Rolling Paper: Analyzing Motion and Energy

[montreal1775]

Homework Statement

A large, cylindrical roll of paper of initial radius R lies on a long, horizontal surface with the open end of the paper nailed to the surface. The roll is given a slight shove (initial velocity is negligible) and begins to unroll. Determine the speed of the center of mass of the roll when its radius has diminished to r. Assume the roll has uniform density.

Homework Equations

$$K_i + U_i=K_f+U_f$$

The Attempt at a Solution

I have set up the problem using conservation of energy:

$$M_g R = (1/2) m v^2 + (1/2) I \omega^2 + m g r$$

I think this is the proper way to set it up, but I don't know how to find the relationship between the initial mass, final mass, and the radius.

 

[Dorothy Weglend]

That looks good to me.

For your question, while the masses will change, the density of the paper will remain the same. Use that fact.

Dorothy

 

[m2003]

Is there a way to determine this without knowing the specific values?

I would first clarify any assumptions that are being made in this problem. For example, is the roll of paper assumed to be a perfect cylinder with no friction or air resistance? Is the surface it is rolling on perfectly smooth? These factors could affect the solution.

Assuming that the roll of paper is a perfect cylinder with uniform density and that the surface it is rolling on is frictionless, we can use the conservation of energy equation you have set up. The initial potential energy is equal to the final kinetic energy and final potential energy, since the paper starts at rest and ends at a lower height. However, we also need to consider the rotational kinetic energy of the roll.

To find the relationship between the initial mass, final mass, and radius, we can use the fact that the volume of the roll remains constant. The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height. Since we know that the height of the roll decreases from R to r, we can set up the equation:

πR^2h = πr^2h'

Where h' is the final height of the roll. We can then solve for h' in terms of R, r, and h:

h' = (R^2/r^2)h

Since the density of the roll is uniform, the mass is directly proportional to the volume. Therefore, we can say that the initial mass (M) is equal to the final mass (m) multiplied by the ratio of the initial and final heights:

M = (R^2/r^2)m

Plugging this into the conservation of energy equation, we get:

Mgh = (1/2)mv^2 + (1/2)Iω^2 + mgr

Substituting in the relationship between M and m, and simplifying, we get:

(R^2/r^2)mgh = (1/2)mv^2 + (1/2)(1/2)mR^2ω^2 + mgr

Solving for v, we get:

v = sqrt(2gh(1-(r^2/R^2)) + (R^2/r^2)ω^2)

This is the speed of the center of mass of the roll when its radius has diminished to r. It includes both the translational and