Horzontal Massive Pulley Problem with Friction

[Kristenx2]

Homework Statement

Consider 2 masses (m₁=13kg, m₂=26kg) connected by a string. The pulley is a uniform disc of radius R=0.05m and mass m=1kg. μ between the horizontal surface and the larger mass is 0.2.
Find Frictional force acting on larger mass.
Find the linear acceleration of the blocks.
Find tension of the string acting on both weights.
Calculate the angular acceleration of the pulley in rad/s².
Calculate the time it will take block M to move 1m when released from rest.
Calculate the number of revolutions made by the pulley during the time period of the precious problem.

Homework Equations

F=μ*m*g
m₂g-T₂=m₂a
α=τ/I
I=1/2mr²
T₂-T₁=1/2m_pa
That's all I could really figure out to use. Would I also use m₁g*μ-T₁=m₁a?

The Attempt at a Solution

I got the friction force fine, 51.012N.

I messed up the next part thinking the pulleys were massless. I did a=(m₂g-μm₁g)/(m₁+m₂ and got 1.962 m/s². Naturally, for the tension part, I ended up with the same tensions of 102N. I know that's not right.

So I did T₂-T₁=1/2m_pa and ended up with 1.94m/s². That seems wrong as well, because my T₁ ended up being 0.646N (mgμ-T=ma). It just seems really off and I am not sure what I am doing wrong. I am almost positive it has something to do with the way I am incorporating the friction, though.

This is what the diagram looks like:

O\_______|m₁(26kg)|____
|//////////////////////////////////
|
|
|M₂(13kg)|

 

[grzz]

Of course you HAVE also to use T$_{1}$ - Frictional force = 26a.

Otherwise you would not have enough equations to solve for the all the unknown physical quantities that you have.

 

[Kristenx2]

Hi, grzz. Sounds good. Does that mean when I go to find the linear acceleration I would use m₂g-m₂a-m₁gμ-m₁a = 1/2m_pa?

 

[grzz]

Yes.

 

[Kristenx2]

So my linear acceleration really is 1.92m/s²?
It just seems strange because when I plug that into (26)(9.81)(0.2)-T=(26)(1.92) I get an obviously ridiculous number, and for the other tension, (13)(9.81)-T=(13)(1.92) I get 102N.

 

[Kristenx2]

Oh! I am doing it backwards. You said T-F_μ=26a, right? So I was actually wrong about (26)(9.81)(0.2)-T=(26)(1.92) and I will not get a low or negative number.

 

[Kristenx2]

For the angular acceleration of the pulley {alpha} I did R₂*T₂-R₁*T₁=I{alpha} and got 105.2 where R=0.05m and I used 102 and 49.4 for my Ts, and 1.94m/s² for my acceleration. However, when I use {alpha}=a/R, using the same values, I end up with 38.8rad/s²... Which one? ;c

 

[grzz]

My values are
T1 = 101.4N
T2 = 102.3N
a = 1.94m/s2