A spherical asteroid with radius r = 123 meters and mass M = 2.00×1010 kg rotate

Question

A spherical asteroid with radius r = 123 meters and mass M = 2.00×1010 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 .

If F = 265 N, how long will it take the tug to rotate the asteroid's axis of rotation through an angle of 13.0 degrees by this method?

What is delta t in hours?


Thank you!

Explanation / Answer

The mass of the asteriod M = 2.0*10^10 kg

the radius of the asteriod R = 123m

then the moment of inertia I = 2/5MR^2

                                            = 2/5 (2*10^10)(123)^2

                                            = 12103.2*10^10 kg.m^2

Now the torque = I

and                 FR = I

then theangular acceleration = FR/I = (265)(123)/12103.2*10^10

                                                         = 2.69*10^-10 rad/s^2

and the angle is = 13deg = 2/360 (13) = 0.2267 rad

From rotational kinematics

                = 1/2t^2

therefore the time

         t = 2/

             = 2(0.2267) / 2.69*10^-10)

              = 0.41*10^5 s or 4.1*10^4s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.