Originally posted by mtthwReally? Strange. I'll have to try it.
Friction isn't propelling anything. Pulling the string is doing the propelling, so the spool moves in that direction. Friction just stops it spinning in place.
I know I'm on pretty safe ground here, because I've just done it with a cotton-reel on the table in front of me. Trust me, the spool winds up 🙂
IF the string rolls itself up, there's more propelling the spool than just the force on the thread. Otherwise, how come the spool moves FASTER than the end you're holding? It has to if it winds up.
I'll have to look into this.
OK, let's try some maths.
Mass of spool is m.
Moment of inertia of spool is I.
External radius (distance from centre to table) of spool is b
Internal radius (distance from centre to the point of contact of the thread) is a
Force pulling the thread is P (and let's assume this is to the right)
Magnitude of frictional force is F
Net force on spool is P - F, so acceleration is (P - F)/m
Net moment @ centre is Pa - Fb (+ve = anti-clockwise), so angular acceleration is (Pa - Fb)/I
If friction is great enough to prevent slipping, then the acceleration of the point of contact with the table is zero.
(P - F)/m + (Pa - Fb)/I = 0
Rearrange to get F = P(I + am)/(I + bm)
Therefore the net force is
P[1 - (I + am)/(I + bm)]
= Pm(b - a)/(I + bm)
> 0 (because b > a)
So there is a net force to the right.
Similarly, the net moment @ the centre is:
P[a - b(I + am)/(I + bm)]
= PI(a - b)/(I + bm)
< 0
So there is a net moment in a clockwise direction - winding the spool up.
Just noticed a couple of errors in my working, although they don't change the conclusions.
Acceleration of the point of contact is zero
=> (P - F)/m + (Pa - Fb)b/I = 0
=> F = P(I + abm)/(I + b^2m)
=> Net force is Pmb(b-a)/(I + b^2m) > 0
and Net moment is PI(a - b)/(I + b^2m) < 0