As mentioned in my earlier post, finding the location of the instant center for link AB at theta = 0 cannot be done by our usual graphical methods. However, knowing the location of this IC can be helpful in understanding the motion of the system at this position.
To encourage you to think about this, I am offering 10 bonus points for this homework problem if you submit a solution showing the IC of link AB at theta = 0. In this solution, I am asking that you do the following:
- Show why the graphical method for finding IC's does not define the IC of AB (that is, use the graphical method and explain the result).
- Provide analysis locating the IC of AB. (HINT: Find the velocities of points A and B in terms of theta_dot using our standard rigid body kinematics equations.)
- Send me a scan of your solution via email prior to the start of class on Friday. I will post all winning entries on the course blog.
- Be sure that what you submit is your own original work.
As 10 bonus points, this will replace your lowest homework score for the semester.
Good luck!
4 comments:
I am confused with the moment of link AB. Is it the moment of the center of mass of the link? In the sketch is shows it relatively low on the link but nowhere in the question does it give any info about it.
Good question.
The answer to this goes back to your days in ME270. M here is a "couple". A couple is actually a pair of two equal but opposite forces acting along parallel lines of action (a "twist"). As you learned in ME270, you can move a couple to any location on a rigid body with the same effect on the body's motion.
In short, the actual location of the point of application is not relevant -- it can be applied anywhere along link AB. When you sum moments on AB about the point of your choice, just add in M*k, where k is the unit vector perpendicular to the page.
That makes sense...but now I am frustrated whether i should do Kinematics (V_a = R_dot*e_r + R*theta_dot*e_theta) or if I should do sum of forces and moments equations. Should I first analyze link AB alone? And should I analyze it at the instant before impact when 'mg' is parallel with the link? Please help!
Recall that the KE T should be written as:
T = 0.5*I_O*omega_OB^2 + 0.5*m*v_G^2 + 0.5*I_G*omega_AB^2
where G is the cm of link AB. Your task with the kinematics is to relate v_G to omega_AB and omega_OB (to reduce your work energy equation to a single unknown).
To do this, write out the rigid body kinematics equation for v_G (look at the earlier post where I presented this equation). The rigid body equations are easier to use than the polar description that you have written.
So in summary, use the work energy equation (not Newton Euler). Do so by treating links AB and OB as a single system. Do all of your analysis at the instant when both links AB and OB are vertical.
Let me know if this does not help.
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