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`        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

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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