Finding the Angular Inertia of a Pulley/Block system.

[GodPlaysDice]

Homework Statement

I have tried this thrice now, to no avail. Please help me. Here is the problem:
"In the figure, block 1 has mass m1 = 430 g, block 2 has mass m2 = 580 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.9 cm. When released from rest, block 2 falls 74 cm in 5.0 s (without the cord slipping on the pulley). What is the pulley's rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s^2."

EDIT:
Both blocks are moving vertically, the lighter one is being raised by the heavier one by means of the pulley.
That centimeter difference is probably throwing everything in my calculations out of wack. I'll see if that was the problem. Thank you.

Homework Equations

F = ma
τ = Iω
a = Rα
Δx = V_ot + 0.5at²
ω = ω_o + αt

The Attempt at a Solution

Here is my attempt (one of many):
(I converted everything into SI base units)
I first found the acceleration:
Δx = 0.75
0.75 = 0.5a(5)
and found that a = 0.3
Next, I found α:

a = Rα
α = 6.1224

Then ω:

ω = 0 + 6.1224(5)
ω = 30.6

Next I drew a free body diagram for mass 2 and an extended FBD for the pulley and wrote the 2nd law equations for each assuming that the direction of the motion of mass 2 is positive.

Pulley:
(tension is T)
T = Iα

Block:
mg - T = ma

I added the two equations, yielding:

mg = Iα + ma

I plugged in the numbers, and solved for I:
I =
but the website I'm using is telling me I'm wrong.
I must know WHY?!
Thank you for any light you may shed on my errors.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[hjelmgart]

Hmm, for starters you wrote Δx = 0.75, but it's 0.74 (not sure if you noticed this).

Also how does the blocks move depending on each other? Is one moving in the horizontal axis and the other on the vertical axis? Or are they both moving in the vertical axis (as I understood it).

Also perhaps I should note, that the first thing that came to my mind, when reading this problem, was conservation of energy. I am so far not sure, if you could actually solve it the way you did though, but thinking about the energies involved here, may be easier.

 

[collinsmark]

Homework Statement

I have tried this thrice now, to no avail. Please help me. Here is the problem:
"In the figure, block 1 has mass m1 = 430 g, block 2 has mass m2 = 580 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.9 cm. When released from rest, block 2 falls 74 cm in 5.0 s (without the cord slipping on the pulley). What is the pulley's rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s^2."

EDIT:
Both blocks are moving vertically, the lighter one is being raised by the heavier one by means of the pulley.
That centimeter difference is probably throwing everything in my calculations out of wack. I'll see if that was the problem. Thank you.

Homework Equations

F = ma
τ = Iω

Red emphasis mine.

I don't think you meant ω there. It should be a different variable.

a = Rα
Δx = V_ot + 0.5at²
ω = ω_o + αt

The Attempt at a Solution

Here is my attempt (one of many):
(I converted everything into SI base units)
I first found the acceleration:
Δx = 0.75

As hjelmgart already pointed out, the distance traveled is 0.74 m, not 0.75. [Edit: unless there's a typo in the problem statement.]

0.75 = 0.5a(5)

Also, you're missing something in the use of this equation above. (Hint: You need to square something.)

and found that a = 0.3
Next, I found α:

a = Rα
α = 6.1224

Then ω:

ω = 0 + 6.1224(5)
ω = 30.6

Next I drew a free body diagram for mass 2 and an extended FBD for the pulley and wrote the 2nd law equations for each assuming that the direction of the motion of mass 2 is positive.

Pulley:
(tension is T)
T = Iα

[Edit: Red emphasis mine again. Don't you mean torque, τ, there?]

Block:
mg - T = ma

Before moving on, you'll need to solve for both tensions. There is a tension on block 1 and a different tension on block 2. They're not the same value.

If the pulley was mass-less, the tensions would be equal (the tension on block 1 would be the same as on block 2). But that's not the case here because the pulley has mass. The two tensions must be calculated separately.

Once you've calculated both the tensions, you can find the net torque on the pulley. And since you already know the pulley's angular acceleration, you can then calculate its moment of inertia (a.k.a. rotational inertia).