How can I solve a dynamics problem involving a pulley, rod, and rotating masses?

[jules.t]

Homework Statement

There is pulley which has a mass hanging from one side, and a rotating rod attached horizontally to the other (both by cables). The pulley has mass; M_p = 9.5kg and radius; R_p =0.2m. The block(A) has mass; M_a=10.9kg. The rod is L = 0.8m long and has mass=M_l7.9kg. The rod is also rotating at w= 2.7 rad/s in the anti clockwise direction. The value for gravity used is g= 9.8m/s/s downwards. Tensions are denoted by T_a and T_b

Homework Equations

Newtons 2nd Law(F=ma), Eulers Equation (M=I\alpha).

The Attempt at a Solution

I have defined my coordinate system as anticlockwise is positive and vertically down is positive.
I_pulley = 0.5MR²
I_{rod at centre of grav} =(1/12)ML²
I have developed some simultaneous equations for the system:
T_a - M_ag = M_aa_a From Newton 2nd Law of Block(A)
a_a = R_p\alpha_p Acceleration relationship
-T_aR_p+T_bR_p=I_p\alpha_p Eulers equation for pulley
T_b - M_rg = M_ra_r Newtons second law on rod.

I also have the following(which i think is where the problem is at):
-T_bL - I_r\alpha_r = -Lw² Eulers equation on the rod at the centre of gravity.
a_r = -L\alpha_r

I solved this system, and got an incorrect answer. As above i think the problem lies in the last two equations.

 

[Simon Bridge]

Welcome to PF;

OK - rod rotating anticlockwise means the pulley is rotating clockwise and the mass is moving up.
The most likely location for an error is a misplaced minus sign as you link the different fbds.

The signs in your equations don't seem to match your stated sign convention.

i.e. $T_a-M_ag=M_aa_a$ gives a negative acceleration if the weight is greater than the tension.

But you've set them up how I would except for the last one, I'd have picked anticlockwise for positive for the motion of the rod... done that way the acceleration (magnitude and sign) of the block A is also the tangential acceleration of the rim of the pulley and the acceleration of point B on the rod.

However - the problem statement lacks a goal.
What are you supposed to be finding out?

 

[jules.t]

My apologies for not including the problems goal.
The problem asks to find the acceleration of pint B in the diagram provided.

 

[haruspex]

I too am confused about the signs.

T_a - M_ag = M_aa_a From Newton 2nd Law of Block(A)

makes up +ve

a_a = R_p\alpha_p Acceleration relationship

makes down and anticlockwise the same sign

-T_aR_p+T_bR_p=I_p\alpha_p Eulers equation for pulley

makes clockwise +ve

T_b - M_rg = M_ra_r Newtons second law on rod.

makes up +ve

-T_bL - I_rα_r = -Lw² Eulers equation on the rod at the centre of gravity.

Makes anticlockwise +ve and ... a couple of problems here.
The rod is rotating about an endpoint so you need to use the parallel axis theorem.
The Lw² term is a centripetal acceleration, which at this point is horizontal. It is supplied by the axle and has nothing to do with the tension.

 

[jules.t]

Thanks guys!
By working through the sign troubles and getting rid of the Lw^2 term, the solution was achieved. Your help is appreciated greatly, and i guess ill need to work on sign conventions before my exam.