# Dynamics - Pulley and Rod

1. Nov 4, 2013

### jules.t

1. The problem statement, all variables and given/known data
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; Mp = 9.5kg and radius; Rp =0.2m. The block(A) has mass; Ma=10.9kg. The rod is L = 0.8m long and has mass=Ml7.9kg. 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 Ta and Tb

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

3. The attempt at a solution
I have defined my coordinate system as anticlockwise is positive and vertically down is positive.
Ipulley = 0.5MR2
Irod at centre of grav =(1/12)ML2
I have developed some simultaneous equations for the system:
Ta - Mag = Maaa From Newton 2nd Law of Block(A)
aa = Rp$\alpha$p Acceleration relationship
-TaRp+TbRp=Ip$\alpha$p Eulers equation for pulley
Tb - Mrg = Mrar Newtons second law on rod.

I also have the following(which i think is where the problem is at):
-TbL - Ir$\alpha$r = -Lw2 Eulers equation on the rod at the centre of gravity.
ar = -L$\alpha$r

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

Last edited: Nov 5, 2013
2. Nov 4, 2013

### 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?

3. Nov 4, 2013

### jules.t

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

4. Nov 5, 2013

### haruspex

I too am confused about the signs.
makes up +ve
makes down and anticlockwise the same sign
makes clockwise +ve
makes up +ve
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.

5. Nov 5, 2013

### 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.

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