Please, still need help with rotating rigid body problem

[gaborfk]

:cry: I am still stuck with this problem. Please, help if you can.
A large, cylindrical roll of tissue paper of initial radius R lies on a long, horizontal surface with the outside end of the paper nailed to the surface. The roll is given a slight shove (V initial = 0) and commences to unroll. Assume the roll has a uniform density and that mechanical energy is conserved in the process. Determine the speed of the center of mass of the roll when its radius has diminished to r = 1.0 mm, assuming R = 6.0 meters.

This is what I got so far.

KE=(1/2)I(Omega)^2

for the cylinder I is :(1/2)MR^2

So mgh=(1/2)I(Omega)^2+(1/2)mv^2

I plugged in I

mgh=(1/2)(1/2)MR^2(Omega)^2+(1/2)mv^2

Then I used (omega)=v/r to get here

mgh=(1/4)MR^2(v/r)^2+(1/2)mv^2 the masses cancel

gh=(1/4)R^2(v/r)^2+(1/2)v^2

Now what? How do I get h? Did I miss anything? Thank you in advance!!

[arildno]

1. At the start, the total system has only potential energy, MgR, where M is the total mass of the paper.
2. At time "t", you should find that the mass of the remaining rolled-up paper, m, is:
m(t)=(\frac{r(t)}{R})^{2}M
3. The potential energy of the total system remaining at time "t" is:
mgr(t)=Mg\frac{r(t)^{3}}{R^{2}}
4. The kinetic energy of the total system at time "t" is:
\frac{1}{2}m(t)v^{2}_{G}(t)+\frac{1}{2}(\frac{1}{2 }m(t)r(t)^{2})\omega(t)^{2}=\frac{3}{4}(\frac{r(t) }{R})^{2}Mv^{2}_{G}(t)
by using the rolling condition to relate the angular velocity to the center of mass of the rolled up paper.
5. Hence, you get:
v_{G}(t)=\frac{\sqrt{4g(R^{3}-r^{3}(t))}}{\sqrt{3}r(t)}