Finding Frictional Force and Time to Stop a Spinning Disk

  • Thread starter Thread starter Destrio
  • Start date Start date
  • Tags Tags
    Disk Spinning
AI Thread Summary
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.
Destrio
Messages
211
Reaction score
0
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
 
Physics news on Phys.org
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top