- #1

Nathanael

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_{0}). Now we allow the carpet to unroll to a radius R

_{f}< R

_{0}.

We can model the mass of the cylindrical part of the carpet as M = kR

^{2}

The differential work done by gravity would be dW = -MgdR = -kR

^{2}gdR

This leads to the total work done by gravity W = ∫dW = -(kg/3)Δ(R

^{3})

But if we consider just the initial and final states: We start with mass M

_{0}=kR

_{0}

^{2}a distance R

_{0}above the ground and end with M

_{f}=kR

_{f}

^{2}a distance R

_{f}above the ground.

So the (negative of the) change in gravitational potential energy is M

_{0}gR

_{0}-M

_{f}gR

_{f}= kg(R

_{0}

^{3}-R

_{f}

^{3}) = -kgΔ(R

^{3}) which is three times what the work done by gravity was.