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Nov 24, 2008

Homework Problem 8/130 - more discussion









There have been several good questions raised in the post below. Maybe I can jump in at this point to say a few things about these questions and some other important points.
  • The EOM that I get for this system is [by summing moments about the no-slip contact point C -- see the lecture example on page 8 of the notes]: (3*m*r/2)*x_dot_dot + k*r*x = 0. Therefore the natural frequency, omega_n, is given by: omega_n = sqrt((2*k)/(3*m)). This natural frequency is NOT equal to sqrt(k/m)!!
  • The actual IC's that you use to get a particular x_max in the response is somewhat arbitrary. That is, you can start the system out with an infinite number of sets of IC's x(0) and x_dot(0) and get the same x_max. [Can you look at the animations above to see what I used for IC's in this simulation?] For your work, I recommended using x(0) as some non-zero number and x_dot(0) = 0, as Phil B explains in a comment in the following post.
  • The amplitude of oscillation for the orange wheel above is roughly twice that of the blue wheel. However, the two wheels take EXACTLY the same amount of time to make one complete cycle of oscillation -- do you know why this is true? Does this agree with your intuition?
  • A friction force is required to prevent slipping as the disk rolls. Watch the friction force (FF) in the above animation as the system moves. The friction force for the orange wheel is larger than the friction force for the blue wheel -- why is this necessary? At what point in one cycle of oscillation is the friction force a maximum? Why does it occur that that time?

3 comments:

Nick Powell said...

i was curious how you get a natural frequency to equal sqrt(2k/3m).

CMK said...

From my FBD of the disk, I used a moment (Euler) equation about the no-slip contact point C:

sum moments about C = 2*((k/2)*r*theta)*r = IC*theta_dot_dot

where theta is measured positively CCW and IC = IG + m*r^2 = (1/2)*m*r^2 + m*r^2 = (3/2)*m*r^2.

Since I chose theta positive CCW, I get theta_dot_dot = - x_dot_dot/r. Substituting this into the above moment equation gives:

(3/2)*m*r*x_dot_dot + k*r*x = 0

The natural frequency is the square root of the coefficient in front of "x" when you divide through the the coefficient of x_dot_dot; that is, omega_n = sqrt(2*k/3*m).

This all comes back to the derivation of the EOM.

Does this help?

Nick Powell said...

yes this helps greatly. i didn't realize the relationship between the EOM and frequency. do the coefficients in front of EOM relate to any other terms such as amplitude or phase?