A spherical asteroid with radius r = 123 m and mass M = 2.50×10^10 kg rotates ab

Question

A spherical asteroid with radius r = 123 m and mass M = 2.50×10^10 kg rotates about an axis at four revolutions per day. A "tug" spaceship attaches itself to a vehicle which follows the asteroid's south pole (as defined by the axis of rotation) and fires its engine, applying a force tangentially to the asteroid's surface keeping the direction of the force in the same plane. The initial situation is shown in the figure. (Figure 1) . Part A If F = 265 N, how long will it take the tug to rotate the asteroid's axis of rotation through an angle of 11.0 by this method?

Explanation / Answer

here,

radius of sphere, r = 123 m
mass of sphere, m = 2.50*10^10 kg
angular speed, w = 4 rev/day = 0.000290 rad/s
( 1 rev/day = 0.0000727 rad/s)

Moment of inertia,
I = (2/5)*m*r^2
I = (2/5)*(2.50*10^10)*(123)^2
I = 1.5129*10^14 kg.m^2

Applied force, F = 265 N

Torsion = F*R = (265)(123) = 32595 N.m

Since Torsion = alpha * I

Solving for angular acceleration,
alpah = torsion/I = 32595/(1.5129*10^14)
alpha = 2.154*10^-10 rad/s^2

From Second eqn of rotational motion,
= o(t) + 0.5*alpha*t²
(11*pi/180) = 0.000290*t + 0.5*2.154*10^-10*t^2

upon solving above quadratic eqn,
time, t = 661.589 s

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