Paper 4, Section II, B

Dynamics and Relativity | Part IA, 2011

The trajectory of a particle r(t)\mathbf{r}(t) is observed in a frame SS which rotates with constant angular velocity ω\omega relative to an inertial frame II. Given that the time derivative in II of any vector u\mathbf{u} is

(dudt)I=u˙+ω×u,\left(\frac{\mathrm{d} \mathbf{u}}{\mathrm{d} t}\right)_{I}=\dot{\mathbf{u}}+\boldsymbol{\omega} \times \mathbf{u},

where a dot denotes a time derivative in SS, show that

mr¨=F2mω×r˙mω×(ω×r),m \ddot{\mathbf{r}}=\mathbf{F}-2 m \boldsymbol{\omega} \times \dot{\mathbf{r}}-m \boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r}),

where F\mathbf{F} is the force on the particle and mm is its mass.

Let SS be the frame that rotates with the Earth. Assume that the Earth is a sphere of radius RR. Let PP be a point on its surface at latitude π/2θ\pi / 2-\theta, and define vertical to be the direction normal to the Earth's surface at PP.

(a) A particle at PP is released from rest in SS and is acted on only by gravity. Show that its initial acceleration makes an angle with the vertical of approximately

ω2Rgsinθcosθ\frac{\omega^{2} R}{g} \sin \theta \cos \theta

working to lowest non-trivial order in ω\omega.

(b) Now consider a particle fired vertically upwards from PP with speed vv. Assuming that terms of order ω2\omega^{2} and higher can be neglected, show that it falls back to Earth under gravity at a distance

43ωv3g2sinθ\frac{4}{3} \frac{\omega v^{3}}{g^{2}} \sin \theta

from PP. [You may neglect the curvature of the Earth's surface and the vertical variation of gravity.]

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