Question

A car is safely negotiating an unbanked circular turn at a speed of 19 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-fourth of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?

Answer 1

V_i (velocity initial) = 19 m/s
u (coefficient of friction) = u/4
V_f (velocity final) = ?

F_netX = F_c + F_f
since F_netX = 0, you can isolate F_c by subtracting F_f

F_c = -(-F_f)
F_c = F_f

F_netY=F_n + F_g
F_netY also equals 0 so you can manipulate to solve for F_n. Fg can also be written as mg and since its going into a negative direction as well as being subtracted, Fg turns into a positive.

F_n = mg

going back to F_c=F_f
we can substitute F_f for umg (coefficient of friction * F_n)
and we can substitute F_c for mv²/r (mass*velocity squared/radius)

Since there are two F_f because there are two different velocities and coefficient of friction, we need to create two different equations

the first equation (still following F_c= F_f) is

m*v_i²/r = umg

and the second is m*v_f²/r = umg/4

For each equation you manipulate to solve for each velocity

v_i = √u*g*r (coefficient of friction*gravity*radius)
v_f = √u*g*r/4

then you take v_f and divide it by v_i, by doing that you also divide √ugr by √ugr/4
(NOTE: √ugr/4 is also √ugr/√4)

this gives you

v_i/v_f = √ugr/4÷√ugr

so √ugr and √ugr cancel out leaving you with 1/√4

v_f/v_i = 1/2

now you have to isolate v_f by multiplying by v_i on both sides

v_f = v_i/2
v_f = (19m/s)/2
v_f = 9.5 m/s