• Support PF! Buy your school textbooks, materials and every day products Here!

Dynamics - Pulley and Rod

  • Thread starter jules.t
  • Start date
  • #1
3
0

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[itex]\alpha[/itex]).


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[itex]\alpha[/itex]p Acceleration relationship
-TaRp+TbRp=Ip[itex]\alpha[/itex]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[itex]\alpha[/itex]r = -Lw2 Eulers equation on the rod at the centre of gravity.
ar = -L[itex]\alpha[/itex]r

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

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,856
1,654
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
3
0
My apologies for not including the problems goal.
The problem asks to find the acceleration of pint B in the diagram provided.
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,463
5,410
I too am confused about the signs.
Ta - Mag = Maaa From Newton 2nd Law of Block(A)
makes up +ve
aa = Rp[itex]\alpha[/itex]p Acceleration relationship
makes down and anticlockwise the same sign
-TaRp+TbRp=Ip[itex]\alpha[/itex]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.
 
  • #5
3
0
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.
 

Related Threads on Dynamics - Pulley and Rod

Replies
4
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
Replies
3
Views
5K
  • Last Post
Replies
2
Views
883
  • Last Post
Replies
9
Views
731
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
8
Views
9K
Top