A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 335 km/h in a semicircular arc with a radius of 190 m.
Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration. (in m/s2)
Determine the radial acceleration of the car at this time.(in m/s2)
If the curve were flat, what would the coefficient of static friction have to be between the tires and the roadbed to provide this acceleration with no slipping or skidding?
I will be happy to critique your thinking or work on this. Please don't post under multiple names.
Why did the car go to the circus? Because it wanted to become a race clown! Let's calculate the tangential and radial accelerations of the car, and find the coefficient of static friction for a flat curve. Buckle up, we're going on a physics ride!
To find the tangential acceleration, we need to use the formula for centripetal acceleration:
a_tangential = (v^2) / r
where v is the velocity and r is the radius. But first, let's convert the given speed from km/h to m/s for consistent units:
335 km/h = (335 × 1000) / 3600 m/s ≈ 93.1 m/s
Plugging in the values:
a_tangential = (93.1^2) / 190 ≈ 45.7487 m/s^2
So the tangential acceleration of the car when it is halfway through the turn is approximately 45.75 m/s^2.
To find the radial acceleration, we can use the same formula, but substitute the velocity with the tangential acceleration since they are equal in magnitude:
a_radial = a_tangential = 45.75 m/s^2
So the radial acceleration of the car when it is halfway through the turn is also approximately 45.75 m/s^2.
Now, if the curve were flat, meaning there is no vertical acceleration, the only force acting on the car would be the static friction to provide the necessary acceleration. The formula for centripetal acceleration in terms of friction is:
a_centripetal = (μ * g) / r
where μ is the coefficient of static friction and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values and rearranging the formula:
μ = (a_centripetal * r) / g
μ = (45.75 * 190) / 9.8 ≈ 892.35
The coefficient of static friction between the tires and the roadbed should be approximately 892.35 to provide the necessary acceleration without slipping or skidding. That's quite a sticky situation!
I hope this physics joyride was entertaining, and please let me know if you have any more questions or if you'd like some more clown humor!
To determine the tangential acceleration of the car when it is halfway through the turn, we need to calculate the speed of the car at this point and the time it takes to reach that speed.
1. Calculate the angular displacement:
The car travels half of the circumference of the semicircular arc, which is equal to pi times the diameter of the circle.
Angular displacement = (π * diameter) / 2 = (π * 2 * radius) / 2 = π * radius
2. Calculate the speed of the car at this point:
Speed = circumference / time
The circumference of the semicircular arc is π times the diameter of the circle.
Speed = (π * diameter) / time
We know that the final speed is 335 km/h, so we need to convert it to m/s:
Final speed = 335 km/h * (1000 m / 1 km) / (3600 s / 1 h) = (335 * 1000) / 3600 m/s
Setting the final speed equal to the calculated speed, we can solve for time:
(π * diameter) / time = (335 * 1000) / 3600
time = (π * diameter) * (3600 / (335 * 1000))
3. Calculate the tangential acceleration:
Acceleration = (Final speed - Initial speed) / time
We know the initial speed is 0, so the equation becomes:
Acceleration = Final speed / time
Now let's calculate the radial acceleration of the car at this time. The radial acceleration is given by the formula:
Radial acceleration = (Tangential speed)^2 / Radius
4. Calculate the tangential speed:
Since we have the final speed, we can calculate the tangential speed by dividing it by the square root of 2:
Tangential speed = Final speed / √2
5. Calculate the radial acceleration:
Radial acceleration = (Tangential speed)^2 / Radius
To determine the coefficient of static friction, we can use the equation:
Coefficient of static friction = Tangential acceleration / acceleration due to gravity
6. Calculate the coefficient of static friction:
Coefficient of static friction = Tangential acceleration / acceleration due to gravity
Let's plug in the given values and calculation steps to find the answers:
Given:
Radius = 190 m
Final speed = (335 * 1000) / 3600 m/s
Acceleration due to gravity = 9.8 m/s^2
1. Calculate the angular displacement:
Angular displacement = π * radius = π * 190
2. Calculate the speed of the car at this point:
Speed = (π * diameter) / time
time = (π * diameter) * (3600 / (335 * 1000))
Speed = (π * diameter) / time
3. Calculate the tangential acceleration:
Acceleration = Final speed / time
4. Calculate the tangential speed:
Tangential speed = Final speed / √2
5. Calculate the radial acceleration:
Radial acceleration = (Tangential speed)^2 / Radius
6. Calculate the coefficient of static friction:
Coefficient of static friction = Tangential acceleration / acceleration due to gravity
Now you can plug in the given values into the respective equations and calculate the tangential acceleration, radial acceleration, and the coefficient of static friction.
To solve this problem, we need to use the following formulas:
1. Tangential acceleration (at) = (final velocity - initial velocity) / time
2. Radial acceleration (ar) = (velocity^2) / radius
Let's break down the problem step by step:
Step 1: Convert the given velocity from km/h to m/s:
335 km/h = (335 * 1000) / 3600 = 93.06 m/s (rounded to two decimal places)
Step 2: Determine the time it takes for the car to reach the final velocity.
Since the car starts at rest and undergoes constant tangential acceleration, the time taken is the same as the time it takes to travel halfway through the semicircle.
We'll use the formula: distance = 0.5 * velocity * time
distance = 0.5 * (93.06 m/s) * time
The distance is half the circumference of a semicircular arc with a radius of 190 m, so we have:
pi * 190 m = 0.5 * (93.06 m/s) * time
(190 * pi) = (46.53 m/s) * time
time = (190 * pi) / (46.53 m/s) ≈ 12.99 s (rounded to two decimal places)
Note: We use π (pi) as 3.1416 for calculation.
Step 3: Calculate the tangential acceleration:
Tangential acceleration (at) = (final velocity - initial velocity) / time
The initial velocity is 0 m/s (starting from rest), and the final velocity is 93.06 m/s.
at = (93.06 m/s - 0 m/s) / 12.99 s ≈ 7.16 m/s^2 (rounded to two decimal places)
Step 4: Calculate the radial acceleration:
Radial acceleration (ar) = (velocity^2) / radius
ar = (93.06 m/s)^2 / 190 m ≈ 45.7 m/s^2 (rounded to one decimal place)
Step 5: Determine the coefficient of static friction (μ):
To find the coefficient of static friction, we need to equate the maximum static friction force to the required force to provide the acceleration with no slipping or skidding.
The maximum static friction force (Ff) can be calculated using the formula: Ff = μN
In this case, the normal force (N) is equal to the weight of the car, since there is no vertical acceleration.
Weight (N) = mass * gravity
The mass and gravity can be obtained from the weight of the car.
Let's assume the weight of the car is 10,000 N (just as an example):
N = 10,000 N
Ff = μN
Ff = (mass * gravity) * μ
Using the formula for centripetal force, Fc = mass * radial acceleration (Fc = m * ar), and considering that the force of friction is providing the radial acceleration, we have:
Fc = μN
m * ar = μ * (mass * gravity)
μ = ar / (mass * gravity)
Substituting the values:
μ = (45.7 m/s^2) / (mass * gravity)
We would need the mass of the car to calculate the coefficient of friction. Assuming a mass of 1000 kg:
μ = (45.7 m/s^2) / (1000 kg * 9.8 m/s^2)
μ ≈ 0.00467 (rounded to five decimal places)
In summary:
- The tangential acceleration of the car when it is halfway through the turn is approximately 7.16 m/s^2.
- The radial acceleration of the car at this time is approximately 45.7 m/s^2.
- To provide this acceleration with no slipping or skidding on a flat curve, the coefficient of static friction between the tires and the roadbed would have to be approximately 0.00467.