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Aiswarya Ram Gupta Grade: 12
        

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??

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

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(vmax)2 = urg


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


 u = uo(1-r/R) so


  (vmax)2 = uog[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 = uog(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


(vmax)2 = uogR/4


 (Vmax) = (uoRg)1/2/2


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


 

6 years ago
Aiswarya Ram Gupta
35 Points
										

thnx

6 years ago
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