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%.

2.1)

LOL - I would guess around mu = 0.96

2.2)
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

2.3)
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 :)

@Damon:

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?

where is the answer for 2.2

To calculate the coefficient of static friction between the car's tires and the road (2.1), you need to use the following formula:

μ_static = (lateral acceleration)/((g) × sin(θ))

where:
- μ_static is the coefficient of static friction
- lateral acceleration is the maximum lateral acceleration of the car (0.96 g's)
- g is the acceleration due to gravity (approximately 9.8 m/s²)
- θ is the angle of the road (assuming it is horizontal, θ = 0°)

Plugging in the values:

μ_static = (0.96 g) / (g × sin(0°))

Since sin(0°) is equal to 0, we divide by 0. Therefore, there is an error in the calculation. Please check the given information and ensure all values are correct.

To determine the minimum radius of an unbanked curve the car can negotiate while traveling at a constant speed of 120 km/h (2.2), you can use the following formula:

radius = (velocity²) / (lateral acceleration)

where:
- radius is the minimum radius of the curve
- velocity is the car's speed (120 km/h converted to m/s)
- lateral acceleration is the maximum lateral acceleration of the car (0.96 g's)

First, convert the velocity from km/h to m/s:
120 km/h = (120 × 1000) m / (60 × 60) s ≈ 33.33 m/s

Now, substitute the values into the formula:
radius = (33.33 m/s)² / (0.96 g)

Plug in the value of g (approximately 9.8 m/s²) and calculate the radius.

To 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% (2.3), you can use a similar approach as in 2.2. However, this time you need to consider the reduced value of the coefficient of static friction.

Let's assume the coefficient of static friction in wet weather is μ_wet. To find the new lateral acceleration (0.8 times the original), use the equation:

lateral acceleration (wet) = lateral acceleration × 0.8

Now, you can use the same formula as in 2.2, but this time instead of the original lateral acceleration, substitute the lateral acceleration (wet). This will give you the maximum lateral acceleration the car can have while negotiating the curve in wet weather.

Once you have the maximum lateral acceleration, you can use the same formula as in 2.2 to calculate the range of speeds at which the car can safely negotiate the curve in wet weather. Instead of the original coefficient of static friction, substitute the μ_wet value you calculated in the previous step.