Asked by Ajay
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.
Answers
Answered by
Damon
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
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
Answered by
Damon correction
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.