#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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??

10 years ago

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...