Respuesta :
When a vehicle is moving on rough unbanked road then the centripetal force on the vehicle is provided by friction force to take a safe turn
now we can write
[tex]F_f = \frac{mv^2}{R}[/tex]
[tex]\mu m g = \frac{mv^2}{R}[/tex]
[tex]v^2 = \mu R g[/tex]
here
v = 25 m/s
R = 170 m/s
now we can find the friction coefficient by above equation
[tex]25^2 = \mu * 170 * 9.8[/tex]
[tex]\mu = 0.375[/tex]
so minimum friction coefficient needed is 0.375
Part b)
now again if we have friction coefficient is half of this what we calculated above
then we can find the safe speed by same equation
[tex]\mu = \frac{0.375}{2} = 0.1875[/tex]
[tex]v^2 = 0.1875* 170 * 9.8[/tex]
[tex]v = 17.67 m/s[/tex]
so here maximum safe speed is 17.67 m/s
Explanation:
Radius of the curve, r = 170 m
Speed of the car, v = 25 m/s
(a) Let [tex]\mu[/tex] is the coefficient of static friction that will prevent sliding. The centripetal force is balanced by the force of static friction as :
[tex]\dfrac{mv^2}{r}=\mu mg[/tex]
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(25)^2}{170\times 9.8}[/tex]
[tex]\mu=0.375[/tex]
(b) Let v' is the maximum speed of the car so that it can round the curve safely if the coefficient of static friction between the tires and pavement is only one-third of what you found in part (a).
It is calculated as :
[tex]v'=\sqrt{\dfrac{1}{3}\mu rg}[/tex]
[tex]v'=\sqrt{\dfrac{1}{3}\times 0.375\times 170\times 9.8}[/tex]
v' = 14.43 m/s
Hence, this is the required solution.
