# Two Masses in Pulley System (Rotational Dynamics)

1. May 5, 2012

### S-Flo

So I've been prepping for my Engineering Physics 1 final, and I've been having trouble with rotational motion problems, but this problem has been driving me crazy for the past two days:

1. The problem statement, all variables and given/known data
(I wanted to include a normal link to an image, but I'm a new member, so here's something)
i.imgur.com/wHAG3.jpg

There are two masses, one on top of a frictionless tabletop (12.0 kg), and one hanging off the edge of the table (5.00 kg), both are connected by an ideal cord wrapped around a frictionless pulley with a radius of 0.26 m and a mass of 2.10 kg. What are the respective tensions of the horizontal and vertical segments of the cord?

Known variables:
mass on top of table = 12.0 kg
mass hanging off edge = 5.00 kg
radius of pully = 0.26 m
mass of pully = 2.10 kg

Unknowns:
Horizontal Tension
Vertical Tension

2. Relevant equations
I = 1/2*m*r2 (moment of intertia)
τ = I*αθ (Torque, and that's angular acceleration on the right)
Krot = 1/2*I*ω2 (rotational kinetic energy)

F = m*a

3. The attempt at a solution

I first attempted to find the torque on the disk and translate that into tangential force on the mass on the table. After that failed I thought that the acceleration for everything in the system should be the same and tried to make acceleration equations for the two masses and the pulley and solve with algebra. Both gave me incorrect answers. I'm not sure whether I'm approaching this incorrectly or if the equations I set up were wrong.

Last edited: May 5, 2012
2. May 5, 2012

### tiny-tim

Welcome to PF!

Hi S-Flo! Welcome to PF!
That should work.

Show us your full calculations, and then we'll see what went wrong.

3. May 5, 2012

### S-Flo

Here's what I tried:

F = ma; so a = F/m

τ = rF = Iaθ; and aθ = ratan;
τ = Iratan; atan = τ/Ir = rF/Ir = F/I
and F should be the force the hanging box exerts on the pully (F = mg), so:
atan = Fg/I

So, if a is the same for all parts of the system, then:

Th/12 = Tv/5 = Fg/I

But Fg = 49.05 and I = 0.07098 which would make atan = 691 m/s2, which is very obviously wrong. And after that, I've had no idea what to do.

4. May 5, 2012

### tiny-tim

Hi S-Flo!
Noooo

raθ = atan

(it might be easier if you use the standard notation of α for angular acceleration )

5. May 5, 2012

### S-Flo

Even then the acceleration comes out to 39.8 m/s2, and the answer is still wrong. I think my whole system is set up incorrectly.

6. May 5, 2012

### tiny-tim

For example, where is the 5g ?

7. May 5, 2012

### S-Flo

It's 5 kg, and it's one of the values given in the initial problem.

Honestly, I don't even know what going on in this problem anymore, and I don't see the point in playing with my setup, since it's obviously wrong.

Last edited: May 5, 2012
8. May 5, 2012

### tiny-tim

no, the gravitational force, 5g

9. May 5, 2012

### S-Flo

I thought that was the force of gravity acting on the 5 kg box. F = ma, so 5g = 49.05
I also thought that was the force acting on the pulley to create torque.

10. May 5, 2012

### tiny-tim

No, Tv - Th is the force creating the torque.

And Tv and 5g are the forces acting on the hanging mass.

11. May 5, 2012

### S-Flo

That explains part of my error, but even then I still can't see how to set this system up.

12. May 5, 2012

### S-Flo

Wait, never mind. I just figured out the problem. Thanks.

13. May 6, 2012

### tiny-tim

now that you've solved it, here's a "cheat" method that's quicker …

(they probably wouldn't like you using it in an exam, but it's still useful as a check on your final figures )

replace the pulley by an ordinary object of "rolling mass" I/r2

then the total mass is 12 + I/r2 + 5 = 12 + 2.1/2*0.262 + 5,

the net force is 5g, so the acceleration is 5g/(12 + 2.1/2*0.262 + 5)

does that give you the right result?​