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Maximum angular velocity of disc skidding across surface

  • Thread starter rickyjoepr
  • Start date
1. Homework Statement
coefficient of kinetic friction between the disk and the surface is 0.42

gravity = 10.6 m/s
disk mass = 1.75x10^9

the skid marks are 1280m long, This is due to the fact that uneven friction had set the saucer in very slow rotation around its principal axis. By approximating the saucer by a solid disk, what is the maximum angular velocity attained during the skid?

2. Homework Equations

Ke = (.5)(I*w^2)


3. The Attempt at a Solution

fk*d = Ke

(.42)(1.75x10^9)(10.6)(1280) = (1/4)(mr^2)*w^2)

masses will cancel

Sqrt[(4)(.42)(10.6)(1280)(1/r^2)] = w

i get .926

but the listed answer is .23

Not sure If I am approaching this correctly
 
got it

Fk*d = ((.5)mVi^2) + (.5)((Iw^2)

eventually simplified to

Sqrt [ (4/r^2)* (.42*10.6*1280 - (1/2)Vi^2) ] = w
 

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