M4 June 2016 Q6
6.
Figure 3 shows a uniform rod \(AB\), of length \(2l\) and mass \(4m\). A particle of mass \(2m\) is attached to the rod at \(B\). The rod can turn freely in a vertical plane about a fixed smooth horizontal axis through \(A\). One end of a light elastic spring, of natural length \(2l\) and modulus of elasticity \(kmg\), where \(k \gt 4\), is attached to the rod at \(B\). The other end of the spring is attached to a fixed point \(C\) which is vertically above \(A\), where \(AC = 2l\). The angle \(BAC\) is \(2\theta\), where \(\dfrac{\pi}{6} \lt \theta \leqslant \dfrac{\pi}{2}\)
Given that there is a position of equilibrium with \(\theta \neq \dfrac{\pi}{2}\)
Given that \(k = 10\)
| Scheme | Marks |
|---|---|
| GPE \(= 4mgl\cos 2\theta + 2mg\times 2l\cos 2\theta\) | M1 |
| \(= 8mgl\cos 2\theta\) | A1 |
| Length of string \(= 2\times 2l\sin\theta\) EPE \(= \dfrac{kmg}{4l}(4l\sin\theta - 2l)^2\) | M1 |
| \(= kmgl(2\sin\theta - 1)^2\) | A1 |
| \(kmgl(4\sin^2\theta - 4\sin\theta + 1) + 8mgl(1 - 2\sin^2\theta) + \text{const}\) | M1 |
| \(= 4mgl\{(k - 4)\sin^2\theta - k\sin\theta\} + \text{const}\) | A1 |
| (6) |
Notes
M1 GPE of rod + particle (relative to a fixed point)
A1 Correct total
M1 Use of EPE \(= \dfrac{\lambda x^2}{2a}\)
M1 Total PE expressed in \(\sin\theta\)
A1 Given answer as printed
| Scheme | Marks |
|---|---|
| \(V' = (4mgl)\{(k - 4)2\sin\theta\cos\theta - k\cos\theta\}\) | M1 |
| \(\Rightarrow V' = 0\quad (k - 4)2\sin\theta\cos\theta - k\cos\theta = 0\) | DM1 |
| \(4\cos\theta\{(2k - 8)\sin\theta - k\} = 0\) | |
| \(\sin\theta = \dfrac{k}{2k - 8}\) | A1 |
| \(\dfrac{\pi}{6} \lt \theta \lt \dfrac{\pi}{2} \Rightarrow \left(\dfrac{1}{2} \lt\right)\dfrac{k}{2k - 8} \lt 1\) | B1ft |
| \(\Rightarrow k \lt 2k - 8\) | M1 |
| \(\therefore k \gt 8\) | A1 |
| (6) |
Notes
M1 Differentiate \(V\) – condone errors but do not accept integration
DM1 Set \(V' = 0\) and solve for \(\text{trig}(\theta) = \mathrm{f}(k)\). Dependent on the first M1
A1 \(\cos\theta = 0 \Rightarrow \theta = \dfrac{\pi}{2}\) need not be seen
B1ft ft on their \(\sin\theta\)
M1 Solve right hand inequality for \(k\)
A1 Given answer
| Scheme | Marks |
|---|---|
| \(V'' = (4mgl)(12\cos 2\theta + 10\sin\theta)\) | M1 |
| \(V'' = 8(mgl)(6\cos 2\theta + 5\sin\theta)\) | A1 |
| \(\sin\theta = \dfrac{10}{12} = \dfrac{5}{6}\qquad \cos 2\theta = 1 - 2\times\dfrac{25}{36} = -\dfrac{7}{18}\) | DM1 |
| \(V'' = 8mgl\left(-\dfrac{42}{18} + \dfrac{25}{6}\right) = mgl\dfrac{44}{3} \gt 0\) | |
| \(\therefore\) (\(V\) min and) equilibrium is stable. | A1 |
| (4) | |
| (16 marks) |
Notes
M1 Substitute \(k = 10\) and find second derivative of \(V\)
A1 Any equivalent form
DM1 Dependent on the previous M1. Substitute their trig. values. Need to be considering the whole of \(V''\)
Accept 14.3 or better
A1 With no errors seen