What is the direction of friction when a spool of wire is rolling on the ground?

[pradeepk]

In class the other day, my professor began talking about spools of wire on the rolling on the ground and when drawing their free body diagrams, he sometimes made the friction acting in the same direction as the motion of the spool. This really confused me, and I was just wondering if anyone could tell me, when friction acts in the same direction as the motion?
Thanks a lot.

 

[P. Brien]

It's actually depends to:
- Inertia of the spool (1/2 MR or the others)
- The location and direction of the tension force to the center mass of the spool
by some mathematics algebra from torque equation, force equation, and the relation of the angular acceleration and the acceleration you will get value of critical radius of spool that the direction of the friction depends on the radius whatever it's more than or less than the critical radius.
However, I'm sorry It's hard to me to write all of the step in here.

 

[rcgldr]

The only cases where rolling friction acts in the direction of motion is when there is deceleration due to an external force, such a gravity on a spool if it's rolling uphill.

 

[tiny-tim]

welcome to pf!

hi pradeepk! welcome to pf!

suppose the wire is being pulled to the right …

if the wire comes out under the spool, then the spool must move to the left, so the direction of friction has to be to the left!

if the wire comes out over the spool, then the spool must move to the right, so that doesn't help us much

so consider whether the torque from the wire makes the bottom of the spool want to move slower or faster than it actually is moving

(btw, the direction of friction on the non-driving wheels of a car always opposes the acceleration of the car)

 

[pradeepk]

hi pradeepk! welcome to pf!

suppose the wire is being pulled to the right …

if the wire comes out under the spool, then the spool must move to the left, so the direction of friction has to be to the left!

if the wire comes out over the spool, then the spool must move to the right, so that doesn't help us much

so consider whether the torque from the wire makes the bottom of the spool want to move slower or faster than it actually is moving

(btw, the direction of friction on the non-driving wheels of a car always opposes the acceleration of the car)

Ok I think I understand now. So in the first example, the force you are pulling with is to the right, however the spool moves to the left, so there must be some force causing it to go in that direction, and that force is friction? In the second example, it moves in the direction of the applied force, so it is just normal circumstances. Is that correct?

 

[rcgldr]

If the wire is being unwound from the top, the direction of the force depends on the relative radius of the wire on the spool and also the ratio of mass of wire to mass of spool. If the radius is zero, it's the same as the spool being pulled from the center and the friction opposes tension. If the radius is large enough, then the friction is in the same direction as the tension.

For this mathematical example, assume the wire is massless, that the spool has the same angular inertia as a solid uniform cylinder, that the spool is rolling to the right, and that positive friction force means to the right.

R = radius of spool
r = radius of wire
t = tension
f = friction force (positive means same direction as tension)
m = mass
a = linear acceleration
α = angular acceleration
τ = torque
I = angular inertia = 1/2 m R²
c = r/R
a = -α R (rolling to right is clockwise)
α = -a / R

a = (f + t) / m
τ = f R - c R t = α I
f R - c R t = α (1/2 m R²)
f R - c R t = -a (1/2 m R)
(2f - 2ct) / m = -a = -(f+t) / m
3f = (2c -1) t
f = (2c - 1) / 3

In this case, if r/R is < 1/2, friction force is to the left (negative), if r/R = 1/2, friction force is zero, and if r/R > 1/2, then friction force is to the right (positive).