Question

3)

A car turning in a circle is acceleratring in the centripetal direction, even if the speed is constant. This centripetal acceleration is the cause of a radially inward directed net force. On a level road this net force is the friction force acting from the road on the tires. You already looked at examples for this.

On racetracks and also highway turns the reliance on friction can be reduced by "banking" the road. This also reduces the risk of the vehicle to roll over. The picture shows two cars and a bus in an extremely banked turn at the Mercedes test track.

Draw a free-body diagram for a car in a banked curve. Then find an expression for the speed at which a car can negotiate the turn without any friction in the radial direction (f=0).

Calculate that speed (in mph) for a 47 m radius turn that is banked by 5 degrees.

4)

You just should have already derived an expression for the "frictionless" speed of a car in a banked turn. Now assume that the coefficient of friction is u_s. Find an expression for the maximum speed at which a car can drive through a turn with radius R and banking angle theta.

Evaluate your answer: test whether in the two limiting cases of "no friction" and "level road" your answer will turn into the two results you already have derived for these cases.

Below answer with the maximum speed (in mph) for a turn with radius 126 m, banking angle 10 degree, and coefficient of static friction 0.71.

Please help I have tried so many times I'm almost out of tries!

Answer 1

3) We have to derive an expression for the speed V, on the banked curved path of radius R and angle .

Please refer to the attached free body diagram (1)

The forces in the vertical direction:

N cos = mg

N = mg / cos (1)

In the horizontal direction:

F(net) = N sin ; putting the value of N from (1) we get:

F(net) = mg (sin/cos) = mg tan

F(net) = mg tan   (2)

This F(net) and F(centripital) balances each other. So,

F(centripital) = F(net)

m V² / R = m g tan   (m gets cancelled both sides)

Solving for V we get:

V = sqrt (R g tan)

given values: R = 47 m and = 5 degree

V = sqrt (R g tan) = sqrt [47 x 9.81 x tan(5)] = 6.35 m/s = 14.20 mile per hr

Hence, V = 14.20 mi/h

4)Now we have to determine the speed V on the same path of radius R and banking angle but with frictional force. The coefficient of friction is .

Refer to the figure (2)

For vertical forces;

N cos = mg + F(f) sin ; F(f) = N

N cos = mg + N sin

N(cos - sin) = mg

N = mg / (cos - sin) (1)

The forces in horizontal direction are:

F(net) = N sin + F(f) cos = N sin + N cos = N (sin + cos)

putting the value of N from (1) we get:

F(net) = [mg/(cos - sin)] x  (sin + cos)

simplifying we get:

F(net) = mg(tan + ) / (1 - tan)

F(net) and F(centripital) balances each other:

F(centripital) = F(net)

m V² / R =  mg(tan + ) / (1 - tan)

m gets cancelled both sides, solving for V we get:

V = sqrt [ R x g(tan + ) / (1 - tan) ]

Given that, R = 126 m ; = 10 and = 0.71; putting this values in above equation to get the velocity:

V = sqrt [ 126 x 9.81 (tan10 + 0.71) / (1 - 0.71 x tan10)]

V = 35.39 m/s = 79.17 mi/h

Hence, V = 79.17 mi/h