It's the O'Reilly four-step plan all over again. Here our goal in the "Solve" step is to arrive at the differential equation of motion (EOM) in terms of x.
- FBD's: above for the bar and disk individually
- Newton/Euler: For the disk, sum M_C = O_x*R = I_C*theta_dot_dot where C is the no-slip contact point. For the bar, sum F_x = -k*x - c*x_dot - O_x = (3*m)*x_dot_dot.
- Kinematics: a_O = a_C + alpha x r_O_C - omega^2*r_O/C. Doing the math gives: theta_dot_dot = +x_dot_dot/R
- Solve: Combine the two Newton-Euler equations (eliminate the reaction O_x) and use the kinematics to produce a single, differential EOM for the system in terms of x.
Let us know if you have questions on this.
3 comments:
the Ic should be 1/2mr^2 + mr^2, correct?
where Ig would just be 1/2mr^2, right?
On the kinematics section of this problem,
x_dot_dot = alpha cross r
= r*alpha (k cross j)
= r*alpha (-i)
but assuming alpha is in the positive-k direction and theta is as defined in the picture, alpha = -theta_dot_dot, so
x_dot_dot = r*theta_dot_dot (positive)
Why do we need to use theta_dot_dot (negative k) instead of alpha (positive k)? And would it be wrong to assume theta in the other direction?
Notice on problem 8-130, theta is defined in the opposite direction, so
x_dot_dot = - r*theta_dot_dot
--to Kyle
Yes, that is correct.
--to Evan
Good questions/observations. There is a direct relationship between the definition of your coordinates and the signs used in the moment and kinematics equations.
* As you noted in this problem I (arbitrarily) chose theta as being positive in the CW sense. Therefore, a positive moment is also CW and x_dot_dot = +r*theta_dot_dot (a CW rotation corresponds to motion of O to the right ... also, that's just the way that the math works out with the cross products).
* In Problem 8/130 I chose theta as positive in the CCW sense. With that choice, a CCW moment is positive and x_dot_dot = -r*theta_dot_dot.
Either choice of sign convention for theta works. Just be consistent between moments and rotations in sign conventions.
Thanks for asking.
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