A motorcyclist is describing a circle of radius 25 m at a speed of 5 m/s . Find the inclination with the vertical and the coefficient of friction between the tyres and ground.

Explanation

Data : v = 5 m/s , r= 25 m , g = 9.8 `m//s^(2) `
(i) `tan tehta= v^(2)/(rg) = ((5)^(2))/(25 xx 9.8) = 1/(9.8) = 0.1021`
`theta = tan^(-1) 0.1021 5^(@) 50` ( inclination with the vertical)
(ii) Centripetal force = Force of friction
` (mv^(2))/r= mu_(s)mg " " (mu_(s)` is the coefficient of friction )
` mu_(s) =v/r = 7/5 =1.4 rad//s OR omega= sqrt(g/r) = sqrt(9.8/5) = sqrt(1.96) = 1.4 " rad/s"`