# Work-Energy Confusion

1. Apr 16, 2012

### eurekameh

Shouldn't there be negative work done by the string?

2. Apr 16, 2012

### collinsmark

As it turns out, no.

There would be [negative] work done by the string if the disk's axle was sliding on the string, causing the system to heat up. But that's not what's happening here. Whenever and wherever the string and disk are in contact, there is no relative motion between the string and disk. Given that fact, how does W = F·s apply?

3. Apr 17, 2012

### eurekameh

Thanks. That makes so much more sense. Am I right in saying that the only thing the string is doing is converting some of the work of gravity into rotational kinetic energy, and if there was no string to begin with, the disk would translate vertically at a faster speed than if the string was there to convert some of the work to rotational kinetic energy?

4. Apr 17, 2012

### collinsmark

I suppose that's a valid way to look at it. The relationship of v = (0.1 m)ω in this case is due to the string, in part. And the tension on the string also contributes to the sum of linear forces acting on the disk (which explain's why the disk's linear acceleration is less than g). The string just doesn't do any work though.

A ball of radius r rolling on a hard, flat surface (ignoring air resistance) will continue rolling indefinitely. No work is being on the ball or by the ball. Yet it is the force of static friction that keeps the ball rolling, as opposed to sliding, and thus plays a role in determining the v = ωr relationship.

Last edited: Apr 17, 2012
5. Apr 18, 2012

### eurekameh

But if it was sliding instead, the force of kinetic friction would be doing negative work?
Also, if there was absolutely no friction at all, the ball would still be going indefinitely, but without rotation, right?

6. Apr 18, 2012

Right.