Suggestions:- FBD
- Work-energy: Since links AB and CD have negligible mass, these links do not contribute to either the potential energy or to the kinetic energy of the system. Therefore, the kinetic energy for the system is T = (1/2)*m*vG^2 + (1/2)*IG*omega^2 for the plate.
- Kinematics: Links AB and CD are always parallel as the system moves, and as a result, vC and vB are always parallel. Based on this, where is the IC for the plate? And, from this, what is the omega for the plate?
- Solve
Let us know if you have any questions on this.
Suggestions:
8 comments:
I was able to find the angular velocity, however I dont really know how to go about solving for the force in the beam.
By determination of the instant center, I get that it is located at an infinite distance. This would mean that the angular velocity of the plate at this instant is zero, correct? (since omega=v_G/r_(to IC). Using this, my answer is close, but not quite the same. Is this right?
For the angular velocity, make sure you take into account that the center of mass G of the block only falls a total of .4 meters.
Also for your kinematics portion, make sure that the distance is only the 'i' component of the links before the object makes contact with the surface. I had made this error, and I hope this helps with anyone that may be stuck.
nick,
This may be too late for you, but I'll give it a shot. I used the Newton-Euler eqns to set up the eqns for the forces in the x-direction and the moments. The forces in the y direction are zero in the members AB and CD because they are two force members so this eqn can be neglected. Once you set up these equations, you still have to solve for aG in the x-direction. You can do this by relating aG to the acceleration at point C. Then using the rigid body eqn, the acceleration at point C can be determined by relating it to the acceleration at point D. Because you only have to worry about the acceleration in the x-direction, the omega^2*rC/D is all that you need. Now, you have two eqns and two unknowns. Hope that helps!
Is Ig=36?
What L do we use ?
--to Nour--
For a rectangle having dimensions of "a" and "b", the centroidal mass moment of inertia is:
IG = (1/12)*m*(a^2 + b^2)
See Appendix D of the textbook.
This may be a simple answer and I could be missing something but when using KE equation 'T = 1/2*m*v_G^2 + 1/2*I_G*omega^2' I don't know how to find v_G in order to find omega. Can I just plug in omega*r_(to IC)into v_G and solve for omega that way?
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