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tpb77
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Homework Statement
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
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 don't see it, or should i need another proof that (e.k) is constant.Thanks in advance for any help you can give.