# Homework Help: Spool of Thread

1. Jun 8, 2004

### e(ho0n3

Here is another problem I'm having trouble understanding:

A narrow but solid spool of thread has radius R and mass M. If you pull up on the thread so that the center of mass of the spool remains suspended in the air at the same place, (a) what force must you exert on the thread? (b) How much work have you done by the time the spool turns with angular velocity w?

How is the center of mass not moving when I'm pulling on a thread and effectively removing thread (a therefore mass) from the spool. And how can it remain suspended in the air? I just can't picture this. How is it possible to do work when the spool hasn't rotated yet?

2. Jun 8, 2004

### TALewis

First, I think the question probably comes with the assumption that the mass of the thread is negligible. Otherwise, this is a more difficult problem, because you are correct that the mass of the spool would be decreasing.

Now, for part (a):

Draw a free-body diagram. There are two forces acting on the spool: the tension in the thread acting upward (T), and the spool's weight acting downward (mg). You want to pull on the thread such that the spool's mass center will remain motionless. Therefore, the acceleration of the center of mass should be zero. Set up your equations of equilibrium in the vertical direction:

\begin{align*} \sum F = ma_G = 0 &= T - mg\\ T &= mg \end{align}

So the tension in the thread must equal the spool's weight. This might seem counter-intuitive, but the beauty of analyzing a problem such as this is that you can consider the linear effect of the forces independently of the rotation they might also cause.

When visualizing this, note that you can't just hold the end of the thread motionless and expect the spool to also remain motionless. In order to supply enough tension to the thread, you must pull the string upward with sufficient acceleration ($$a = mgR^2/I_G$$, I believe, which equals 2g for *edit* a cylindrical spool).

Part (b):

The center of mass doesn't move, so there is no change in potential energy and no change in linear kinetic energy. The initial rotational kinetic energy is zero. We spin up to an angular velocity $\omega$. We must do work (U) on the spool to change its kinetic energy. Therefore:

$$U &= \frac{1}{2}I_G\omega^2$$

I could be neglecting something--others please correct me if I have made an error--but I believe the situation is as straightforward as the above.

Last edited: Jun 8, 2004
3. Jun 8, 2004

### e(ho0n3

Unfortunately this question didn't come with any assumption. If I would have known this little detail, I wouldn't have posted this question in the first place. I'm starting to get annoyed with my physics book now.

Now you've confused me. Where did you get that acceleration from? I guess it would be better if I think of the spool as a top. I can tie a string around the top, throw it in the air and make it stay at a certain position in the air as I pull the string from it (the good old day...).

Looks right to me.

4. Jun 8, 2004

### TALewis

That acceleration of the thread came from a more complicated expression that I neglected to show because it wasn't necessary to determine the tension in the thread. I'll show it here to back my claim up:

The sum of the moments (torque) about the mass center equals the moment of inertia times the angular acceleration. On the spool, we have the thread tension T acting through a moment arm R:

$$\sum M = TR = I_G\alpha$$

However, we showed that T = mg previously, so we have:

$$mgR = I_G\alpha$$

Additionally, if the mass center is motionless, then the tangential acceleration of the point T where the thread comes off the spool is equal to the radius times the angular acceleration:

\begin{align*} a_T = R\alpha\\ \frac{a_T}{R} = \alpha \end{align}

So the angular acceleration of the spool is equal to the tangential acceleration of point T (which is also the linear acceleration of the thread) divided by the radius. Substituting that into the previous expression and solving for aT:

\begin{align*} mgR &= \frac{I_Ga_T}{R}\\ a_T &= \frac{mgR^2}{I_G} \end{align*}

Furthermore, the moment of inertia of a cylinder about its axis is equal to $$(1/2)mR^2$$. Substituting that into the previous:

$$a_T = 2g$$

I believe that is correct, but I'm still going over it to try to catch any inconsistencies, since it's easy to make mistakes in a problem like this.

Edit: I guess the spool is not necessarily cylindrical, so if you were to calculate the acceleration, you would use the appropriate value for the moment of inertia corresponding to the spool's shape. That isn't given in your problem, so it's unimportant. The thread acceleration for a cylindrical spool is left above for illustration.

Last edited: Jun 8, 2004