Question

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a)
Calculate the ideal speed in (m/s) to take a 120 m radius curve banked at 15°.
(b)
What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 15.0 km/h?

Answer 1

Hi there!

(A)

For a banked curve, we know that:

∑F = MgsinФ (force due to gravity)

And:

∑F = mv²/r

The cosine of the centripetal force, mv²/r, is involved in the summation of forces, so:

(mv²/r)cosФ = mgsinФ

Simplify the equation:

(v²/r)cosФ = gsinФ

Divide both sides by cosФ:

(v²/r) = gtanФ

Solve for v:

v = √grtanФ

Plug in given values. g ≈ 9.8 m/s²

v = √(9.8 · 120 · tan(15)) = 17.75 m/s

(b)

Incorporating friction, we now can rewrite the sum of forces. Let's first determine the frictional force acting on the car.

As the horizontal component of the centripetal force is the net force, the vertical component works in the direction of the normal force, so we must include it in the calculation of the frictional force.

First, convert 15 km/h to m/s:

15km / h * 1000 m / 1km * 1 h / 3600 sec = 4.167 m/s

Thus:

∑F = mgsinФ - μ(mgcosФ + (mv²/r)sinФ)

mv²/r = mgsinФ - μ(mgcosФ + (mv²/r)sinФ)

Divide all terms by the mass:

v²/r = gsinФ - μ(gcosФ + (v²/r)sinФ)

Solve for μ:

v²/r - gsinФ = - μ(gcosФ + (v²/r)sinФ)

-(v²/r - gsinФ)/(gcosФ + (v²/r)sinФ) = μ

Plug in the given values:

μ = 0.25