How Does a Decreasing Radius Affect the Speed of a Rotating Cylinder?

[gaborfk]

Please, help with rotating rigid body problem

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!

 

[wais]

To solve this problem, we need to use the conservation of energy principle, which states that the total energy in a system remains constant. In this case, the initial energy of the system is solely potential energy, given by mgh, where m is the mass of the roll, g is the acceleration due to gravity, and h is the height of the roll.

As the roll unrolls, the potential energy is converted into rotational kinetic energy and translational kinetic energy, given by (1/2)I(omega)^2 and (1/2)mv^2, respectively.

Since the roll is unrolling on a horizontal surface, there is no change in height (h) and thus no change in potential energy. This means that the initial potential energy is equal to the final kinetic energy, given by (1/2)I(omega)^2+(1/2)mv^2.

We can rewrite this equation as mgh = (1/2)I(omega)^2 + (1/2)mv^2, and since the mass cancels out, we can solve for v to find the speed of the center of mass of the roll when its radius has diminished to r = 1.0 mm.

v = sqrt((gh - (1/2)I(omega)^2) / (1/2)r^2)

To find the moment of inertia (I) of the roll, we can use the formula for a solid cylinder, given by (1/2)MR^2. Plugging this into the equation, we get:

v = sqrt((gh - (1/4)MR^2(omega)^2) / (1/2)r^2)

Now, we need to find the value of (omega) at r = 1.0 mm. Since (omega) = v/r, we can plug in v = 0 (since the roll is initially at rest) and r = 6.0 m (given in the problem).

This gives us (omega) = 0/6.0 = 0 rad/s.

Plugging this into the previous equation, we get:

v = sqrt((gh - 0) / (1/2)(0.001)^2)

Simplifying, we get:

v = sqrt(2gh * 10^6)

Finally, we can plug in the given values of g = 9.8 m/s^2 and