Force to stop a rotating object

[Phys185Help]

Homework Statement

The 3.2kg , 37-cm-diameter disk in the figure is spinning at 350 rpm .

How much friction force must the brake apply to the rim to bring the disk to a halt in 2.4s?

Homework Equations

F = m * a
I = (1/2) * mass * radius² (I probably need to use this but I can't figure out how it works in)
rotational speed = rotations/sec * circumference

The Attempt at a Solution

I've tried going about it multiple ways, but I think I'm just leading myself in circles

Going to try to solve using F = m * a

Solving for a:
350 rpm / 60 seconds = 5.83 rotations/second
5.83 rotations / second * (2π * .37m / 2) m/rotation = 6.77 m/s

If I want to go from that to 0 m/s in 2.4s then it's (6.77m/s) / 2.4s = 2.82 m/s²

So since F = m * a = 3.2kg * 2.82 m/s² = 9.03 N

Except that answer is wrong.

I would love to use the different rotational formulas but I can't figure out what I'm supposed to use and when to switch from radians to meters and such.

 

[Phys185Help]

Also tried this:

Net torque = I * α
I = (.5 * 3.2kg * .185m²) = .0548 kg*m²
α = ((350rpm / 60 s) - 0) * 2π / 2.4s = 15.27 rad/s

Net torque = .0548 * 15.27 = .8363 Nm

Net torque = F * radius
.8363 = F * .185

F = 4.52 N

This one seems more correct but I've done it wrong so many times I don't know if I'm losing my mind.

Edit: Also this one is exactly 1/2 my other answer so I assume in the other one or in this one I divided/multiplied by 2 somewhere when I shouldn't have.

 

[CWatters]

#2 is on the right track.

Force = mass * acceleration ... is the linear case
Torque = Moment of inertia * angular acceleration ... is the same thing but for rotation.

Net torque = I * α
I = (.5 * 3.2kg * .185m2) = .0548 kg*m2

You didn't post a diagram but I assume it's a disc so I = 0.5*m*r²

Where do you get 0.185m² ?

α = ((350rpm / 60 s) - 0) * 2π / 2.4s = 15.27 rad/s

That should be rad/s² because α is the angular acceleration.