In the figure provided with this problem statement, the author has provided us with two sets of coordinate axes: x-y-z and x'-y'-z'. If you want to keep the notation in this problem consistent with what we have used in lecture: replace x-y-z axes with X-Y-Z (and make these fixed axes) and replace x'-y'-z' with x-y-z (with these being attached to the disk).
Other suggestions:
- With x-y-z axes and observer attached to the disk, the omega vector has two components: omega_x about the FIXED X-axis and omega-z about the MOVING z-axis.
- When differentiating omega to find alpha, take particular note of which rotation axis is fixed and which rotation axis is moving.
- Note that point O can be considered to be on the same rigid body as the disk (O is a point on the centerline of the rotation axis of the disk around the horizontal shaft). Therefore, in the acceleration equation a_B = a_O + (a_B/O)_rel + alpha x r_B/O + 2*omega x (v_B/O) + omega x (omega x r_B/O), you have a_O = 0.
- With the observer attached to the disk, you have (v_B/O)_rel = (a_B/O)_ rel = 0.
2 comments:
I'm having trouble relating my little x to my big X. Can you just say that big x is 0.2 multiplied by little x?
At the instant shown, the xyz axes and XYZ axes are aligned. That is: i = I, j = J and k = K.
Therefore, to relate big x to little, you simply need to say: i = I (not I = 0.2*i). The DIRECTIONS of the unit vectors are what is important, not where they lie on the page.
Let me know if that does not help.
Post a Comment