# Tissue paper roll problem

1. Mar 17, 2004

### Houyhnhnm

OK here's a question which I really, really can't understand.

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 (initial velocity is about zero) and commences to unroll. Determine the speed of the center of mass when its radius has diminished to r = .14cm assuming R is 6.1cm.

Somebody told me to use angular momentum, but how can I when the initial velocity is zero? I'm quite confused. Also, they give that g = 9.80 m/s^2, why would we need that?

2. Mar 17, 2004

### NateTG

Do you know about conservation of energy?

3. Mar 17, 2004

### Houyhnhnm

Is there such thing as rotational potential energy?

4. Mar 17, 2004

### ShawnD

More info is needed. You can't solve this without at least knowing the initial velocity (after it's pushed).

Last edited: Mar 17, 2004
5. Mar 17, 2004

### Houyhnhnm

Last edited by a moderator: Apr 20, 2017
6. Mar 17, 2004

### NateTG

$$\lim_{\vec{v}_0 \rightarrow \vec{0}}\vec{v}$$ exists, so you can just assume that the inital velocity is arbitrarily small.

No, there is however, rotational kinetic energy --
$$KE_{rot}=\frac{1}{2}I\omega^2$$

The moment of inertia of the cylinder is $$\frac{1}{2}MR^2$$

One of the problems is that the problem is unclear about which center of mass it wants to know about (whether to include the dropped paper or not).

Leaving the paper behind makes things significantly more complicated, so I'll use the other answer.

This leads to the following equation:
$$mhg=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2$$
which has two unknowns, so we need a second equation which relates the velocity of the center of gravity to the angular speed.
If we assume that the toilet paper has negligible thickness, then we get:
$$\omega=\frac{v}{r}$$
Applying the formula for moment of inertia yieds
$$I=\frac{1}{2}mr^2=\frac{1}{2}M\frac{r^3}{R^2}$$
The change in height of the center of gravity:
$$h=R-\frac{r^3}{R^2}$$
If we plug things back in we get:
$$Mg(R-\frac{r^3}{R^2})=v^2(\frac{r^2}{4R^2}M+\frac{1}{2}M)$$
the $$M$$'s cancel
$$g(\frac{R^3-r^3}{R^2})=v^2(\frac{r^2+2R^2}{4R^2})$$
The denominators cancel
$$\sqrt{\frac{g(R^3-r^3)}{r^2+2R^2}}=v$$
This is the velocity of the center of gravity. The velocity of the center of the cylinder is
$$\frac{R^2}{r^2}\sqrt{\frac{g(R^3-r^3)}{r^2+2R^2}}$$

I come up with roughly $$40 m/s$$ for the small cylinder, and roughly $$2 m/s$$ for the center of gravity of the whole roll. YMMV, and you should check to make sure what I did makes sense, and whether you can catch the math errors I snuck in.