# Dynamics - Pulley and Rod

## 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; 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

## 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.
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:

Related Introductory Physics Homework Help News on Phys.org
Simon Bridge
Homework Helper
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?

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

haruspex
Homework Helper
Gold Member
I too am confused about the signs.
Ta - Mag = Maaa From Newton 2nd Law of Block(A)
makes up +ve
aa = Rp$\alpha$p Acceleration relationship
makes down and anticlockwise the same sign
-TaRp+TbRp=Ip$\alpha$p Eulers equation for pulley
makes clockwise +ve
Tb - Mrg = Mrar Newtons second law on rod.
makes up +ve
-TbL - Irαr = -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.

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.