Torque question, Newton's second law for rotaion?

Question

Auniform horizontal disk of mass M and radius R is rotating about the vertical axis through its center with an angular velocity ώ. when it is placed on a horizontal surface, the coefficient of kinetic friction between the disk and the surface is μk . (a) Find the torque dτ exerted by the force of friction on a circular element of raduis r and width dr. (b) Find the total torque exerted by friction on the disk. (c) Find the time required for the disk to stop rotating.

Answer 1

For a disk uniformly resting on a surface consider the infinitesimal element in polar coordinates

rdrdθ. If the disk has a mass M than the normal force on this area is M/(πR^2)rdrdθ. The friction force resisting this motion is μN or
Mμ/(πR^2)rdrdθ. Infinitesimal torque dτ=rMu/(πR^2)rdrdθ

total torque is found by integrating this expression from r=0 to R and for θ=0 to 2π

∫∫Mu/(πR^2)r^2drdθ=2/3uMR

for a disk of mass M, I=MR^2/2

to get the stopping time first find the angular deceleration,

the force of deceleration has to equal the friction force.

Iα=2/3uMR

α=(2/3uMR)/I=(2/3uMR)/(MR^2/2)=4/3 u/R

The time to stop rotating t=ω/α=3Rω/(4u)

Answer 2

That's easy, read the book...