Angular Acceleration of a Pulley with Mass

[OlicityFangirl]

Homework Statement

A pulley hangs of mass, m, and radius, R, hangs from the ceiling. Two blocks of masses, m₁ and m₂ are connected by a massless, non-stretchable rope on the pulley (assume no slipping). What is the angular acceleration of the pulley and what is the ratio of the tension forces on the vertical portions of the rope while the blocks are moving?

Homework Equations

τ=Iα
F=ma
I_cyl=0.5mR²
a_tan=αR

The Attempt at a Solution

Because the pulley has a mass, the tension forces on each side of the rope are different (if the pulley were massless then the tensions on each side would be the same, to my knowledge). The torque determines the angular acceleration:

τ=Iα

Since the tension work in opposite directions, the tensions, T₁ and T₂, act in opposite directions (where T₁ is on the side of block 1 and T₂ is on the side of block 2).

τ=Iα=T₁-T₂

I can use Newton's Second Law to try and find the two unknown tension forces:

F₁=m₁a=m₁g-T₁
F₂=m₂a=T₂-m₂g

Solving for T:

T₁=m₁g-m₁a
T₂=m₂g+m₂a

Plugging this into Iα=T₁-T₂:

m₁g-m₁a-m₂g+m₂a=Iα

Here is where I start to get more unsure. I still have the unknown linear acceleration, a. I think I can relate it to α using a=αr since there is no slipping. Then my equation would be:

m₁g-m₁αR-m₂g+m₂αR=0.5mR²α

Then, I can isolate and solve for α. I like to try and see if my answers make intuitive sense, and I'm not sure this one does:

α= g(m₁-m₂) / 0.5mR²+m₁R-m₂R

It seems angular acceleration increases as the mass of block 1 increases. This makes sense because if the weight of block 1 is greater, the pulley will rotate faster. However, the denominator also increases as m₁ increases and decreases as m₂ increases, which seems counterintuitive to me.

The ratio of tensions (T₁/T₂) is as simple as m₁(g-a)/m₂(g-a). I'm not sure how to think about this, but I'm confident in my work on this part.

Thanks for checking over my work!

 

[Doc Al]

Since the tension work in opposite directions, the tensions, T₁ and T₂, act in opposite directions (where T₁ is on the side of block 1 and T₂ is on the side of block 2).

τ=Iα=T₁-T₂

You are on the right track, but fix that equation. (You left out a factor of R on the right hand side. You need torques on both sides of the equation.)

 

[OlicityFangirl]

Oops right I forgot that τ=F_tanR.

When fixing this, my equation becomes:

R(T₁-T₂)=Iα

Plugging in my tension values from Newton's Second Law and assuming a=Rα:

R(m₁(g-a)-m₂(g+a))=0.5mR²α

I can cancel out an R. I also seem to have forgotten to distribute the negative through T₂ in my original post.

m₁g-m₁Rα-m₂g-m₂Rα=0.5mRα

Isolating and solving for angular acceleration now gets:

α= g(m₁-m₂) / R(0.5m+m₁+m₂)

My ratio of tensions should be unchanged.

 

[Doc Al]

Isolating and solving for angular acceleration now gets:

α= g(m₁-m₂) / R(0.5m+m₁+m₂)

Looks good to me, assuming you can treat the pulley as a solid disk.

My ratio of tensions should be unchanged.

The ratio of tensions (T₁/T₂) is as simple as m₁(g-a)/m₂(g-a).

One of those minus signs should be a + sign. You can also express this ratio in terms of the known masses, but the expression might be messy. (I didn't work it out.)

 

[OlicityFangirl]

Oh oops, I just made a typo. It should be:

T₁/T₂ = m₁(g-a) / m₂(g+a)

We always treat pulleys as solid cylinders. Thanks so much for your help!