The Daytona International Speedway shown below has the following track statistics: Bank turn of 31.0o, a radius of curvature of 1000 ft, and a total length of 2.50 mi. Given that the typical weight of a NASCAR car is 3000 lbs, and that the friction coefficient between the tires and the track is around 0.9-1.3 (use 0.900), estimate the maximum speed of the race car in a banked turn.
I guess we are doing feet and pounds and such.
force down on car = 3000 lbs = m g
car mass = 3000/g
normal force of weight on track = 3000cos 31
so max friction force down tangent to track = 3000 cos31*0.9
component of weight down tangent to track = 3000 sin 31
acceleration tangent to track = v^2/R * cos 31 down track
so
3000 * .9 * cos 31 + 3000 sin 31 = m v^2/R = 3000/g * v^2/R
.9 cos 31 + sin 31 = (v^2/R) / g = ratio of centripetal to gravitational acc
(of course in the end the weight does not matter :)
R = 1000
g = 32 ft/s^2
solve for v in feet/second
Forgot to continue component of centrripetal down slope
acceleration tangent to track = v^2/R * cos 31 down track
so
3000 * .9 * cos 31 + 3000 sin 31 = m v^2/R = 3000/g * v^2/R cos 31
.9 cos 31 + sin 31 = (v^2/R) cos 31 / g = ratio of centripetal to gravitational acc
(of course in the end the weight does not matter :)
R = 1000
g = 32 ft/s^2
solve for v in feet/second
To estimate the maximum speed of the race car in a banked turn, we can use the concept of centripetal force.
Centripetal force is the force that keeps an object moving in a curved path. In this case, it is provided by the friction between the tires of the race car and the track surface.
The centripetal force required to keep the car moving in a circular path can be calculated using the formula:
Fc = m * v^2 / r
Where:
- Fc is the centripetal force
- m is the mass of the car (3000 lbs)
- v is the velocity of the car
- r is the radius of curvature of the turn
In a banked turn, the vertical component of the normal force will provide part of the centripetal force, while the horizontal component of the normal force will provide the rest. The normal force is given by:
N = m * g
Where:
- N is the normal force
- m is the mass of the car
- g is the acceleration due to gravity (32.2 ft/s^2)
The vertical component of the normal force can be calculated using the formula:
N vertical = N * sin(θ)
Where θ is the bank angle (31.0o). The horizontal component of the normal force can be calculated using the formula:
N horizontal = N * cos(θ)
Since the friction coefficient between the tires and the track is given as 0.900, we can calculate the maximum value of frictional force using the formula:
F friction = μ * N horizontal
Finally, we can equate the frictional force to the required centripetal force:
F friction = Fc
By rearranging the equation, we can solve for the maximum velocity (v):
v = sqrt( r * μ * g * tan(θ) )
Let's calculate the maximum speed.
Given data:
- Mass of the car (m) = 3000 lbs
- Friction coefficient (μ) = 0.900
- Radius of curvature (r) = 1000 ft
- Bank angle (θ) = 31.0o
- Acceleration due to gravity (g) = 32.2 ft/s^2
First, convert the mass of the car from pounds to slugs:
Mass (m) = 3000 lbs * (1 slug / 32.2 lbs) ≈ 93.17 slugs
Next, convert the radius of curvature from feet to miles:
Radius (r) = 1000 ft * (1 mi / 5280 ft) ≈ 0.1894 mi
Now, substitute the values into the equation:
v = sqrt(0.1894 mi * 0.900 * 32.2 ft/s^2 * tan(31.0o))
v ≈ sqrt(2.39)
v ≈ 1.546 mi/s * (3600 s/h) * (1 h/60 min)
v ≈ 5581.6 mi/h
Therefore, the estimated maximum speed of the race car in the banked turn is approximately 5581.6 miles per hour.