- #1
- 1,650
- 246
Suppose we have a carpet rolled (along it's length) into the shape of a cylinder (of radius R0). Now we allow the carpet to unroll to a radius Rf < R0.
We can model the mass of the cylindrical part of the carpet as M = kR2
The differential work done by gravity would be dW = -MgdR = -kR2gdR
This leads to the total work done by gravity W = ∫dW = -(kg/3)Δ(R3)
But if we consider just the initial and final states: We start with mass M0=kR02 a distance R0 above the ground and end with Mf=kRf2 a distance Rf above the ground.
So the (negative of the) change in gravitational potential energy is M0gR0-MfgRf = kg(R03-Rf3) = -kgΔ(R3) which is three times what the work done by gravity was.
We can model the mass of the cylindrical part of the carpet as M = kR2
The differential work done by gravity would be dW = -MgdR = -kR2gdR
This leads to the total work done by gravity W = ∫dW = -(kg/3)Δ(R3)
But if we consider just the initial and final states: We start with mass M0=kR02 a distance R0 above the ground and end with Mf=kRf2 a distance Rf above the ground.
So the (negative of the) change in gravitational potential energy is M0gR0-MfgRf = kg(R03-Rf3) = -kgΔ(R3) which is three times what the work done by gravity was.