Dynamics - Pulley System with Rotating Rod

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bonannic
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Homework Statement


At the instant shown, the rod R is rotating about its centre of rotation with ω=3.8rad/s.

mA=10kg;

The pulley, with mP=8.7kg and RP=0.2m, may be modeled as a uniform disc.

The rod, with mR=4.1kg and L=0.8m, may be modeled as a thin beam rotating about one end.

g=9.8m/s ².

What is the magnitude of the acceleration of point B at this instant?

Homework Equations


ΣF=ma (N2) ΣM=Iα (Eulers equation)

The Attempt at a Solution


IP=(1/2)MR2
IRod at centre of rotation=(1/3)ML2

I defined upwards and anticlockwise to be positive and thus derived the following equations:
ΣFA=TA-mAg=mAaA
ΣMP at centre=RpTA-RpTB=IPαP
ΣMRod at end=-LTB+(1/2)LMRg=IRαR
where TA= Tension force acting between A and pulley and TB=Tension force acting between rod and pulley

I then found these constraints on aB in terms of aAPR
-aB=aA
5aBP
(-5/4)aBR
assuming that aB is acting upwards

Then, by subbing aB into the three original equations, I got the following system of equations:
TA+mAaB=mAg
RpTA-RpTB-5IPaB=0
-LTB+(5/4)IRaB=-(1/2)LMRg

However, when I solve this system of linear equations I get the wrong answer. I have a feeling this is because I ignored the angular velocity of the rod but I can't see that would affect the acceleration of B.
 

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If you ignored everything (gravity and tension) except for the fact that the rod is rotating. Would B have zero acceleration?
 
The question does not specify. All the information in the problem statement is all the information that the question gives.
 
It is not a question about the problem, it is a question to you.
 
Ah ok. So the end of the rod would be accelerating towards the pivot point as well.
I took this into consideration and got the right answer. I guess I should have studied the end of the rod more closely. Thank you so much for your help :)