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 110 m radius curve banked at 15° m/s (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 10.0 km/h?

## Answer

(a)

The centripetal force when negotiating curve is m * v * v /

r

where m is mass, v is velocity and r is the radius of the

curve

The force due to the weight of the car along the plane of the

bank

is m* g * sin theta

In order for the car to successfully negotiate the curve, these

two

forces must be equal

This implies

m * v * v / r = m * g * sin theta

we can cross out m so our equation becomes

v * v / r = g sin theta

=> v * v / 110 = 9.8 sin 15

=> v^2 = 110 * 9.8 * sin 15

=> v^2 = 279

Taking square roots gives

v = **16.7
m/s**

(b)

v = 10 km/h = 10/3.6 m/s = 2.78 m/s

μs = (g * sin(θ) – Ac * cos(θ) ) / (Ac * sin(θ) + g * cos(θ)

)

With Ac = V^2 / r = 0.07 m/s^2

μs = (9.8 * sin(15) – 0.07 * cos(15) ) / (0.07 * sin(15) + 9.8 *

cos(15) )

= **0.26**