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A stationary upright cone has a taper angle theta=45^@, and the area of the lateral surface S_0=4.0m^2. Find: (a) its taper angle, (b) its lateral surface area, in the reference frame moving with a velocity v=(4//5)c along the axis of the cone. |
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Answer» Solution :In the frame K in which the CONE is at rest the coordinates of A are `(0,0,0)` and of B are `(h, h tan theta, 0)`. In the frame `K^'`, which is moving with velocity v along the axis of the cone, the coordinates of A and B at time `t^'` are `A:(-VT^',0,0), B: (hsqrt(1-beta^2)-vt^', h tan theta, 0)` Thus the TAPER angle in the frame `K'` is `tan theta'=(tan theta)/(sqrt(1-beta^2))(=(y_B'-y_A')/(x_B'-x_A'))` and the lateral surface area is, `S=pih'^(2) SEC theta' tan theta'` `=pih^2(1-beta^2)(tantheta)/(sqrt(1-beta^2))sqrt(1+(tan^2theta)/(1-beta^2))=S_0sqrt(1-beta^2cos^2theta)` Here `S_0=pih^2sectheta tan theta` is the lateral surface area in the rest frame and `h'=hsqrt(1-beta^2)`, `beta=v//c`. |
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