Paper 4 , Section II, B

Dynamics and Relativity | Part IA, 2013

(i) An inertial frame SS has orthonormal coordinate basis vectors e1,e2,e3\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}. A second frame SS^{\prime} rotates with angular velocity ω\boldsymbol{\omega} relative to SS and has coordinate basis vectors e1,e2,e3\mathbf{e}_{1}^{\prime}, \mathbf{e}_{2}^{\prime}, \mathbf{e}_{3}^{\prime}. The motion of SS^{\prime} is characterised by the equations dei/dt=ω×eid \mathbf{e}_{i}^{\prime} / d t=\boldsymbol{\omega} \times \mathbf{e}_{i}^{\prime} and at t=0t=0 the two coordinate frames coincide.

If a particle PP has position vector r\mathbf{r} show that v=v+ω×r\mathbf{v}=\mathbf{v}^{\prime}+\boldsymbol{\omega} \times \mathbf{r} where v\mathbf{v} and v\mathbf{v}^{\prime} are the velocity vectors of PP as seen by observers fixed respectively in SS and SS^{\prime}.

(ii) For the remainder of this question you may assume that a=a+2ω×v+ω×(ω×r)\mathbf{a}=\mathbf{a}^{\prime}+2 \boldsymbol{\omega} \times \mathbf{v}^{\prime}+\boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r}) where a\mathbf{a} and a\mathbf{a}^{\prime} are the acceleration vectors of PP as seen by observers fixed respectively in SS and SS^{\prime}, and that ω\omega is constant.

Consider again the frames SS and SS^{\prime} in (i). Suppose that ω=ωe3\omega=\omega \mathbf{e}_{3} with ω\omega constant. A particle of mass mm moves under a force F=4mω2r\mathbf{F}=-4 m \omega^{2} \mathbf{r}. When viewed in SS^{\prime} its position and velocity at time t=0t=0 are (x,y,z)=(1,0,0)\left(x^{\prime}, y^{\prime}, z^{\prime}\right)=(1,0,0) and (x˙,y˙,z˙)=(0,0,0)\left(\dot{x}^{\prime}, \dot{y}^{\prime}, \dot{z}^{\prime}\right)=(0,0,0). Find the motion of the particle in the coordinates of SS^{\prime}. Show that for an observer fixed in SS^{\prime}, the particle achieves its maximum speed at time t=π/(4ω)t=\pi /(4 \omega) and determine that speed. [Hint: you may find it useful to consider the combination ζ=x+iy\zeta=x^{\prime}+i y^{\prime}.]

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