My analysis for the alpha vector agrees with Emily's. Her result is correct.
My analysis also shows the velocity of B to be in the negative y direction. This makes sense because, although A is moving in the positive y direction, arm AB is rotating about the negative x axis (swinging backwards). The velocity due to AB swinging backwards is greater than the positive velocity of A. Therefore, B has a velocity in the negative y direction.
Am I understanding your questions correctly? If not, let me know.
I'm getting a negative number for Vb as well but I do not see my error.
ReplyDeleteIf you are using the same coordinate axis as defined in the problem picture you should get alpha to be in the negative j direction.
If you differentiate your omega equation dWz/dt, dTheta(dot)/dt and dk/dt are zero. But there is still one term left, -theta(dot)(di/dt).
di/dt=(W(total)Xi)
W(total)Xi=WzkXi=Wzj
Plug this into the differentiated omega equation and:
alpha=-theta(dot)Wzj
I'm still unsure as to why Vb is in the negative j direction.
My analysis for the alpha vector agrees with Emily's. Her result is correct.
ReplyDeleteMy analysis also shows the velocity of B to be in the negative y direction. This makes sense because, although A is moving in the positive y direction, arm AB is rotating about the negative x axis (swinging backwards). The velocity due to AB swinging backwards is greater than the positive velocity of A. Therefore, B has a velocity in the negative y direction.
Am I understanding your questions correctly? If not, let me know.
I found my error. I forgot to include a negative sign in front of the resultant j component from the cross product of omega and r(o to b).
ReplyDeleteDouble check your signs and see if that fixes your problem.