Finding Frictional Force and Time to Stop a Spinning Disk

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The discussion revolves around calculating the frictional force and the time required for a spinning disk to come to rest on a flat surface. The frictional force is determined using the equation Ff = μN, where N is the normal force, which equals the gravitational force acting on the disk. Participants express confusion about the direction of forces, particularly regarding the angular velocity (ωi) and angular acceleration (α), and whether they can assume specific directions for these variables. The calculations presented involve torque (τ), moment of inertia (I), and the relationship between friction and angular motion, leading to a derived formula for time to stop the disk. The final formula suggests that the time to stop depends on the initial angular speed, the radius of the disk, and the coefficient of kinetic friction.
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A uniform disk of radius R and mass M is spinning with angular speed ωi. It is placed on a flat horizontal surface; the coefficient of kinetic friction between the disk and surface is μk.
a) Find the frictional force on the disk.
b) How long will it take for the disk to come to rest.

I'm confused about the directions of the forces.
I know mg will be down. And there will be no forces in the x direction.

Ff = μN

Will the normal force be up, opposite and equal to the force of gravity, or will the rotation change that?

Since we don't know the direction the disk is spinning can we find the direction of ωi and α?

Can we just choose for the direction of ωi and α to be downwards from the torque of the disk?

Thanks
 
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If I have τ = Iα
Can I assume α is down?

so I can have:
ΣF = τ - f

or would it be:
ΣF = τ + mg - f - N

Iα + mg = f + N
I = ω/R
ωα/R + mg = f + N
 
another try:

τ = f x R
τ = Iα
τΔt = I(ω - ωi)
fRΔt = (1/2)MR^2(ω - ωi)
fΔt = (1/2)MR(ω - ωi)

Δt = (1/2)MR(ω - ωi) / μkFn
Δt = MR(ω - ωi) / 2μkMg
Δt = R(ω - ωi) / 2μkg

is this correct?

thanks
 
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