Solving the Rotational Motion of Asteroid

[TyloBabe]

Homework Statement

A spherical asteroid with radius r = 123m and mass M = 2.50×10^10kg rotates about an axis at 4 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 (at the south pole). If F = 265 N , how long will it take the tug to rotate the asteroid's axis of rotation through an angle of 12.0 degrees by this method?

Homework Equations

delta L/delta t = F*R=I*alpha
omegaf^2=omegai^2+2alpha*theta
L=I*omega

The Attempt at a Solution

First I of course change 12 degrees into radians. I then used the fact that F*R=I*alpha to solve the for angular acceleration. Using this angular acceleration, I solved for the final angular velocity of asteroid after it has reach the 12 degree orientation. I then stated that the change in L was equal to the Final L minus the Initial L. In this case, the initial L just equals I*omega1, or the rotation of the asteroid initially. I then said that the final L was equal to the square root of L1 and L2, or Lf=sqroot((I*w1)^2 + (I*w2)^2). So then I just used (Lf-Li)/(F*r)=delta t to solve for the time it took to change the asteroids orientation. But obviously I'm not getting the correct answer...


The main thing that confuses me is the fact that while the ship is rotating the asteroid, the asteroid itself is already rotating. So doesn't that mean that at some point, the ship will be facing in the opposite direction as it was initially, and therefore actually pushing the asteroid back to its original orientation?

Here's a picture from my book http://i.imgur.com/xzj9o.jpg

 

[anathema]

Thank you for your question. Your approach seems to be on the right track, but there are a few things that need to be clarified.

First, when calculating the change in angular momentum, it is important to note that the force applied by the tug spaceship is only acting for a certain amount of time, not continuously. Therefore, the change in angular momentum would be equal to the impulse (F * t) instead of just the final angular momentum minus the initial angular momentum.

Secondly, your equation for the final angular velocity (omega2) is incorrect. The correct equation is:

omega2 = sqrt(omega1^2 + 2 * alpha * theta)

where omega1 is the initial angular velocity (in this case, 4 revolutions per day), alpha is the angular acceleration (which you have correctly calculated using F*R=I*alpha), and theta is the angle through which the asteroid's axis of rotation is being rotated (in radians).

Finally, to address your concern about the tug spaceship facing in the opposite direction, you are correct that at some point, the tug will be facing in the opposite direction as it was initially. However, this does not mean that the asteroid will be pushed back to its original orientation. This is because the force applied by the tug is tangential to the surface of the asteroid, meaning that it is always perpendicular to the axis of rotation. Therefore, the force will always contribute to the change in angular momentum in the same direction, regardless of the orientation of the tug spaceship.

I hope this helps clarify your approach. Let me know if you have any further questions. Keep up the good work!