Moment of Inertia of a Disk and Pulley System

[ChiHawksFan]

Homework Statement

Hi all,
I just had a quick question regarding a problem that I have to write a lab report on. The problem involves solving for the moment of inertia of a system that involves a pulley with a hanging mass attached to a spool that turns a flywheel. The solution is supposed to be in terms of the mass the hanging weight and the acceleration of the weight. In lab we were eventually given the equation that could be used to solve for the moment of inertia. It is as follows: I={[r²(mg-ma)]/α} My question involves how to derive this. I know that it is a sum of all forces problem, but I can't seem to get the right answer. Here is a link to the problem. It is on the second page and is titled "Moment of Inertia Problem".

Homework Equations

∑F=ma
∑τ=Iα
a=rα, α=a/r
m_h = The mass of the hanging weight
M_d = The mass of the disk, ring and spool configuration
r = the radius of the spool

The Attempt at a Solution

Here is a drawing I made of the situation:

My attempt at the solution is as follows:

∑F_y = m_ha
m_hg - τ = m_ha
τ = m_hg - m_ha
∑τ = Iα ----------> I = τ/α
I = (m_hg - m_ha)/α
α = a/r
I = [r(m_hg - m_ha)/a]

I am unsure where to get the other r from.

Thanks for any help!

 

[SammyS]

Homework Statement

Hi all,
I just had a quick question regarding a problem that I have to write a lab report on. The problem involves solving for the moment of inertia of a system that involves a pulley with a hanging mass attached to a spool that turns a flywheel. The solution is supposed to be in terms of the mass the hanging weight and the acceleration of the weight. In lab we were eventually given the equation that could be used to solve for the moment of inertia. It is as follows: I={[r²(mg-ma)]/α} My question involves how to derive this. I know that it is a sum of all forces problem, but I can't seem to get the right answer. Here is a link to the problem. It is on the second page and is titled "Moment of Inertia Problem".

Homework Equations

∑F=ma
∑τ=Iα
a=rα, α=a/r
m_h = The mass of the hanging weight
M_d = The mass of the disk, ring and spool configuration
r = the radius of the spool

The Attempt at a Solution

Here is a drawing I made of the situation:
View attachment 82722

My attempt at the solution is as follows:

∑F_y = m_ha
m_hg - τ = m_ha
τ = m_hg - m_ha
∑τ = Iα ----------> I = τ/α
I = (m_hg - m_ha)/α
α = a/r
I = [r(m_hg - m_ha)/a]

I am unsure where to get the other r from.

Thanks for any help!

You are using the variable, τ, for two different things. (It looks like maybe tau, rather than letter, t .)

The τ in the force equations is tension. Usually we use uppercase T .

The one I highlight in red above, τ, is torque.

How is torque, τ, related to tension, T, in this case?

 

[ChiHawksFan]

You are using the variable, τ, for two different things. (It looks like maybe tau, rather than letter, t .)

The τ in the force equations is tension. Usually we use uppercase T .

The one I highlight in red above, τ, is torque.

How is torque, τ, related to tension, T, in this case?

Actually, I did mean for τ to be torque in each equation. I realize now that I totally forgot to add tension into my ∑F_y equation. I will try that and see if I can get a better result. Thanks for the help!

 

[ChiHawksFan]

I tried something else, and I think I am a little closer to a solution.

τ = Fdsinθ, τ = Iα, α = a/r, d = r = the radius of the spool, F = m_hg - m_ha

Fr = Iα
Fr = I(a/r)
Fr² = Ia
(Fr²/a) = I
I = [r₂(m_hg - m_ha)]/a

This solution is closer, but instead of 'α' in the denominator, there is just 'a'. Perhaps I wrote something down wrong in my lab notebook?

 

[ChiHawksFan]

I tried something else, and I think I am a little closer to a solution.

τ = Fdsinθ, τ = Iα, α = a/r, d = r = the radius of the spool, F = m_hg - m_ha

Fr = Iα
Fr = I(a/r)
Fr² = Ia
(Fr²/a) = I
I = [r₂(m_hg - m_ha)]/a

This solution is closer, but instead of 'α' in the denominator, there is just 'a'. Perhaps I wrote something down wrong in my lab notebook?

Yep, I did write something down wrong. It is supposed to be 'a' in the denominator. Thanks for the help Sammy, and sorry for the error!