Suggestions:
- FBD
- Work-Energy
- Kinematics
- Solve
This problem is very similar to the example problem on page 71 of the lecture notes that was worked out in class. Use that problem as a guide (particularly for the kinematics that caused your instructor problems during lecture...). In particular, use the result discussed in lecture that omega_AB = -omega_OB.
Let us know if you have any questions.
5 comments:
Just as an FYI for anyone else who didn't know this, the work done by a couple is integral(M dtheta). It is on page 471.
so for this problem, would the work be M*theta or 2*M*theta. I was figuring it would be the former since it is acting on the one member which only changes a single value of theta, but I wasn't sure since the same moment is also causing the movement of the other arm theta degrees as well.
I am with you, I think it is just M*theta because if you look at the diagram they have separated the theta into two for you already, so they are taking the mirrored movement into account that way. Just my thoughts, hope they help.
but it acts on AB from A to O which would be 2 theta, right? Since we are interested in A as it strikes O?
I think omega_AB causes bar AB to only rotate theta, and omega_OB causes bar OB to rotate theta as well, giving a total of 2(theta). Since M is applied to bar AB, I think it should only cover an angle of theta.
Post a Comment