Aaron Klapheck's Website

1st. Write the coordinate system. In the solution the coordinate system is written in the upper left-hand-side of the page. See the image below for the zoom in of the coordinate system.

You many notice in addition to the three axes that there are also three curved arrows (hover your mouse over the curved arrows to make sure you know which ones I am talking about). These arrows indicate the sign (positive or negative) of a cross product.
For instance: k hat cross i hat = +j hat. j hat is positive in this case because travel from k had to i hat is in the positive direction.
j hat cross i hat = -k hat. k hat is negative because travel from j hat to i hat is in the negative direction.
Remember: cross products are not commutative, that is order is important. a x b does not usually equal b x a.

2nd. Find the Ground Point (Gr).

For the diagram given, point A would be a good choice for a ground point. We could choose any other point on the ground (for instance the base of the second "tower") because any two points on the ground will have the same velocity and acceleration relative to each other; that is zero. In this case a velocity was given and a velocity is asked to be found. Because no reference point is given it is assumed that the ground point is the reference. After choosing a reference. point (Gr), that point should be indicated in the coordinate system diagram as shown above.

3rd. Restate the given and found in terms of the reference point and the coordinate system used.

The ground point was chosen in the previous step so simply state all velocities with reference. to the ground point A. v_B/A means the velocity of B with respect to A.

4th. Find the position vectors r.
Notice the notation used (r_A-->B). See the Dynamics home page for an explanation of this notation.

5th. Find the velocity vectors v and omega.

You may notice that omaga_A/B is not needed (I solved for this only to demonstrate my method further).
Because we are dealing only with rigid bodies the rotational reference. frame is not needed. Therefore we can use the simplified version of the equation.
Because we are trying to find v_C/A we can break it up into its components, v_A/B + v_B/C, and solve them independently.

The Given and Find section start out correct.

The main change is that once a student gets good at identifying the ground point they can start writing the given and find velocities and accelerations with respect to it in the first place.
Once this is done all that is left is to find the position and velocity vectors. Many of the in-between steps are skipped due to the understanding of how they work.