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Relativistic Mechanics

Problem 1.343

A stationary upright cone has a taper angle θ=45∘\theta=45^{\circ}, and the area of the lateral surface S0=4.0 m2S_{0}=4.0 \mathrm{~m}^{2}. Find: its taper angle; (b) its lateral surface area, in the reference frame moving with a velocity v=(4/5)cv=(4 / 5) c along the axis of the cone.

Reveal Answer
1.343. (a) tan⁡θ′=tan⁡θ1−β2.\tan \theta^{\prime}=\frac{\tan \theta}{\sqrt{1-\beta^{2}}} . Hence θ′=59∘\quad \theta^{\prime}=59^{\circ}; (b) S=S= =S01−β2cos⁡2θ=3.3 m2=S_{0} \sqrt{1-\beta^{2} \cos ^{2} \theta}=3.3 \mathrm{~m}^{2}. Here β=v/c\beta=v / c
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