1. The problem statement, all variables and given/known data A pendulum consists of a particle of mass m at the end of a light rigid rod of length l, the other end of the rod being freely attached to a stationary point 0. Let e(t) be a unit vector pointing along the rod, so that the position vector relative to 0 of the particle at time t is le(t). (a) Show from the principle of linear momentum that e x e'' - (g/l) e x k = 0 where g is the acceleration due to gravity, and k is a unit vector pointing vertically upwards. ( e'' represents the second derivative of the unit vector wrt time ) (b) If e(t) is confined to a vertical plane containing 0, the pendulum is called simple. Suppose that e(t) = sin θ(t) i - cos θ(t)k, where i is a horizontal unit vector. Deduce that θ'' + (g/l) sinθ = 0 (c) If e(t) rotates about k, the pendulum is called conical. Suppose that e' = w(t) (k x e(t)) Deduce that t →w(t) and t →e(t).k are constant throughout the motion. If w(0) = w > 0 and e(0).k = -cosθ with 0 < 0 < pi/2, show that w = sqrt( g secθ/l) Hint. The following identity holds for the triple vector product: u x ( v x w) = (u.w)v + (u.v)w 3. The attempt at a solution First two parts are alright, e x e'' + g/l e x k = 0 so e'' must be = - g/l k so le'' = -g k a = -gk so F = -gm k which is just the force due to gravity, so the original statement is true The second part i can do just by working through the differentiation of sinθ i - cosθj, then subbing in and cancelling all the terms. the last bit though i think should be e' = w( k x e) multiply both sides by e so e x e' = w( e x (k x e)) = 1/m ( e x p) = w( e x ( k x e)) the d/dt of the left is = 0, since the angular momentum is constant, then w( (e.e)k -(e.k)e) from the hint must be constant in time so d/dt [wk - w(e.k)e ]= 0. I can see how this would imply that dw/dt = 0, but not how d(e.k)/dt must also be 0 Have i done enough and just dont see it, or should i need another proof that (e.k) is constant. Thanks in advance for any help you can give.