A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate `(dv)/(dt)=a`. The friction coefficient between the road and the tyre is `mu`. Find the speed at which the car will skid.

Explanation

This is the case of non-uniform circular motion. Here the friction will be inclined to velocity. One component of the friction will provide necessary centripetal force and another component will provide tangential force.
`f sin theta=F_(c)=(mv^(2))/(R )`
`f cos theta=F_(t)=m a_(t)=m(dv)/(dt)=ma_(0)`
`f=sqrt(F_(c)^(2)+F_(t)^(2))=sqrt(((mv^(2))/(R ))^(2)+(ma_(0))^(2))`
`f=sqrt_(max)=muN=mumg`
for no slipping or skidding. we have
`flef_(max)`
`sqrt(((mv^(2))/(R ))^(2)+(ma_(0))^(2))lemumg`
`(v^(4))/(R^(2))+a_(0)^(2)lemu^(2)g^(2)`
`vle[(mu^92)g^(2)-a_(0)^(2))R^(2)]^(1//4)`
Maximum speed for no skid
`v_(max)=[(mu^(2)g^(2)-a_(0)^(2))R^(2)]^(1//4)`
in the context of tangential acceleration:
`V^(2)=mu^(2)+2a_(1)s=0+2a_(0)s`
`s=(v_(max)^(2))/(2a_(0))=(R )/(2)[((mug)/(a_(0)))^(2)-1]^(1//2)`