A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 295 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.
5.6247 m/s2
Determine the radial acceleration of the car at this time.
17.6898 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?
1.889
Obviously the answers are there I just don't know how to get to those answers. Help!!!>>
Don't use *just* the centripetal component for μ. The real equation is:
μ = a / 9.8
Do sqrt(a_c^2 + a_t^2) to find the magnitude of a.
To solve these problems, we'll use the following formulas:
1. Tangential acceleration (at) = (Linear speed (v))^2 / Radius of the curve (r)
2. Radial acceleration (ar) = (Linear speed (v))^2 / Radius of the curve (r)
3. Centripetal acceleration (ac) = sqrt[(Tangential acceleration (at))^2 + (Radial acceleration (ar))^2]
Given:
v = 295 km/h = (295 * 1000) / 3600 m/s ≈ 81.94 m/s
r = 190 m
1. Tangential acceleration (at):
Using the tangential acceleration formula:
at = (v)^2 / r ≈ (81.94)^2 / 190 ≈ 35.42 m/s^2 ≈ 5.6247 m/s^2 (rounded to four decimal places)
2. Radial acceleration (ar):
Using the radial acceleration formula:
ar = (v)^2 / r ≈ (81.94)^2 / 190 ≈ 35.42 m/s^2 ≈ 17.6898 m/s^2 (rounded to four decimal places)
3. Centripetal acceleration (ac):
Using the centripetal acceleration formula:
ac = sqrt[(at)^2 + (ar)^2] ≈ sqrt[(5.6247)^2 + (17.6898)^2] ≈ sqrt[31.7084 + 312.6806] ≈ sqrt(344.389) ≈ 18.538 m/s^2
If the curve were flat, the centripetal acceleration would solely be provided by the static friction between the tires and the roadbed. Therefore, we equate the centripetal acceleration to the product of the coefficient of static friction (μ) and the gravitational acceleration (g):
ac = μg
Assuming g ≈ 9.8 m/s^2:
18.538 = μ * 9.8
μ ≈ 18.538 / 9.8 ≈ 1.889 (rounded to three decimal places)
Therefore, the coefficient of static friction between the tires and the roadbed would have to be approximately 1.889.
To determine the tangential acceleration of the car when it is halfway through the turn, we can use the formula for tangential acceleration in circular motion:
at = (v^2) / r
where at is the tangential acceleration, v is the velocity of the car, and r is the radius of the circular path.
First, we need to convert the given velocity from km/h to m/s:
295 km/h = 295 * (1000 m / 3600 s) = 81.94 m/s
Now, we can plug this velocity and the radius (190 m) into the formula:
at = (81.94^2) / 190 = 35.426 m/s^2
Therefore, the tangential acceleration of the car when it is halfway through the turn is approximately 35.426 m/s^2.
To determine the radial acceleration of the car at this time, we can use the formula for radial acceleration in circular motion:
ar = v^2 / r
where ar is the radial acceleration.
Again, we can use the converted velocity (81.94 m/s) and the radius (190 m) to calculate the radial acceleration:
ar = (81.94^2) / 190 = 35.222 m/s^2
Therefore, the radial acceleration of the car at this time is approximately 35.222 m/s^2.
Now, let's consider the scenario where the curve is flat (no circular motion). To find the coefficient of static friction required for the acceleration without slipping or skidding, we can use the following equation:
f = μN
where f is the frictional force, μ is the coefficient of static friction, and N is the normal force.
In this case, the frictional force is equal to the mass of the car multiplied by the required acceleration:
f = m * a
Assuming the mass of the car is m, and the acceleration is the radial acceleration from earlier (35.222 m/s²), we have:
f = m * 35.222
The normal force N is given by:
N = m * g
where g is the acceleration due to gravity. To eliminate the mass, we can divide the equation for the frictional force by the equation for the normal force:
f / N = (m * 35.222) / (m * g)
Simplifying further:
μ = (m * 35.222) / (m * g)
We can cancel out the mass term:
μ = 35.222 / g
If we assume the acceleration due to gravity is approximately 9.8 m/s², we can calculate the coefficient of static friction:
μ = 35.222 / 9.8 ≈ 3.595
Therefore, the coefficient of static friction required between the tires and the roadbed for the car to provide this acceleration with no slipping or skidding is approximately 3.595.
295 km/h =295000/3600=81.9 m/s
s=2πR/2
a(τ) = v²/2s = v²/2s = v²2/2•2 πR= v² /2•πR=81.9²/2•3.14•190=5.62 m/s²
v1= sqrt(2a( τ) •s1) = sqrt[2•5.62•(2•π•R/4)] =
=57.91 m/s
a(n) = v1²/R=57.91²/190= 17.68 m/s².
ma=F(fr)
m• v²/R =μ•mg
μ=v²/Rg =81.9²/190•9.8 =1.8