Algebra of pulley-mass energy equation

[PacFan01]

Homework Statement

[/B]
A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disc of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.30 m off the ground. What speed does this block have when it hits the ground?

Homework Equations

ΔE = 0
E_i = E_f (closed system)
E = K + U^G + E_th + E_s
U_i^G = K_transf + K_rotf
K_trans = ½mv²
K_rot = ½Iω²
I_{Solid Cylinder} = ½mr²
ω = v/r

The Attempt at a Solution

[/B]
First, I step up the energy equation and substitute in my energy expressions. I've used the subscript "1" to represent the 0.20-kg mass, the subscript "2" to represent the 0.25-kg mass, and the subscript "p" to represent the pulley.

U_1i^G + U_2i^G = U_1f^G + U_2f^G + K_1f + K_2f + K_pf

m₁gh_1i + m₂gh_2i = m₁gh_1f + m₂gh_2f + ½m₁v₁² + ½m₂v₂² + ½Iω²

I know what the next step is, but I'm not sure how to get there algebraically.

I should end up with:

(m₁ - m₂)gΔh = ½(m₁ + m₂)v_f² + ¼m_pv_f²

I understand that v₁, v₂, and v_p are equal, so no problem there. I see that how we substitute ½mr² for I, and v/r for ω. I don't understand how we end up with (m₁ - m₂)gΔh. I understand that we pull the common variable g out, and I can kind of understand how the h-values become Δh. However, it looks like this when I set it up:

m₁gh_1i + m₂gh_2i - m₁gh_1f - m₂gh_2f = ½m₁v₁² + ½m₂v₂² + ½Iω²

Which becomes:

g(m₁h_1i + m₂h_2i - m₁h_1f - m₂h_2f) = ½m₁v₁² + ½m₂v₂² + ½Iω²

Any help is greatly appreciated.

 

[ehild]

The length of the rope does not change. If the 0.25 kg block descends by 0.3 m, the other block rises by 0.3 m.
(h_1f-h_1i= h_2i-h_2f)

 

[PacFan01]

Thanks for the reply, ehild.

I initially responded with another question, but I think I understand now.