Because of this, my new homework hint is to use point O for the KE of OB and to use the cm G for the KE of link AB. Therefore, the KE for the system is:
T = 0.5*I_O*omega_OB^2 + 0.5*m*v_G^2
+ 0.5*I_G*omega_AB^2
You will then need to use kinematics (Step 3) to relate v_G to omega_AB. That is you need to use:
v_G = v_B + omega_AB x r_G/B
where
v_B = v_O + omega_BO x r_B/O = omega_BO x r_B/O
omega_BO = - omega_AB (since the angular speeds of the two links are the same, theta_dot, but of opposite signs)
You will then need to use kinematics (Step 3) to relate v_G to omega_AB. That is you need to use:
v_G = v_B + omega_AB x r_G/B
where
v_B = v_O + omega_BO x r_B/O = omega_BO x r_B/O
omega_BO = - omega_AB (since the angular speeds of the two links are the same, theta_dot, but of opposite signs)
Please add comments and questions to this post.
Apologies for the implied misdirection in lecture on this!
6 comments:
When A strikes O, is link OB stationary? (Is the kinetic energy of OB zero when A hits O?)
In this problem, you are asked about the situation immediately BEFORE A strikes O (so its not a question on impact, etc).
We will need to rely on kinematics in order to relate the motion of OB and AB right before contact occurs.
When finding the velocities is it correct to use the given moment as the omega value for the rod AB?
how to find U(1-->2)? does the moment do work?
--to dmill--
Not sure of what you are asking here.
The moment acting on rod AB does work, so the moment should be included in the expression for work, U(1->2). This moment is not the same as the omega for the bar.
Let me know if I missed the point of your question.
--to wendy--
Recall from lecture that the work due to a couple M is"
U(1->2) = +-integral(M*d_theta)
where theta is the angle of rotation of the body on which M acts and with "+" used when M has the same sense as the rotation (or "-" when M has the opposite sense as the rotation).
Post a Comment