posted by Chelsea on .
I need to check my answers for the following:
An Aston Martin V8 Vantage sports car has a maximum “lateral acceleration” of
0.96 “g’s”, after which the car begins to skid out of its turn.
2.1 Calculate the coefficient of static friction between the car’s tyres and the road.
2.2 Determine the minimum radius of unbanked curve the car can negotiate while travelling at
a constant 120 km/h.
2.3 If the road is banked at 15.0°, determine the range of speeds at which the car can safely
negotiate the curve in wet weather, when the coefficient of static friction is reduced by 20%.
LOL - I would guess around mu = 0.96
120,000 m/h * 1 h/3600 s = 33.33 m/s
m v^2/R = .96 m g
R = v^2/(.96 g) = 118 meters
mu = .8*.96 = .768
Normal F = m g cos 15 = .966 m g
friction force = mu Normal = .768*.966mg =.742 mg
component of weight down slope = m g sin 15 = .259 mg
Total down slope = (.742 + .259)mg
component of centripetal acceleration down slope = (v^2/R)cos 15
= .966 v^2/R
force down slope = mass * acceleration down slope
(.742+.259)mg = m * .966 v^2/R
v^2/R = 1.04 g
if R = 118
v^2 = 118 *1.04 * 9.81 34.7 m/s
now reverse mu to - .768 and do this again to find minimum speed with slipping downhill. It might be zero :)
I don't understand why that is the answer for 2.1
weight down = m g
max friction force = mu m g
centripetal force = m (lateral acceleration)
mu m g = m a = m (.96 g)
a = .96 g
We call "lateral acceleration" centripetal acceleration :)
2.3) The question says "range of speeds", so i was expecting a quadratic formula.( What do you mean by reversing mu?
@Damon what happens to the sin component of the centripetal force?