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
