What did I do wrong in this college-level physics problem?

Question

Hi,
I'm in the middle of a physics homework due tomorrow, and I'm stuck on this specific problem. The result I'm getting doesn't make any physical sense, afaik.
The problem is the following:
There is a 1.0 kg mass at the edge and on top of a rotating disk. The disk is revolving at 40 rpm and has a diameter of 1.2 m. What is the minimum coefficient of friction required for the mass to remain on the disk?
Here's what I've written so far: http://i.imgur.com/LgmCygE.jpg
Here's how I defined the variables I use in the above document:
F_fic is the fictitious force that pulls the mass opposite the center of the the disk,
f is the friction,
N is the normal force,
µ is the coefficient of friction (this is what we're looking for),
m is the mass that's on top of the disk,
g is the standard gravity,
a_c is the centripetal acceleration,
a_ref is the acceleration of the reference frame,
v is the tangential speed,
R is the radius of the disk,
ω is the angular speed.
In a nutshell, here's my reasoning:
The normal force is equal and opposite the weight of the mass, therefore:
N = mg
The friction is proportional to the normal force:
f = µN
In reference to the disk, the mass doesn't accelerate, therefore:
ma = ∑F ⇒ 0 = µmg - fictitious force ⇒ µ = F_fic / mg
The fictitious force is equal and opposite the mass multiplied by the acceleration of the referential, the latter being the centripetal acceleration. Therefore :
F_fic = -m*a_ref = -m*a_c = -mv²/R
The value of the tangential speed isn't given, but we know the value of the angular speed, therefore:
v = Rω, and ω = 40 rpm*2π/60s = 4π/3 rad/s
If we put all this together we get this:
µ = F_fic/mg = (-mv²/R)/mg = (-m(Rω)²/R)/mg = -Rω²/g
If we put the numbers in:
µ = -Rω²/g = -(0.6 m)(4π/3 rad/s)²/(9.8 m/s²) = -1.07
When I put the numbers in the formula, I get a result of -1.07, which seems impossible for a coefficient of friction (how can it be negative?). Strangely enough, several other classmates got the same answer through different methods and no one could figure out what was wrong.
Could someone please explain what I don't get or what I did wrong?
Thanks :)

? · Accepted Answer

Rather than trying to figure out what you did wrong (why in the world are you using a fictitious force here--I don't see any way in which it is helping) I'll show you how I would do it.  If I get the same answer, you'll know you're right, and if I get a different one, I'll let you figure out what the difference in our approaches is.

Use Newton's Second Law.  Sketch a Free Body Diagram of the mass.  It will have a gravitational force mg downward, a normal force n upward, and a frictional force f toward the center of the disk.  Choose a coordinate system. I strongly suggest choosing one axis to point in the direction of the known acceleration--the other must be perpendicular.  I'll let the "c" (for center) axis point toward the center of the disk, and I'll let the y axis point upward.

Write out Newton's Second Law for each axis, referring to the FBD to get the correct signs.  I'll show those signs explicitly, so symbols represent magnitudes.  First, the y direction.

ΣF_y = n - mg = m(a_y) = 0

n = mg

So far, you and I agree :)  Next, the "c" direction.

ΣF_c = f = m(a_c) = m v²/r = m rω²

The friction force is proportional to the normal force,

μ mg = m rω²

so

μ = rω²/g

Our signs are different, because ... OK, here's your sign error:

ma = ∑F ⇒ 0 = µmg - fictitious force ⇒ µ = F_fic / mg
 The fictitious force is equal and opposite the mass multiplied by the acceleration of the referential, the latter being the centripetal acceleration. Therefore :
 F_fic = -m*a_ref = -m*a_c = -mv²/R

when you said "µmg - fictitious force", you accounted for the direction of the fictitious force by showing the sign explicitly.  When you later say "F_fic = -m*a_ref", you shouldn't have included a negative sign again--you'd already accounted for the direction of the fictitious force.

Anyway, substitute known values.

μ = rω²/g = (0.6 m) [ (40 rev/min)(2π rad/rev)(1 min / 60 s) ]² / (9.81 m/s²) = 1.07

which does seem unusually large, but it isn't actually impossible.  You may recall that the coefficient of static friction between automobile tires and dry asphalt is about 1.

? · Answer

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What did I do wrong in this college-level physics problem?

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