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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The question does not specify. All the information in the problem statement is all the information that the question gives.
 
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 :)