A spherical asteroid with radius r = 123 m and mass M = 2.00×10^10 kg rotates ab
ID: 1698845 • Letter: A
Question
A spherical asteroid with radius r = 123 m and mass M = 2.00×10^10 kg rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to the asteroid's south pole (as defined by the axis of rotation) and fires its engine, applying a force F tangentially to the asteroid's surface as shown in the figure
Please show all steps and reasoning. Will rate immediately!
A spherical asteroid with radius r = 123 m and mass M = 2.0010^10 kg rotates about an axis at four revolutions per day. A tug spaceship attaches itself to the asteroid's south pole (as defined by the axis of rotation) and fires its engine, applying a force F tangentially to the asteroid's surface as shown in the figure uploaded image If F = 265^N, how long will it take the tug to rotate the asteroid's axis of rotation through an angle of 14.0^degree by this method? Please show all steps and reasoning. Will rate immediately!Explanation / Answer
Let a = Acceleration produced by the space ship From Newton's second law of motion, F = m a 265 = (2.0 x10^10) * a a = 132.5 x10^-10 m/s² Angular acceleration, alpha = a / r = 132.5 x 10^-10 / 123 = 1.08 x10^-10 rad/s² no. of revolutions per day = 4 Angular velocity, W = 4(2 Pi )/(24x60x60) = 2.91x10^-4 rad/s Angular displacement, theta = w t + (1/2) * alpha * t^2 14 = ( 2.91 x 10^-4 t ) + (1/2) * 1.08 x10^-10 * t^2 (1/2) * 1.08 x10^-10 * t^2 + ( 2.91 x 10^-4 t ) - 14 = 0 5.38 x 10^-11 * t^2 + ( 2.91 x 10^-4 t ) - 14 = 0 By solving the quadratic equation, t = 47689.5 s = 13.25 hours
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