gaborfk
Nov14-04, 02:52 PM
I am kind of stuck with this problem. Any takers?
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? Thank you in advance!!
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? Thank you in advance!!