Question

A car rounds a banked curve as we will discuss in class on Tuesday. The radius of curvature of the road is R and the banking angle is θ. (a) In the absence of friction, what is the safe speed for the car to take this curve? (b) Now assume the coefficient of friction between the car’s tires and the road is µs. Determine the range of speeds the car can have without slipping up or down the road. (c) What is the minimum value of µs that makes the minimum speed zero? (d) If θ = 25.0 ◦ , for what values of µs can the curve be taken at any speed? Note: The upper limit of µs you will find is practically impossible to achieve for the car’s tires and the road.

Answer 1

A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) µ_s = tan θ

D) µ_s = 0.4663

Step-by-step explanation:

A) The forces acting on the car will be;

Force due to friction; F_f

Force due to Gravity; F_g

Normal Force; F_n

Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.

Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n

Thus, sum of forces about the vertical j^ direction gives;

ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0

Since F_f = µ_s × F_n ;

F_n•cos θ − mg + (µ_s × F_n × sin θ) =0

F_n = mg/[cos θ + (µ_s•sin θ)]

Also, sum of forces about the centre i^ direction gives;

ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r

Plugging in formula for F_n gives;

ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r

Making v the subject gives;

v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B) What we got in a above is the minimum speed the car can have while going round the turn.

The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.

Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;

v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

Thus the range is;

√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;

ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0

Thus;

mg(sin θ - µ_s•cos θ) = 0

Making µ_s the subject gives;

µ_s = sin θ/cos θ

µ_s = tan θ

D) If θ = 25.0°;

Thus;

µ_s = tan 25

µ_s = 0.4663