A cyclist rides along a circular path in a horizontal plane where the coefficient of friction varies with the distance from the centre O of the path as u=u₀(1-r/R) where R is the maximum distance upto which surface is rough. Find the radius of circle with the centre O at which the cyclist can ride with maximum velocity and what is the velocity??

here in this case , if centrifugal force balances the friction then cycle will not slip but

after a particular value of velocity it starts slipping...

m(v_max)² = urg

  (vmax)² = urg/m       ................1

 u = u_o(1-r/R) so

  (vmax)² = u_og[r(1-r/R)]     ..........2

 from above equation we can say , vmax is a function of radius of path ...

now using concept of minima maxima ,

1)diffenertiating the above eq wrt r

2)put d/dr (vmax)² = 0

  after diffenentiating , RHS = u_og(1-2r/R)

   after putting it to 0 , we get r = R/2

this is the maximum possible radius of path for cycle not to slip ...

now , at r = R/2 eq 2 becomes

(v_max)² = u_ogR/4

 (V_max) = (u_oRg)^1/2/2

this is the maximum velocity with which cycle can move...

thnx