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Feb 11, 2008

Problem 7/27

I was able to find both the Vb and alpha, but both had the wrong sign. I am guessing that is why my Ab is wrong. Can someone help?
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3 comments:

Emily said...

I'm getting a negative number for Vb as well but I do not see my error.

If 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.

February 11, 2008 at 10:07 PM
CMK said...

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.

February 12, 2008 at 12:26 AM
Emily said...

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).

Double check your signs and see if that fixes your problem.

February 12, 2008 at 12:44 AM

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