# Math20512

### Section A

##### A1
• Many of you wrote V = -x2 - y2, instead of V = -xy. The question had F = yi + xj, not F = xi + yj.
• A few even gave V to be a vector !!! (where did you get that idea from?)
• And just a few seemed to have forgotten that $$T = \textstyle\frac12(\dot x^2 + \dot y^2)$$.
##### A2
• Most got the expression for the centre of mass correct, but many then treated it is if it was a single particle at that point (a partial mark was given for this, but really the whole point was to show that the internal forces all cancelled).
##### A3
• How many times do some of you have to be told to use brackets! A surprisingly large number of you wrote, correctly, that $$p=\dot q -2$$ so that $$\mathcal{L} = \frac12(p+2)^2-2(p+2)-\sin q$$ and then deduced that
$$H = p\dot q - \mathcal{L} = p\dot q - \frac12(p+2)^2-2(p+2)-\sin q$$
Aaaaggh!
##### A4
• Many did this well, but in the proof that Ω is skew-symmetric, a few forgot that (AB)T = BTAT (and not ATBT)
##### A5
• In this question (about the robotic arms), many seemed to assume the system was planar (for which you would have got partial marks), even though the question explicitly mentioned moving in space. Generally I gave partial marks if it was clear there was some indication the student had an idea about what finding a system of coordinates means. The answer is simply that the end B moves on the surface of a sphere (requiring 2 coordinates), and for each position of B, the free end C can move on a sphere so requires an additional 2 coordinates. Thus in all it is a 4 degree of freedom system.

### Section B

##### B6
• (a) Most did this well, except for the general solution. See the notes (Chapter 3) to see how to get the general solution from the normal modes.
• One amusing thing: a few of you correctly found ω2 = 5 or 1, so deduced (correctly) ω = √5 or √1, and then went on using √1 as if it was something different from 1, so the period was given as T = 2π/√1. (I was tempted to deduct a mark as it seemed to suggest a lack of understanding, but I didn't!)
• (b)(i) Mostly ok, using F = ma and F = -dV/dx.
• (b) (ii) A few got the equilibria to satisfy x3 = μ x, and then divided by x to get x = ±√μ, so losing the x = 0 solution — you should have got out of that habit at A-level!
• b(ii) A few of you lost half a mark (!) for saying that the solutions x = ±√μ was unstable when μ<0: what you should have realized was that it doesn't even exist in that case!
##### B7

Each part of this question was marked out of 4 marks.

• (a) In this system, gravity acts on the mass M but not on m. Also, the mass M can move so contributes to the kinetic energy and many ignored the resulting $$\frac12 M\dot r^2$$ term in the kinetic energy. Others had a potential energy with an mg cosθ term, which would require the table top to be vertical! In fact only the mass M contributes to the potential energy, and the correct potential energy was Mgr (or -Mg(l-r) — they agree up to a constant).
• (b) Most did this ok. Method marks were given even if you had the wrong answers in part (a) (provided the method was correct, of course!).
• (c) This seemed to challenge many of you. If M is stationary, then r is constant so $$\dot r=0$$, and if m rotates uniformly then $$\dot\theta$$ is constant (=ω). Put these into the equations of motion and you get the relation between r and ω that was asked for.
• (d) This is of course new as you hadn't seen the Routhian before; of course, there were enough details in the question to be able to work it out. Quite a few did this ok, but several wrote μ in terms of $$\dot\theta$$ so ending up with an expression involving $$\dot\theta$$ and not μ. Actually you should have used the expression $$\mu = mr^2\dot\theta$$ to eliminate $$\dot\theta$$ and end up with an expression involving only μ. (Just like getting the Hamiltonian from the Lagrangian: one eliminates $$\dot q$$ by writing it in terms of p). One ends up with
$$\mathcal{L}_\mu(r,\dot r) = \frac12 (m+M)\dot r^2 - \frac{\mu^2}{2mr^2}-Mgr$$
The first term is then T, and the second two make $$-V'_{\mu}$$.
• (e) Those who managed part (d) did this ok; it's just a 1 degree of freedom system governed by potential energy $$V_\mu(r) = \frac{\mu^2}{2mr^2}\mu(r)+Mgr$$.
##### B8

• (a) Several confused conserved quantity with conservative force (which will have lost you 2 marks). The right answer is that f(p(t),q(t)) is constant as a function of t when (p(t), q(t)) is a solution to Hamilton's equations. If you just wrote $$\frac{df}{dt}=0$$, I accepted that and gave you 2 marks, but if you wrote $$\frac{\partial f}{\partial t}=0$$ I only gave 1 mark — it's actually wrong isn't it! (So I was being generous)