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[Sixteen]

Sorry for making an entire thread for this, but I'm completely stuck and didn't think this fit in the Math help thread.

The question asks: Given a 4.0 kg block 5.0 m above the ground and a 2.0 kg block on the ground connected by a rope hanging from a pulley with radius 0.16 m and moment of inertia 0.56 (kg*m^2), find the speed of the 4.0 kg block the moment it hits the ground using energy methods

What I've got so far:
PE_i = PE_f + KE + RE
m₁gh = m₂gh + (1/2)m₁v² + (1/2)m₂v² + (1/2)I(omega²)
4*5*9.81 = (2*5*9.81) + ((1/2)*(4)*(v^2) + (1/2)*(2)*(v^2)) +  ((1/2)*I*(omega^2)) (v^2 is the same for both blocks)
10g = (3v²) +  ((1/2)*I*(omega^2))

The problem I'm having is with the angular velocity omega. Using the equations:
omega² = 0² + 2 * angular acceleration * delta theta
delta theta = 5 / r
angular acceleration = ||torque|| / I
||torque|| = ||<-r,0,0> x <0,-4g,0> + <-r,0,0> x <0,-2g,0>|| = 2rg

I get:
omega² = 2*(2rg/I)*(5/r) = 20g/I

Problem is, plugging that value back into the conservation of energy equation gives:
10g = (3v²) + (1/2)*I*(20g/I), or
10g = (3v²) + 10g
which doesnt work out.

I've found the answer in the back of the book, and it's 2.65 m/s, but I have no clue how they got this answer.
By converting angular velocity to tangential velocity (since the rope doesn't slip), I got 2.99, which is pretty close, but still wrong, and I still have no clue where to go.

Could anyone help me out? Thanks for your time.

[Vector]

Where's your diagram?

[Sixteen]

From masteringphysics, just in case they have something about reposting their content.

If you meant my free-body diagram, in terms of the pulley I have the tension of the rope on the left = 4*9.81 and the tension of the rope on the right = 2*9.81, pointing downwards on opposite sides. Gravity shouldn't affect the pulley itself, so I left that out. I have counterclockwise being positive torque and the origin being the ground at the left.

I have the feeling I'm missing something to do with signs, but I've redone the equations several times and I still can't find it.

[Draignean]

Uhhh... *Eyes glaze over*

You did NOT have to do that much work. I have no idea what you did wrong in your math, but I can show you a faster, much easier, path to the correct answer.

Recall that I = M(r^2)*0.5 (For a disc)

And Omega = v/r

so I*Omega^2 = (M(r^2)*0.5)*((v^2)/(r^2)) And after cancellation = M(v^2)*0.5

Since we have the radius and the moment of inertia, we can find the mass. It's about 43.75, if you don't mind rounding errors.

Plugging in M(v^2) for I*Omega we get

4*5*9.81 = (2*5*9.81) + ((1/2)*(4)*(v^2) + (1/2)*(2)*(v^2)) +  ((1/4)*43.75*(v^2))

10g = 3v^2 + 10.94v^2

(10g/13.94) = v^2

After multiplication and division we get  sqrt(7.04) = 2.65 m/s

EDIT: Why did you take a cross product for your torque when all you needed was magnitude? It looks fine, (That part. But I generally dislike torques.) but it seems like you're doing a lot of unecessary work.

[MonkeyHead]

^^^ The above method is correct, and also identifies where you went wrong.

[Sixteen]

Huh. Thank you so much, I ran in to a bunch of similar questions and I was approaching those the wrong way too, kept on looking for where I went wrong in the equation itself.

For the cross products, I just went for magnitude my first time through, but after the results came out wrong I went back through all the parts that I simplified looking for what could have messed it up.

Again, thanks so much, it would've taken me forever to get it right on my own.

[Zrk2]

Just always set up your FBDs, and it should all flow from there.