M4 January 2005 Q6
6.
A smooth wire \(PMQ\) is in the shape of a semicircle with centre \(O\) and radius \(a\). The wire is fixed in a vertical plane with \(PQ\) horizontal and the mid-point \(M\) of the wire vertically below \(O\). A smooth bead \(B\) of mass \(m\) is threaded on the wire and is attached to one end of a light elastic string. The string has modulus of elasticity \(4mg\) and natural length \(\tfrac{5}{4}a\). The other end of the string is attached to a fixed point \(F\) which is a distance \(a\) vertically above \(O\), as shown in Fig. 1.
| Scheme |
|---|
| Take \(O\) as zero p.e. |
| Mechanical potential energy \((mgh) = -mga\cos 2\theta\) |
| Elastic potential energy \(\left(\dfrac{\lambda x^2}{2l}\right) = \tfrac{1}{2} \times \dfrac{4mg}{\tfrac{5}{4}a}\left(2a\cos\theta - \tfrac{5}{4}a\right)^2\) |
| Total p.e. \(= -mga\left(2\cos^2\theta - 1\right) + \dfrac{8mg}{5a}\left(\dfrac{8a\cos\theta - 5a}{4}\right)^2\) |
| \(= -2mga\cos^2\theta + mga + \dfrac{mga}{10}(8\cos\theta - 5)^2\) |
| \(= \dfrac{mga}{10}(8\cos\theta - 5)^2 - 2mga\cos^2\theta + c \qquad\) (change of constant with referral of p.e. to any other zero position.) |
Notes
The published mark scheme for this paper is a set of worked answers: no mark allocation is printed.
(Corrected from the printed mark scheme: the extension in the elastic potential energy is printed as \(2a\cos 2\theta - \tfrac{5}{4}a\); the string length is \(FB = 2a\cos\theta\), as used in the next line.)
| Scheme |
|---|
| Equilibrium when p.e. is max/min so \(\dfrac{\mathrm{d}E}{\mathrm{d}\theta} = 0\) |
| \(\dfrac{\mathrm{d}E}{\mathrm{d}\theta} = \dfrac{mga}{10} \times 16 \times (8\cos\theta - 5)(-\sin\theta) + 4mga\cos\theta\sin\theta = 0\) |
| \(mga\sin\theta\left(-\tfrac{64}{5}\cos\theta + 8 + 4\cos\theta\right) = 0\) |
| \(mga\sin\theta\left(8 - \tfrac{44}{5}\cos\theta\right) = 0\) |
| \(\sin\theta = 0,\ \ \theta = 0\) |
| or \(\ \cos\theta = \tfrac{10}{11},\ \ \theta = 24.6^\circ\) |
| Scheme |
|---|
| \(\dfrac{\mathrm{d}^2E}{\mathrm{d}\theta^2} = mga\sin\theta\left(\tfrac{44}{5}\sin\theta\right) + mga\cos\theta\left(8 - \tfrac{44}{5}\cos\theta\right)\) |
| When \(\theta = 0\), \(\ \dfrac{\mathrm{d}^2E}{\mathrm{d}\theta^2} = mga\left(8 - \tfrac{44}{5}\right) = -\tfrac{4}{5}mga\) which is \(\lt 0\) so max \(E\) so unstable. |
| When \(\theta = 24.6^\circ\), \(\ \dfrac{\mathrm{d}^2E}{\mathrm{d}\theta^2} = mga\left(\tfrac{44}{5}\left(1 - \left(\tfrac{10}{11}\right)^2\right) + \tfrac{10}{11}\left(8 - \tfrac{44}{5} \times \tfrac{10}{11}\right)\right) = \tfrac{84}{55}mga\) |
| which is \(\gt 0\) so min \(E\) so stable. |