Three equal uniform rods, each of mass \(m\) and length \(2a\), form the sides of a rigid equilateral triangular frame \(ABC\). The frame is free to rotate in a vertical plane about a fixed smooth horizontal axis \(L\) which passes through \(A\) and is perpendicular to the plane of the frame.
The frame is held with \(AB\) horizontal and \(C\) below \(AB\), and released from rest.
Given that the centre of mass of the frame is two thirds of the way along a median from a vertex,
| Scheme | Marks |
|---|---|
| \(\text{MI} = 2 \times \dfrac{4ma^2}{3}, + \left(\dfrac{ma^2}{3} + m(2a\cos 30^\circ)^2\right)\) | M1 A1, A1 |
| \(= 6ma^2\) GIVEN ANSWER | A1 |
| (4) |
Notes
M1 for attempt at MI, dimensionally correct, with correct no. of terms
First A1 for first two terms
Second A1 for third term unsimplified
Third A1 for correct GIVEN ANSWER
| Scheme | Marks |
|---|---|
| \(X - 3mg\cos 60^\circ = 0\) since \(\dot{\theta} = 0\) | M1 |
| \(X = \dfrac{3mg}{2}\) | A1 |
| \(3mg\sin 60^\circ \pm Y = 3m\cdot\dfrac{2}{3}2a\cos 30^\circ\ddot{\theta}\) | M1 A1 A1 |
| \(M(A)\), \(\ 3mga = 6ma^2\ddot{\theta}\) OR \(6a\ddot{\theta} = 2\sqrt{3}g\sin 60^\circ\) | M1 A1 |
| Eliminating \(\ddot{\theta}\), | DM1 |
| \(Y = \dfrac{mg\sqrt{3}}{2}\) | A1 |
| \(R = \sqrt{X^2 + Y^2} = \dfrac{mg}{2}\sqrt{3 + 3^2} = mg\sqrt{3}\) | M1 A1 |
| (11) | |
| (15 marks) |
Notes
First M1 for equation of motion inwards with usual rules and RHS = 0
First A1 for a correct value of \(X\)
Second M1 for equation of motion tangentially with usual rules
Second and Third A1 for a correct equation. A1A0 if one error.
Third M1 for equation of rotational motion about \(A\) to give equation in \(\ddot{\theta}\) only
Fourth A1 for a correct equation
N.B. If \(m\) is used consistently in all 3 equations, treat as only one error and penalise in first equation.
Equation of rotational motion could be obtained by differentiating an energy equation and putting \(\theta = 0\):
\(3mg\cdot\dfrac{2}{3}2a\cos 30^\circ\cdot(\sin(\theta + 30^\circ) - \sin 30^\circ) = \dfrac{1}{2}6ma^2\dot{\theta}^2\)
\(2mga\sqrt{3}(\sin(\theta + 30^\circ) - \sin 30^\circ) = 3ma^2\dot{\theta}^2\)
Differentiating,
\(2mga\sqrt{3}\cos(\theta + 30^\circ) = 6ma^2\ddot{\theta}\)
When \(\theta = 0\),
\(2mga\sqrt{3}\cos 30^\circ = 6ma^2\ddot{\theta}\)
\(6a\ddot{\theta} = 2\sqrt{3}g\cos 30^\circ\)
Fourth DM1, dependent on previous 2 M’s, for eliminating to give \(Y\) in terms of \(mg\)
Fifth M, independent, for finding magnitude using their \(X\) and \(Y\), in terms of \(mg\) only
Fifth A1 for correct answer