Two Masses in Pulley System (Rotational Dynamics)

[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:

Homework Statement

(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

Homework Equations

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

F = m*a

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.

Thanks in advance!

 

[tiny-tim]

Welcome to PF!

Hi S-Flo! Welcome to PF!

… 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.

That should work.

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

 

[S-Flo]

Here's what I tried:

F = ma; so a = F/m

τ = rF = Ia_θ; and a_θ = ra_tan;
τ = Ira_tan; a_tan = τ/Ir = rF/Ir = F/I
and F should be the force the hanging box exerts on the pully (F = mg), so:
a_tan = F_g/I

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

T_h/12 = T_v/5 = F_g/I

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

 

[tiny-tim]

Hi S-Flo!

a_θ = ra_tan

Noooo …

ra_θ = a_tan​

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

 

[S-Flo]

Hi S-Flo!


Noooo …

ra_θ = a_tan​

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

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.

 

[tiny-tim]

I can't follow your equations.

For example, where is the 5g ?

 

[S-Flo]

I can't follow your equations.

For example, where is the 5g ?

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.

 

[tiny-tim]

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

no, the gravitational force, 5g

 

[S-Flo]

no, the gravitational force, 5g

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.

 

[tiny-tim]

I also thought that was the force acting on the pulley to create torque.

No, T_v - T_h is the force creating the torque.

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

 

[S-Flo]

No, T_v - T_h is the force creating the torque.

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

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

 

[S-Flo]

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

 

[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?​